Bulls and Cows

Bulls and Cows is a two player game. First player thinks of a N digit secret number and second player guesses that number. This is done by second player making a guess and first player telling the number of bulls and cows in the guess. If the digits of guess matches secret and they are in right position, it's a bull, if they match but on different position, they are cows. For e.g. Let secret be 0123 and guess be 0012. The guess has 1 bull, 2 cows. The game continues, until second player gets a response of N bulls.

The problem is to write a computer solver for bulls and cows that would try to guess the secret making only a small number of attempts.

Solution: show

Let D be the unique digits that can be put in the secret. Let N be the length of the secret number. The total number of secrets is D^N.

A naive solution is to try all D^N guesses until N bulls is the response. This approach makes D^N attempts in worst case. Not good.

We can do better than this. Let us make a random first guess and we get a response r. We create a pruned set by selecting all numbers (from D^N possible) that have response r with the guess. Secret must be in the pruned set. We can pick a number from this set as our guess and repeat the logic until we hear N bulls.

There are three ways to pick the next number
a) Pick the first number from the pruned set
b) Pick the number randomly from the pruned set
c) Use a heuristic to pick a number (Note this number can be outside the pruned set).

Our experiments show that b) is better than a). For N = 4, D = 10 with distinct digits in secret (5040 possible secrets) and initial guess of 0123, the pick first strategy makes 6.0051 attempts on average and 9 attempts in a worst case. On the other hand, randomly picking from the pruned set makes 5.9136 attempts on average and 9 attempts in a worst case.

Note that optimal strategy for N=4, D=10 with distinct digits in secret (5040 possible secrets) takes an average of 5.2131 attempts with 7 attempts in the worst case (technical paper). It is also guaranteed that 6 attempts is the worst case lower bound for any algorithm. Since 5.2131 is theoretical best, 5.9136 achieved by random is not bad.

Let us now explore option c. There are (N+1)*(N+2)/2 - 1 response classes starting from (0 bulls, 0 cows) to (N bulls, 0 cows). Note the -1 is there because if the response is N-1 bulls then we cannot have 1 cow. The response to a guess, partitions the D^N numbers (as shown in Fig 1).

Fig 1. Response partition of 10^4 numbers for the secret 0123

Now we know that the secret lies in the pruned set. We need to pick our next guess that partitions the pruned set in the best possible manner. One way to judge the goodness of partition is by information gain [sum_i ni * log (ni)]. The issue now is that to compute information gain, we need to compute responses for D^N * length(pruned set) pairs.

Note that the best guess can be outside the pruned set, that is why we need to compute D^N * length(pruned set) responses and not length(pruned set)^2. To see this consider that pruned set is [089, 189, 289, 389, 489, 589, 689, 789]. If our secret is 789 we need 8 guesses but if we try 012 as guess 1 then we get pruned set [389, 489, 589, 689, 789], then we try 345 as our guess 2 to get a pruned set [689, 789] and now two more guesses are required. So we solved it in 4 attempts instead of 8.

But the bigger issues is that this technique can only be applied if length of pruned set is small, otherwise it can be computationally prohibitive and memory intensive. Our strategy is as follows:

step 1. Pick distinct digits to make first guess.
step 2. Select random guess from the pruned set.
step 3. Perform step 2, three or four times.
step 4. Pick guess using information gain measure.
step 5. Perform step 4, until game is solved.

For N = 4, D = 10, (5040 possible secrets), this strategy makes 5.7201 attempts on average with 8 attempts in a worst case. Summary of algorithms is as follows:

D=10, N=4, 5040 secrets
algo	average #attempts	worst #attempts
pick first	6.0051	9
pick random	5.9136	9
heuristic pick	5.7201	8
optimal	5.2131	7

D=10, N=4, 10000 secrets
algo	average #attempts	worst #attempts
pick first	8.0291	13
pick random	6.6323	11
heuristic pick	6.2001	8

Source code: bullscows.py

Technical resources: optimal strategy, bullscows.pdf

Number of Paths in a Rectangular Grid

Problem 1: Consider a rectangular grid of n x m rectangular cells. The problem is to find the number of shortest (or monotonic in this case) paths along the edges of the cell that start at (0,0) and end at (n,m). How do we print all such paths?

A monotonic path consists of moving up or right but not down or left. Figure 1 illustrates such a path for a grid of size 5 x 3.

Figure 1

Solution: show

To go from (0,0) to (n,m), we need to make n moves to the right and m moves up. These moves can be performed in any order and still we reach the point (n,m). So the total number of ways is simply the total number of ways n red balls and m green balls can be arranged in a straight line. This is simply ^n+mC_m.

We can print all such paths using recursion. The following code does this.

Function printPath(n, m, path)
   if n == 0 and m == 0
      print path
      return
   end

   if n > 0
      printPath(n-1, m, path + "right")
   end

   if m > 0
      printPath(n, m-1, path + "up")
   end
end

Problem 2: Now we need to print all monotonic paths that do not go above the diagonal y = x. Figure 2 shows such a path. Note that the path in figure 1 goes above the diagonal, hence not desirable in this case. How many such paths exist for n x m grid (n >= m)?

Figure 2

Solution: show

Let us first discuss how to print all monotonic paths that do not go above the diagonal. In order for a path to be valid, at all points (x,y) on the path, y should be less than or equal to x. We can simply ensure this constraint while printing paths, using the following code.

Function printPath(x, y, n, m, path)
   if x == n and y == m:
      print path
      return
   end

   if x < n
      printPath(x+1, y, n, m, path + "right")
   end

   if y < m and y < x
      printPath(x, y+1, n, m, path + "up")
   end
end

The total number of such paths are [(n-m+1)/(n+1)] * ^n+mC_m. For n = m, this simplifies to [1/(n+1)] * ²ⁿC_n which is known as Catalan number. The wiki page also contains other interesting problems that have Catalan number as their solution.

Now, In order to obtain the number of paths, we need to subtract the number of paths that cross the diagonal from the total number of paths ^n+mC_m.

Consider a path that crosses the diagonal. Let that path cross the diagonal for the first time at point (x,y). Since this is the first point after crossing the diagonal, so y should be equal to x+1. So the path has taken x right and y up moves. It is yet to take n-x right and m-y up moves to reach (n,m). If we flip the n-x right to n-x up and m-y up to right, then starting from point (x,y), we will reach the point (x+m-y,y+n-x). Setting y = x+1, we get that the resulting end point is (m-1,n+1). Since m <= n, the resulting end point is above the diagonal. So we can do surgery on all paths that cross the diagonal to make them reach (m-1,n+1). So the total number of paths that cross the diagonal must be the total number of paths from (0,0) to (m-1,n+1). This is ^n+mC_m-1.

So the desired number is = ^n+mC_m - ^n+mC_m-1  = [(n-m+1)/(n+1)] * ^n+mC_m. This is a proof based on Andre's reflection method.

Alternately:
Andre's reflection method was used to originally solve a very interesting problem called Ballot problem. The ballot problem states that in a two candidate election, candidate A polled p votes and candidate B polled q votes (p > q). What is the probability that candidate A was leading over candidate B throughout the counting.

Well the probability is simply (p-q)/(p+q).

We can use the Ballot theorem to find the number of monotonic paths below the diagonal (that do not even touch the diagonal). We can assume right move to be candidate A and up move to be candidate B and the two polled n,m votes respectively. So the probability of number of right leading number of up is (n-m)/(n+m). So the number of monotonic paths that are below the diagonal (do not even touch) is [(n-m)/(n+m)] * ^n+mC_m.

Now since we are allowed to touch the diagonal, we can assume instead going to (n+1,m). This is because the first two moves would be right moves (otherwise we might touch the diagonal). Then we eliminate the first right move and allow a diagonal touch. So the total number of ways is [(n+1-m)/(n+1+m)] * ^n+m+1C_m = [(n-m+1)/(n+1)] * ^n+mC_m.

Problem 3: Now let us consider that apart from up and right moves we can take up-right (diagonal) moves. Figure 3 shows such a path. How many such paths exist for n x m grid?

Figure 3

Solution: show

This problem is similar to problem 1, except that paths can contain diagonal moves. One diagonal move consumes 1 right and 1 up move.

So let us assume that we make r diagonal moves, where r <= min(n,m). In this case, we need to make n-r right moves and m-r up moves apart from r diagonal moves to reach (n,m). The number of ways, we can do this is = (n+m-r)! / [ (n-r)! * (m-r)! * r! ]. If we set r = 0, then we get the solution to problem 1. But in this case r can go from 0 to min(m,n). So the total number of ways is =

Sum_r (n+m-r)! / [(n-r)!*(m-r)!*r!], where r in [0, min(m,n)]

Alternately, we can derive the recurrence relation for the total number of paths. Let D(n,m) be the total number of paths from (0,0) to (n,m) with up, right and diagonal moves. Clearly, we can reach (n,m) from (n-1,m) by a right move, from (n,m-1) by a up move, and from (n-1,m-1) by a diagonal move. These are the only three atomic steps. So, D(n,m) = D(n-1,m) + D(n,m-1) + D(n-1,m-1) for n,m > 0. D is called as Delannoy number.

The code to print all paths has a small addition to the code for problem 1.

Function printPath(n, m, path)
   if n == 0 and m == 0
      print path
      return
   end

   if n > 0
      printPath(n-1, m, path + "right")
   end

   if m > 0
      printPath(n, m-1, path + "up")
   end

   if n > 0 and m > 0
      printPath(n-1, m-1, path + "diagonal")
   end
end

We can extend this problem further by asking the number of paths that do not cross the diagonal (same constraint as problem 2). The number of paths in this case is called Schroder number.

Source Code: grid_paths.py

Finding Top K Elements in Large Dataset

Problem: Consider the problem of finding top K frequent numbers from a file with N numbers. Now consider that N is very large such that it cannot fit in the memory M available to the program. How do we find the top K frequent elements now with the assumption that K < M < N.

Deterministic solution: show

The idea is similar to solving the problem for small N (N < M) (see here). For large N, divide the problem into chunks of size <= M, then solve it.

In order to divide the problem, consider a uniform hashing function H that takes a key and returns an integer from the set {1,2,....,ceil(N/M)}. So we get N/M chunks with roughly M unique keys per chunk. Since the number of unique numbers (U) is less than N, we expect each chunk to have less than M unique numbers.

Now for each chunk, we compute the frequency of numbers contained in that chunk. This frequency computation can be done in memory and we can maintain a MIN-HEAP of size K which can be directly updated (follow the steps presented here). As a result only two reads of the dataset and one disk write is required to get the top K frequent items. The complexity of the algorithm is O(N log K).

Probably for a small K (K < M^2/N), we can compute the top K per chunk in O(K log M) and combine the top K for all chunks N*K/M (<M) to get the overall top K. The total complexity of the algorithm is O(N + M log M).

Alternate method:
We assume that the disk has a huge capacity which allows us to make a disk based hash table. Imagine that we create a very large file with holes and divide the file into B blocks with a capacity to hold more than N/B numbers and their integer counts. Using a uniform hash function H which takes as input an arbitrary number x and return H(x): the block number from the set {0, 1, ..., B-1}, we can store the numbers and their frequencies on disk map. H would give us the offset to seek within the file and we write number (and their frequency) sequentially once in correct block (or via another hash function H1).

Now we proceed as usual, start counting the numbers in memory using a hash table (former approach). When we encounter a new number that could not be put in memory, we purge some entries from the hash table. The purged entries are written to the disk map. Now to purge the entries, we maintain an array which counts the frequency of the keys that lie within a block. Keys that belong to the most infrequent blocks can be purged first (or blocks that are least recently used or that lead to least disk access, etc).

The following code gives a very basic detail of this approach

BlockFreq = array(B)
NumberFreq = hashtable(M)
diskwrite = 0
for i = 1:N
   x = A[i]
   BlockFreq[H[x]] += 1

   if NumberFreq.haskey(x)
      NumberFreq[x] += 1
      continue
   end

   if NumberFreq.hasspace()
      NumberFreq[x] = 1
      continue
   end

   if DiskMap.haskey(x)
      DiskMap[x] += 1
   else
      DiskMap[x] = 1
   end

   if diskwrite == 10
      purge(NumberFreq, BlockFreq)
      diskwrite = 0
   else
      diskwrite += 1
   end
end

Here purge is a procedure to purge some set of keys from NumberFreq based on the BlockFreq. Note that this code omits several key details of this process, so the idea presented here is quite crude.

Single pass probabilistic solution: show

Solution 1 is quite efficient as it requires only two disk reads of the dataset, but the bottleneck can be the disk writes during the initial chunk formation. We can reduce that bottleneck by considering a data-structure called Bloom filters.

So consider that we have B uniform hash functions H1, H2, ..., HB and each hash function converts a key to a range {1,2,...,R}. Now imagine an array C of size B x R (<M) that represents count of how many times each key is seen. For each number (say x) that we read from the dataset, compute Hi[x] and increment C[i,Hi[x]] by 1. So we maintain B counts of x in different R buckets. We can say that the true count of x is less than min(C[1,H1[x]], ..., C[B,HB[x]]).

Now if the query is to get all the elements with frequency greater than some threshold then we can use bloom filters to get all such numbers (with some false positives though, which can be filtered using another pass on the dataset). This can save a complete write of the data to the disk. (see the paper: Computing Iceberg Queries Efficiently).

But in our case, we are interested in finding the top K frequent numbers. Following modification can be used to estimate the frequency of each number.

MH = MIN-HEAP(K)
for i = 1:N
   x = A[i]
   for b = 1:B
      C[b,Hb(x)] += Sb(x)
   end

   if contains(MH,x)
      increment count of x in the heap
   else 
      f = median(Hi(x) * Si(x), \forall i)
      if f > min(MH)
         remove-min(MH)
         insert(MH, (x,f))
      end
   end
end

The Sb functions is a {-1,+1} hash function and this data-structure is called CountSketch. More details of the method is available in the paper: Finding Frequent Items in Data Streams.

Note that the above algorithm takes a single passes on the dataset (and no disk write) but it is not guaranteed to give the top K frequent items. It can make some mistakes for some less frequent items. In practice the choice of the hashing functions can be critical for the performance of the algorithm.

Finding Top K Frequent Items

Problem: Consider a file containing N numbers. For e.x. {2,3,4,3,1,78,78,3} is a file containing N=8 numbers. Assume that N is small enough to fit in the computer's memory M. So in this case N << M. Now we need to find top K frequent numbers in the file. For the above example, output should be {3,78} for K = 2.

Running Time: Time complexity should be either O(N), O(N log K), O(N + K log N)

Solution: show

Finding top K frequent elements is a classical database and data streaming problem and there are several solutions to it ranging from deterministic to probabilistic. I'll illustrate the three obvious ones here.

The first step is to count how many times each number appears in the file. If the file is pre-sorted then we need a single scan over the file.

Function COUNTS:
   A = load_file_in_array

   last = A[1]
   ctr = 1
   for i = 2:N
      if last == A[i]
         ctr += 1
      else
         CountMap{last} = ctr;
         last = A[i]
         ctr = 1
      end 
   end

   // required for the last element of the array
   CountMap{last} = ctr
end

Note that CountMap is a hashmap that stores counts of all the elements. The procedure COUNTS is quite efficient if the file is pre-sorted. Now if the file is not pre-sorted then sorting increases the time complexity to O(N log N). In that case, we can do better by directly using hashmap to maintain current count of each number without sorting the file as follows:

Function EFFICIENT_COUNTS:
   A = load_file_in_array
   
   for i = 1:N
      CountMap{A[i]} += 1
   end
end

The above procedure obtains counts of each element in a single scan of the file. Hence it runs in O(N) time. So now we have all the numbers along with their frequencies in an array (can be extracted by enumerating all keys of the CountMap or by another scan of the file!!). So for the example we have in the problem statement, we get the following tuple: {(2,1), (3,3), (4,1), (1,1), (78,2)}.

The problem that remains is to find the most frequent K numbers from the array. The naive approach would be to sort numbers on their frequencies and pick the top K. This would take O(U log U) time where U (=5) is the number of unique elements in the array. If we consider U = O(N) then the time complexity of this sorting is O(N log N). We can do better than that as follows:

Approach 1: O(N) time
Use selection algorithm to find the Kth most frequent number (on the second element of the tuple) using the Selection Algorithm in O(U) time. The Kth most frequent element partitions the array in two parts: first part containing top K most frequent elements and second part containing bottom U-K-1 frequent elements. So we get the top K most frequent elements in no particular order in O(N) time (assuming U = O(N)). They can be sorted in O(K log K) if needed. Note that although this approach runs in O(N) time, the constants hidden in the O-notation can be large. So in practice this approach can be slower than the two approaches described below.

Approach 2: O(N log K) time
Pick first K tuples and put them on MIN-HEAP, where a tuple (x,y) is less than a tuple (a,b) if y is less than b. The time complexity to make the min heap of size K is O(K).

Then for the remaining U - K elements, pick them one by one. If the picked element is lesser than the minimum on the heap, discard that element. Otherwise remove the min element from the head and insert the selected element in the heap. This ensures that heap contains only K elements. This delete-insert operation is O(log K) for each element.

Once we are done picking all the elements, the elements that finally remain in the min-heap are the top K frequent items which can be popped in O(K log K) time. The overall cost of this approach is O(K + (U-K) log K + K log K) = O(K + U log K). Since K < U and U = O(N), we get the time complexity of O(N log K).

Approach 3: O(N + K log N) time
This approach is similar to approach 2 but the main difference is that we make a MAX-HEAP of all the U elements. So the first step is to make the max heap of all the elements in O(U). Then remove the maximum element from the heap K times in O(K log U) time. The K removed elements are the desired most frequent elements. The time complexity of this method is O(U + K log U) and by setting U = O(N) we get O(N + K log N).

Let us stop for a moment and contrast approach 2 from 3. For simplicity let T2 = K + N log K be the time complexity of approach 2 and T3 = N + K log N be the time complexity of the third approach. Figure below plots the ratio T2/T3 for N=100 and for different values of K. We observe that approach 3 is considerably better for small values of K whereas approach 2 is better for large values of K. Though actual difference depends on the constants involved we can still see the merit of one approach over another.

Problems solvable using Hashtable

Hashtable are extremely useful data-structure as they provide storage and retrieval in O(1) time (amortized). Several problems of algorithms can be very efficiently solved using hashtables which otherwise turn out to be quite expensive. In this post, we consider some of these problems:

Problem 1: Remove duplicate elements from an unsorted array of size N.

Solution: show

Lets first see the solution without hashtable. Naive solution would be to compare each pair of elements. If they are same then drop one of them. This solution would take O(N^2) time in average case and O(N) time in best case, when all the elements are same. The following code does that:

for i = 1 to N-1
  if a[i] == NaN: continue
  print a[i]
  
  for j = i+1 to N
    if a[i] == a[j]: a[j] = NaN

Above code prints all non-duplicates from the array. Naive solution is average case O(N^2). In practice if you believe that array consists of few numbers duplicated lots of times, say K numbers duplicated N/K times, then the running time of above algorithm is O(KN).

We can do better than worst case O(N^2) time. If we sort the elements in O(N logN) time and compare adjacent elements in O(N), we can solve the above problem in O(N logN) time.

sort(a) // O(N log N) time, O(1) space using Heap Sort

k = a[1]
for i = 2 to N
  if a[i] == k: a[i] = NaN
  else: k = a[i]

Using heap sort, we can make the above algorithm run in O(N log N) time in worst case and it would take O(1) extra space (apart from N for the array). Using hashtable, we can solve this problem in O(N) time (amortized) and O(N) space (for the hash table). The algorithm looks like following:

for i = 1 to N
  if hash.has_key(a[i]) == false:
    print a[i]
    hash.put(a[i])

Above algorithm stores an unseen element in hash-table and ignores the seen element. To check if an element is seen or not, hash table provides storage/retrieval methods in O(1) time.

Problem 2: Find intersection of K unsorted array of N elements each. Intersection consists of elements that appear in all the K arrays.

Problem 3: How to make a linked list support operations in O(1) time. The operations on linked list can be insertion after any arbitrary valued node, deletion of any arbitrary valued node.

Problem 4: Find all unique pairs of element in an array that sum to S. For ex. If array = {2,4,6,4,6} and S = 8 then answer is {(2,6), (4,4)}

Problem 5: Consider an array containing unique elements. Find a triplet of elements in the array that sum to S (extension of problem 4). Can hash-tables improve the running time of your algorithm.

Problem 6: Consider two strings of size M, N. Perform string matching in size O(M+N).

Dragon and Knight

Problem: A dragon and knight live on an island. This island has seven poisoned wells, numbered 1 to 7. If you drink from a well, you can only save yourself by drinking from a higher numbered well. Well 7 is located at the top of a high mountain, so only the dragon can reach it.

One day they decide that the island isn't big enough for the two of them, and they have a duel. Each of them brings a glass of water to the duel, they exchange glasses, and drink. After the duel, the knight lives and the dragon dies.

Why did the knight live? Why did the dragon die?

Solution: show

Dragon knows that knight cant reach well 7. So he thinks that if i give knight water from well 6 or 7, knight would die for sure. So he would get water from well 6 or 7 for knight.

Knight knew that no matter from which well he gives the water, dragon will go and drink from well 7 and live. So he got normal water for dragon and before duel drank water from well 1.

When they exchanged glasses and drank water, dragon rushed to well 7 and drank poison from it, thinking that it would cure the poison he just drank (but he drank normal water), so dragon died. Knight on the other hand already had poison from well 1 so what dragon gave him effectively cured him. So he lived.

Problem: Now consider that Dragon and Knight are equally intelligent then who is expected to die.

Answer: Both survive the battle.

Solution: show

After the battle, dragon drinks from well 1 in order to guarantee that he is poisoned. He then drinks from well 7 to cure himself. Even if knight gets normal water or poison from well 1-6, dragon is now cured and survives.

As for the knight, he drinks from well 1 before the duel. If dragon has got poison for him then he is cured. But if dragon got normal water then he has to ensure that he cures by drinking poison from higher than well 1. So after the battle, knight re-drinks from well 1. This guarantees that he is poisoned from well 1. Then he goes and drinks from well 2 to cure himself.

So both dragon and knight survive the battle.

Number and Age of David's Kids

Problem: The product of the ages of David's children is the square of the sum of their ages. David has less than eight children. None of his children have the same age. None of his children is more than 14 years old. All of his children is at least two years old. How many children does David have, and what are their ages?

Answer: (12,6,4,2)

Solution: show

A way to mathematically analyze this problem is to write the general form of equations, differentiate and maximize w.r.t to largest age. That leads to a Fibonacci number series. But after that complex heuristics have to be employed to devise the solution (and we cannot guarantee if that is the unique).

This is a constraint satisfaction problem. The best way to solve it is write the code. The following code solves the problem in general sense:

Function find_ages(kidAges)  
   s = sum(kidAges)
   p = product(kidAges)
   if s*s == p
      print kidAges

   l = length(kidAges)
   if l >= 7
      return
   else if l == 0
      x = 14
   else
      x = kidAges[l]
   end

   for i = 2 to x-1
      kidAges1 = clone(kidAges)
      kidAges1.add(i)
      find_ages(kidAges1)
   end
end

// call
find_ages([])

After running the code, we see the following solution

[12, 6, 4, 2]

Note that in the above pseudo code, we cloned the kidAges array. If we dont clone then it wont work properly. Down side of cloning is that now we use a huge amount of memory. How do we write code without cloning? Check the python script attached below.

Source Code: david_kids_ages.py

Largest Sum of Consecutive Numbers

Problem: Given an array of N integers (both positive and negative), find the sub-sequence with largest sum.

For ex: Let A = {1 2 -5 4 5 -1 2 -11} then largest sum is 10 (start = 4, end = 7)

Solution: show

The naive solution to this problem can be formulated and coded in O(N^3) time. Consider every possible pair of start and end index and calculate sum of the elements within those index and keep track of the max sum. The following code performs that

max = - Inf
imax = jmax = -1
for i = 1 to N
  for j = i to N
    
    // get sum of elements enclosed in index (i,j)
    s = 0
    for k = i to j
      s += A[k]

    if s > max
      max = s
      imax = i
      jmax = j

Clearly the above algorithm is O(N^3). We can definitely improve the running time of the above algorithm. After observing carefully, we can see that an operation that is run several times (O(N^2) times) is computing the sum of elements between indices i,j. Can we do it quickly. Indeed we can.

If we save the cumulative sum of elements in an array CSUM, then CSUM[i] indicates the sum of elements enclosed in indices (1,i), CSUM[j]-CSUM[i-1] would then indicate the sum of elements enclosed in indices (i,j). The following algorithm captures this intuition.

csum[0] = 0
for i = 1 to N
  csum[i] = csum[i-1] + A[i]

max = - Inf
imax = jmax = -1
for i = 1 to N
  for j = i to N
    s = CSUM[j] - CSUM[i-1]
    if s > max 
      max = s
      imax = i 
      jmax = j

This is the best we can do with this approach. This problem can be done in O(N) time using a clever approach. Consider a sub-sequence with sum s > 0. Let the next element n is such that s + n < 0. Clearly this sub-sequence cannot be part of largest subsequence that contains s and n as removing s,n from that sub-sequence we get a larger sum. Easy to prove using contradiction. This idea is captured in the following O(N) algorithm

max = -Inf
imax = jmax = -1
i_temp = -1
s_temp = 0
for i = 1 to N 
  s_temp += A[i]

  if s_temp > max
    // keep track of max so far
    max = s_temp
    imax = i_temp
    jmax = i

  if s_temp < 0
    // abort this sub-sequence
    s_temp = 0
    i_temp = i + 1

  

Problem: Given an array of N integers (both positive and negative), find the sub-sequence with largest absolute sum. This problem differs from the one at the top as we want the absolute sum (taking mod)

For ex: Let A = {1 2 -5 4 5 -1 2 -11} then largest absolute sum is 11.

Solution: show

Simply keep track of max and min at the same time in the above O(N) solution.

Assuming that we have to use the above algorithm without modification, then we can take these steps to get max absolute sum in O(N) time:

1. Run O(N) algorithm on array A to get the max
2. Create array B by multiply all numbers in A by -1
3. Run O(N) algorithm on array B to get the max
4. Pick the max between the output of two runs of the algorithm

Instead of trying the above two methods, we can try a very nice and simple approach

// build csum matrix as before
csum[0] = 0
for i = 1 to N
  csum[i] = csum[i-1] + A[i]

answer = max(csum) - min(csum)

The answer is maximum element of csum minus the minimum element of csum.

Polynomial Evaluation

Problem: Given a polynomial with degree bound of n=m^r. How do we evaluate the polynomial at n different points such that the polynomial can be recovered from these n points (i.e. the coefficients can be calculated back using the points).

Note: We do not want to recover the coefficients but cleverly choosing the points as that impacts the running time of the algorithm.

Solution: show

Let the polynomial be A(x) = a0 + a1 x + a2 x^2 + ... + a(n-1) x^(n-1).

A(x) has a degree bound of n. Its actual degree can be lower than n-1 if a(n-1) equals zero but that doesn't matter here.

Using Horner's rule, we can compute a polynomial in O(n) time by arranging the terms as follows:

A(x) = a0 + x (a1 + x (a2 + ... + x (a(n-2) + x a(n-1))))

So it will take O(n^2) time to compute polynomial at n different points. But we can use divide and conquer to do better than that i.e. in O(n lg n). We can create "m" different polynomials such that each term of ith polynomial has degree mod m = i.

=> A(x) = [ a0 + a(m) x^m + a(2m) x^2m ... ] +
                   [ a1 + a(m+1) x^(m+1) + a(2m+1) x^(2m+1) ... ] +
                   .
                   .
                   .
                   [ a(m-1) x^(m-1) + a(2m-1) x^(2m-1) + ... ]

=> A(x) = A[0](y) + x A[1](y) + ... + x^i A[i](y) + ... + x^(m-1) A[m-1](y)

where y = x^m, and
            A[i](x) = a(i) + a(m+i) . x + ... + a(n-m+i) . x^(n/m-1)

Still the recurrence for evaluation each point is T(n) = m T(n/m) + O(m) = O(n). So evaluating polynomial at n points leads to O(n^2) which is not great.

Now we can choose the n points cleverly. Lets set w as principal nth root of unity.
=> w^n = 1
=> w = e^(2 . pi . i / n)
          = cos (2 . pi /n) + i . sin(2 . pi / n)

The two interesting properties of w are
1. w^(n/2) = -1
2. [w^(kn/m + p)]^m = w^(kn + mp) = w^(mp)

So the n points on which the polynomial can be evaluated are w^0, w^1, ..., w^(n-1). Now we need to evaluate lower order polynomials of size n/m at only first m powers of w as afterwards it cycles again and we can reuse the computation. This comes from the observation that

A(w^p) = A[0](w^(mp)) + w^p A[1](w^(mp)) + ... + w^(m-1) A[m-1](w^(mp))

Also,
A(m^(kn/m+p)) = A[0](w^(mp)) + w^(kn/m+p) A[1](w^(mp)) + ... + [w^(kn/m+p)]^(m-1) A[m-1](w^(mp))

So A[0], ... A[m-1] are evaluated at p = { 0, 1, ..., m-1 }, and the computation is used for all the other n-m powers of w.

The general recurrence of this algorithm is:

T(n) = m T(n/m) + O(n)
=> T(n) = O(n lg n)

m = 2 works well in practice and is employed by Fast Fourier Transform in order to intrapolate the coefficients of the polynomial. This is generally used for multiplying two polynomials to get higher order polynomial. The steps are to first evaluate the lower order polynomials at n points and then use the multiplication of the point values to generate the coefficients of the higher order polynomial.

Longest Monotone Sequence and Palindromes

Problem: Given an array of integers, find the longest subsequence of elements which monotonically increases. for ex. array = {1 4 8 2 5 7 3 4 6}, the longest subsequence = {1 2 3 4 6}

Solution: show

Let us construct the solution using the LCS problem. LCS is longest common subsequence, which says that given two string X, Y find the longest common subsequence between them, for ex. if X = {a b c a b c} and Y = {b a c b a c} then a possible LCS(X, Y) = {b c b c}. LCS problem can be solved using dynamic programming in O(n^2).

We can solve the longest monotonically increasing sequence problem using the solution to LCS. Let X = original string. Let Y = sort X (increasing order). LCS(X, Y) gives the desired answer (easy to observe).

There is a faster O(n log n) algorithm (for details check here)

Problem: Given a string, find the longest size palindrome in the string. for ex. string = "aaabbccaccbaaa", solution = "bccaccb"

Solution: show

The trivial solution is to check if a palindrom of all possible size starts from a given index. For a index i in array, the possible possitions to test are i < j <= n. Total time taken = O(n^3). We can solve this problem using Longest Common Substring and suffix trees to solve this problem in O(n) time.

Loops in Linked List

Problems with linked lists occur mainly while building or using a bad memory manager. Let us discuss two well known linked list problems.

Problem: Find if two linked lists short at some node. Find the node at which the two lists short.

Solution: show

Subtract the length of smaller list from the larger one. Move a pointer ahead on larger list by the difference of lengths. Now put a pointer on head of smaller list. More two pointers together. The point at which they meet is the point where they short. The following code gives the idea.

d = length(head1) - length(head2)
l1, l2 = head1, head2
if d < 0
  d *= -1 
  l1, l2 = head2, head1

ptr1 = l1
for i <- 1 to d
  ptr1 = ptr1->next

ptr2 = l2
while ptr1 != null
  if ptr2 == ptr1
    return ptr1

ptr1 = ptr1->next
ptr2 = ptr2->next

We can use this idea to find the shorted node to find the solution to next problem.

Problem: A linked list might contain a loop. How do we detect existence of the loop and find the node from which loop starts. Propose an O(n) time algorithm that doesn't take extra space and doesn't modify the linked list.

Solution: show

To find if the linked list contains a loop, run two pointers over the linked list, one that move one node at a time and other that moves two nodes at a time. Both of these pointers meet if the linked list contains a loop, otherwise the pointer moving with twice speed reaches the end while the slower pointer reaches midway.

fastptr = head
slowptr = head

while fastptr != null and fastptr->next != null
  fastptr = fastptr->next->next
  slowptr = slowptr->next
  if fastptr == slowptr
    return "loop exists"
return "loop doesnt exist"

If loop exists, then fast pointer meets the slow pointer before slow pointer completes one rotation of the loop. This fact can be easily proved. Once the pointers meet, we can count the number of nodes in the loop.

loopNodeCount = 0
do 
  loopNodeCount += 1
  slowptr = slowptr->next
while slowptr != fastptr
return loopNodeCount

Now using the trick used in the two linked list shorting problem above, we start two pointers, one from the head and other after advancing "loopNodeCount" nodes ahead. The node at which both the pointers meet is the loop node.

advptr = head
for i in range(1, loopNodeCount)
  advptr = advptr->next

normalptr = head
while normalptr != advptr
  normalptr = normalptr->next
  advptr = advptr->next
return advptr

Monty Hall Problem

This is a very famous probability problem. This problem illustrates that probability is sometimes more convincing than "what we perceive as obvious". I'll post another problem that defeats "obvious reasoning".

Monty Hall problem: Suppose you're on a game show, and you're given the choice of three doors: Behind one door is a car; behind the others, goats. You pick a door, say No. 1, and the host, who knows what's behind the doors, opens another door, say No. 3, which has a goat. He then says to you, "Do you want to pick door No. 2?" Is it to your advantage to switch your choice?

Solution: show

Well for a mathematician stuck on the game show, I would suggest tossing the unbiased coin. The reason being that Game Show host might be clever and seeing that mathematician got the prize behind door 1, would play this card in order to entice the mathematician to switch. Mathematician can be well aware of this trick of Game show host and might not switch. But game show host thought so .. errr. This is a recursive logic and would go to infinity.

Under the assumption that game show host has no hidden motive and performs this step always, the best choice is to switch. That increases the chances by 33.33%. There is a very simple explanation to this. The prize could be behind any doors. Since you pick door 1. Assume game show hosts says that either keep door 1 or take both door 2 and 3, you would go for both door. That doubles up wining chances to 66%. He opens door 3 for you and you open door 2 when you say switch.

Red Green Cards: A box has three cards. First card has both sides green, second card has both sides red, third card has one side green and one side red. A card is picked at random and its one side is observed to be green. What is the probability that the other side of the card is also green.

Solution: show

Well it seems that answer should be 1/2. But its 2/3. For explanation, let us label card 1 as g1|g2, second card as r1|r2, third card as g3|r3. Now since one side of the card is green. It can either be g1, g2, g3. Three events and two point to first card.

Finding Second Smallest Element

Jargon: Order Statistics problem is to find the kth smallest element in an unsorted array A[1..n]. This problem can be solved in O(n) time in average case using randomized select algorithm or in O(n) time in worst case time using an algorithm that chooses the pivot element carefully.

Problem: How to find second smallest element in an array A[1..n] of size n. We would like to do in much less than 2n comparisons.

Answer: Time Complexity = n + ceil(log n) - 2

Solution: show

Finding the smallest element is trivial. The following code would find the smallest element using n-1 comparisons.

min = A[1]
for i = 2 to n
  if A[i] < min 
    min = A[i]
return min

The trivial algorithm to find the second minimum is to keep another variable (smin) along with min and if an element knocks out smin then check if it knocks out min and accordingly do the book-keeping. It can be done using 2n-3 comparisons and takes constant space. A point to observe is that the second smallest element is knocked out by the smallest one at some stage. If we preserve which element is knocked out by whom then finding the second smallest element would just require us to find the smallest amongst the elements knocked out by the smallest element. But in order to do that we need to build the algorithm in a way that we are playing a pair wise tournament. Lets say we play a tournament and knockout n/2 elements after n/2 comparisons. The recurrence relation looks like this:

T(n) = T(n/2) + n/2

Solving this recurrence gives that T(n) = n - 1. The height of the recurrence tree is lg n. So if we just check these elements knocked by root (min) we get the second minimum.For. E.x.

1   5 2   6  
\ /   \ /  
 1     2
  \   /
    1

As soon as we build the tree using n-1 comparisons, we can check the second largest amongst the number knocked out by 1. i.e. 2, 5. so the second minimum is 2. The number of comparison is n + ceil(log n) - 2, where log is taken to the base 2. We implement this algorithm by maintaining a list of indices knocked by an element. The python code is towards the end of this post.

Another algorithm that is very easy to implement but sadly takes 2n - 2logn - 1 in terms of array element comparisons. Simply build a min heap (rooted at 1). Return the smaller of numbers at index 2 or 3 in the array. The cost of min heap building is = 2 * [n/4 + 2n/8 + ...... + (logn - 1) * n / 2^logn]. The solution to this arithmetic - geometric series is 2n - 2logn - 2. Comparing elements at 2nd and 3rd index adds one to the cost. Its very easy to implement though.

Code for n + log n - 2 algorithm: second_smallest.py
Alternate implementation using linked list: second_smallest_linkedlist.py

String Permutations

Printing all permutations of a string is a very common interview question. We'll discuss this problem and some interesting variations of it. The original problem of string permutation says, "print all permutations of a string". As an example, if the string is "abc" there are 6 permutations {abc, acb, bac, bca, cab, cba}. Assume that string contains no duplicate elements.

Solution: show

There are many ways in which this problem can be solved. The most easiest of the ways is to code it using simple yet powerful recursion technique.

function permute(str, d)
  if d == length(str)
    print str
  else
    for i <- d to length(str)
      swap(str[d] <-> str[i]) // swap character for permutation
      permute(str, d + 1)
      swap(str[d] <-> str[i]) // undo swap for parent call

Above code performs the swapping operation twice. First time to generate all possible permutations of character at that level and second time to restore to the original string. It is important to restore to the original string (as passed to this call), otherwise the algorithm ends up printing redundant permutations. Additionally, this code assumes that there are no duplicates in the string, else it fails.

There are other ways in which string permutation can be printed that require circular shifting, but it cant get any simpler than above.

Source Code: permute_unique_chars.py

Lets us assume that string can contain duplicates. How do we print all non-redundant permutations of the string. For ex. If string is "abab" the permutations are {baab, abba, abab, baba, bbaa, aabb}

Solution: show

Clearly the solution to first problem fails here. This is due to fact that we generate all permutations of the string after the swapping step. Same character could get swapped to dth spot again, leading to redundant permutations. A very simple trick solves the game. The trick is to pre-sort the string on alphabetical order. Now we need to ensure that while we are swapping, to generate permutation for that particular character, if that matches the last swapped character we ignore generating permutations. Effectively, we are doing a cyclic shift and before the end of the call we reset the cyclic shift. The following code implements the same:

function permute(str, d)
  if d == length(str)
    print str
  else
    lastSwap = Nil
    for i <- d to length(str)
      if lastSwap == str[i]
        continue
      else
        lastSwap = str[i]

      swap(str[d] <-> str[i]) // swap character for permutation
      permute(str, d + 1)

    for i <- d to length(str)-1
       str[i] = str[i+1]
    str[length(str)] = last

If the string is pre-sorted alphabetically then the above algorithm works magically. Note that in this case, after permute is called, we do not undo the swap. This ensures that we are cyclic shifting and hence the intermediate str from d to l is also sorted. After the for loop call, we reverse the cyclic shift. The reason is that sorting ensures that if there are duplicates then the second occurrence of the duplicate element is picked immediately after we finish generating permutations for the first occurrence of the duplicate element. If sorting is not allowed, then you need to maintain a local hashtable (local to each function call), that maintains characters swapped to dth spot and ensures that duplicate candidates are discarded. Presort and make life simple.

Source Code: permute_chars.py

Some other constraints could be added to make the problem more interesting. One such constrain could be of partial ordering of characters. For Ex. 'a' should appear before 'c', and 'b' should appear before 'c' in all permutations. There can exist ordering that leads to no permutation of the string such as 'a' precedes 'b', 'b' precedes 'c', 'c' precedes 'a'. Additionally implementing the permutation algorithm in these cases require checking for ordering violation at each permutation level or just at the leaf of recursion, depending on whether checking for ordering violations is more costly or generating all permutations.

An interesting problem, that talks of special kind of ordering and can be very efficiently computed is as follows: Assume that the string is made up of equal number of '{' and '}'. You need to print all permutations of this string which could satisfy a C compiler i.e. an ending bracket is balanced by an opening bracket. For ex. for '{{}}' the permutations are [{{}}, {}{}].  

Solution: show

The hard question is how many such permutations exist. For N open and N close brackets, the number of such permutations is Nth Catalan number = 1/(n+1) * factorial(2*N)/factorial(N)^2. The solution is derived from the problem: Number of paths in rectangular grid.

Permutation problems can be easily solved using recursion. The trick is to formulate all the possible cases at a particular level and code according to the formulation. The steps used below illustrate this approach. Let's formulate the problem in general sense. Assume that at a given step of recursion, we have n open brackets and m closed brackets that are not used so far. Now we can formuate each and every case for a given (n,m) and code the possible substeps.

Problem Formulation:
(n,m) ---> NOT POSSIBLE,   if m < n
(n,m) ---> PRINT AND DONE, if m = 0 and n = 0
(n,m) ---> (n-1, m)        if n > 0   
              +
           (n, m-1)        if m > 0 and m > n

The above formulation is generic and takes into account all possible input values of (n,m). If we put following constraint on the initial value of n and m such as following, then actual implementation can omit some of the above cases:

Constraints: 
n > 0
m == n

The actual implementation that follows above constraints and problem formuation:

function permute(str, n, m)
  if m == 0
    print str
  else
    if n > 0
      permute (str + '{', n - 1, m);

    if m > n
      permute (str + '}', n, m - 1);

To enforce the above mentioned constraints, permute function should be kept as inner/private function and the actual function that should be exposed is as follows:

function GenerateCStyleBrackets(N)
  if N <= 0
    return

  str = new String[2 * N]
  permute(str, N, N) // str is passed as pointer.

The memory cost is O(N). The maximum stack depth is also O(N). So total space cost is O(N).

Source Code: permute_brackets.py

Party Friends

Problem: In any party, there exists two people with same number of friends. Is is true or false. Assume that party has more than one people.

Solution: show

Yes, The statement is true. We can prove it by contradiction.

Let there be N people at the party.
Let all the N people have different number of friends.

Since a person can have maximum of N-1 friends and a minimum of 0 friends. Hence the possible number of friends for a person is 0, 1, 2, ..., N-1. Since, we are assuming that there are no two people with same number of friends, so everyone has different number of friends. This means that someone has 0 friend, someone has 1 friend, so on and someone has N-1 friends. But 0 & N-1 cannot coexist. This is because if a person has N-1 friends then that means he is friend with every one, including the one who has 0 friends, which is a contradiction. Hence by pigeon hole principle at least two people should have the same number of friends.

Innocents and Criminals: Finding Minority Entity

Problem: There are two types of people in a particular city, innocents and criminals. All you know is that innocents are in majority and would like to get rid of criminals and criminals would like to protect themselves from persecution. You can ask any number of questions, with yes/no type answer, from any person in the city. Propose an algorithm which requires asking the minimum number of questions.

Hint: Trivial solution requires asking O(n^2) questions. You can do it in less than 2n questions.

Solution: show

The trivial solution is very simple and is in-fact O(n^2) time algorithm. Round up every person and ask him about the status of everyone else. People with majority vote of criminal are criminals and people with minority vote of criminal are innocents. Following code solves it:

for i = 1 to N
   for j = 1 to N
      if person[i].isCriminal(j)  // doesnt matter if i == j
         vote[j] += 1
      else
         vote[j] -= 1

for j <- 1 to n 
   if vote[j] > 0
      person[j].persecute()

The most efficient algorithm requires asking only 2N-2 questions. The key to solution of this problem lies in the solution to the problem of Finding Majority Element. If we can find one innocent person in the city, then he can label everyone else truthfully. We know for sure that an innocent will tell the truth. Criminals can lie or say truth depending on circumstances.

Assume that we formulate the problem this way. Let us consider a pool of people, who claim to be innocents. Also let us say that we declare that we will pick the first person in the pool to be our innocent man and he will reveal the identity of everyone else. Then definitely both innocents and criminals would try their best to capture the first spot in the pool.

Now, we play a game like this. We choose a person randomly to represent the first person in the pool. Now we select a new person randomly and ask him should the last person inducted in the pool is Innocent. If he says Yes, then we add him to the pool. If he says No then we remove him and the last person from the pool and discard them from further selection. If the pool is empty we select a person not selected previously to get the first spot in the pool.

We repeat the process until we don't have anymore people to consider. The intuition is that even if all Criminals join the pool initially (by lying), then innocents would boot them out as they are in majority. If more innocents join the pool initially then criminals will not be able to boot all innocents out. Hence the first person remaining in the pool is indeed an innocent. We can see this argument working inductively as well.

The following code implements above logic with the help of a stack:

stack.push(person[1])
for i <- 2 to N
   if !stack.isEmpty() && person[i].isCriminal(stack.top())
      stack.pop()
   else
      stack.push(person[i])
return stack.bottom()

Finding First Non-Zero Digit of N Factorial

Find the first non-zero digit of N!. Assume N! to be extremely large.

Solution:
If we assume that the product doesnt contain either of 2's and 5's, then finding last digit is trivial. Simple multiply the result with next number and take its mod 10. The problem is complicated as 2's and 5's lead to trailing 0's. So the first step is to remove 2's and 5's. Then solve the problem. The cancel 5's with 2's and multiply back the left over 2's.

For Ex. 7! = 1 * 2 * 3 * 4 * 5 * 6 * 7. We extract 2's and 5's so the product is 1 * 3 * 3 * 7. The trailing digit is 3. We have extracted four 2's and one 5. So we get 3 extra 2's. So multiply to 3 we get 4 as the answer. To check 7! = 5040

d = 1 
n2 = 1

for i = 3 to n
  j = i

  while j%10 == 0 //remove trailing zeros
    j /= 10

  while j%2 == 0 //count of 2's
    j /= 2
    n2 += 1

  while j%5 == 0 //cancel a 5 with a 2
    j /= 5
    n2 -= 1    

  d = (d * j) % 10

d = (d * 2^n2) % 10 //multiply remaining 2's to d
return d

Count Trailing Zeros in N Factorial

How many trailing zeros does N factorial contain, for ex. 5! = 120 so ans is 1. Propose an algorithm to count trailing zeros. Assume that N! is very large and does not fit into word size of the computer.

Solution:
This problem is straight forward. N! = 1*2*3*4*5* .... *N. If we count all 5's in this product, then that many trailing zeros are there. This is because a 5 is countered by 2 to produce a zero. Additionally, there are more 2's than 5's so counting 5's gives the right answer. The following pseudo-code does that:

count = 0
i = 5
while i <= n
  j = i
  while j%5 == 0
    j = j/5
    count += 1
  i += 5
return count

This algorithm takes n/5 outer loop iterations and is bounded by log n [base=5] inner loop iterations, hence it takes O(n log n) time. We can clearly do better than this.

Actually, we can solve this problem in O(log n) time. For a given i the term i/5 gives total number of 5's that come before it. For ex. 17/5 is 3, there are 3 numbers that contain fives which are less than 17 i.e. 5, 10, 15. So we count number of 5, then number of 25, then so on.

count = 0
d = r = 1
while r > 0
  d *= 5
  r = [n/d] // [x] is greatest integer less than x
  count += r
return count

This algorithm runs in O(log n) time and an asymptotic improvement on the previous algorithm.

Picking k Elements Randomly from Stream

Consider the problem of picking K elements randomly from a stream of N elements. Solve the problem for known and unknown (but finite N).

Solution:
For K=1, the solution is posted in my previous blog post Picking an Element Randomly from Stream.

Case 1: N is known
In this case problem can be solved in O(N) time using O(K) space. The idea is that pick the first element with probability K/N. If the element is picked then pick the next element with probability (K-1)/(N-1) otherwise pick it with probability K/(N-1). To see that the second element is still picked with probability K/N. There are two cases.

1. First element is picked. So second element is picked with probability = K/N * (K-1)/(N-1).
2. First element is not picked. So second element is picked with probability = (1-K/N) * K/(N-1).

So the selection probability of second element is = K/N*(K-1)/(N-1) + (1-K/N)*K/(N-1) = K/N. Similar logic extends further. The following code implements this approach.

for i = 1 to N
   if rand <= K/(N+1-i)
      print A[i]
      K -= 1
      if K == 0
         break
      end
   end
end
if K != 0
  print A[N-K:N]
end

Case 2: N is unknown.
If we do not know N beforehand, and yet we want to guarantee that all elements are selected with probability K/N, we can use Reservoir Sampling algorithm.

for i = 1:K
   S[i] = A[i]
end

for i = K+1:length(A)
   j = random(1,i)
   if j <= K
      S[j] = A[i]
   end 
end

The first K elements of the stream are automatically selected. So for N = K this solves the problem. Now consider N = K+1. For i <= N-1 all elements are selected with probability 1 so far. The last element should be selected with probability K/N and so should others be. The index j is less than or equal to K with probability K/(K+1), so last element is rejected with probability 1/(K+1). For the elements in the array at index j they can be rejected if last element is accepted and its index is j. The chances of that is K/(K+1) * 1/K = 1/(K+1). So all elements are rejected with 1/(K+1) probability. Now consider N = K+2. Until index i=K+1 all elements are rejected with probability 1/(K+1). Using same logic as we applied above, we can see that an element gets rejected with probability 2/(K+2).

Optimal Coin Change Problem

Given an unlimited supply of coins of denominations C1, C2, ..., CN we wish to make change for a value V. Give an algorithm for producing change with minimum number of coins.

Complexity: Depending on denomination of coins, we can either use fast greedy algorithm or dynamic programming.

Solution:
Consider two specific types of denominations of coins
a) 1, C, C^2, C^3, ..., C^(N-1)
b) 1, C, 2C, 3C, ...., (N-1)C

For both of these arrangements, the problem can be solved using a greedy approach. We see that if we first pay with the highest denomination coin, the value (for which change is required) reduces greatest (as compared to any lower denomination). Additionally we see that optimal solution would require us to use as many highest denomination coin as possible. We can prove this claim using contradiction. The greedy algorithm that is pretty fast and runs in O(N), is as follows:

change = 0
for j = N to 1
  change += int(value/C[j])
  value = value % C[j]
return change

The generic problem of coin change cannot be solved using the greedy approach, because the claim that we have to use highest denomination coin as much as possible is wrong here and it could lead to suboptimal or no solutions in some cases. For ex, if coin denomination are 1, 4, 6 and we want change for 8. Using one coin of denomination 6 forces us to use 2 coins of denomination 1, but clearly the optimal solution is to use 2 coin of denomination 4 each. The key observation to make is the optimal structure of the sub-problem. If we assume for the time being that we have optimal change for all values < i. To produce change for value i, we need to add a coin to some previous solution with value j < i. We need to choose j that minimizes the value of change[i], the O(V*N) algorithm looks like the following:

change[0] = 0 // no coin required.
for i = 1 to V
  change[i] = INF
  for j = 1 to N
    if i >= C[j] and change[i] > change[i - C[j]]
      change[i] = 1 + change[i - C[j]]
return change[n]

To print the choice of coins, we can keep a separate array to track which coin, cj was chosen for a given i.

Source Code: coin_change.py

Picking an Element Randomly from Stream

How to pick a number randomly from a stream (or a linked list) with a finite but unknown length N? Solution must use constant space and guarantee that number is picked with probability = 1/N.

Solution:
If we know the length of the stream, then problem is trivial. Simply generate a random number r between 1 and N. Declare the number at index r as the desired random number.

For an unknown but finite N, we need that elements are selected with equal probability. Let us first build the loop invariant property for the solution. Say we have seen k elements so far, the selected number (say n) must be selected with probability 1/k. As we see a new element (say n1), we should set n=n1 with probability 1/(k+1). This ensures that n1 is selected with probability 1/(k+1). In the other case, n (the one NOT replaced by n1) is selected with probability = 1/k * (1-1/(k+1)) = 1/(k+1). Hence this strategy ensures that a number a selected with probability 1/N for N>=k. This loop invariant is adhered by the following code:

n=0, N=0
while Stream.hasElement()
  ++N
  if rand <= 1/N 
    n = Stream.nextElement()
return n

After the algorithm processes all elements, n would contain the randomly selected element and N would contain the number of elements in the stream.

Generating Random Numbers

How to generate random integer between A and B, using an unbiased coin. For ex. If A = 3 and B = 6 then your algorithm should generate 3, 4, 5, 6 with equal probabilities.

Solution:
Without loss of generality, we consider the problem of generating random number r in range [0,n] with n=B-A. We can return r+A as the solution to input problem. The number n can be represented using k bits where k = 1 + [log n] ([x] indicates the largest integer smaller than x). Now the k-bits can be generated using the unbiased coin trivially in O(k). The generated number r can be greater than n with probability = 1-(n+1)/2^k. If that happens then repeat the experiment. Pseudo code for the algorithm looks like the following:-

function random(A, B)
  n = B - A
  k = 1 + int(log n)
  r = 0
  for i = 1 to k
    bit = coin_toss(); // head means 1, tail means 0
    r += bit
    r <<= 1
    if r > n
      return random(A, B)
  return r

This function can go into infinite loop. Let us calculate the expected number of times random(A,B) is called. It is called once by the the actual caller and called again if r > n. Let us assume that the expected number of calls be e. The probability that r <= n is given as Pr(r <= n) = (n+1)/2^k.

=> e = 1 * Pr(r <= n) + (1 + e) * Pr(r > n) => e = 2^k/(n+1) 

In the worst case, n=2^m+1. In this case, k=1+m. We get

==> e = 2 * 2^m/(2^m + 2) < 2

So expected number of internal calls is at-most 2. This gives a good assurance against going into infinite loops. Note that e can be estimated by using the observation that random variable e - 1 is a geometric distribution. Another way to reach above conclusion is by observing probability of failure, p = 1-(n+1)/2^k < 1/2. Since the probability of failure is less than 1/2, the chances that we run into infinite loop is quite low.

The above solution can be explained in simpler terms as well. Assume that we have people numbered A to B and we are playing a tournament. In each round of the tournament, two players are paired up, they call on the coin toss, player who wins goes to next level. In case if pairs can not be formed we add dummies to form pairs. If a normal player wins we are done, but if dummy wins we repeat the experiment again. So for example if we have 3, 4, 5, 6, 7 as initial number, and we add 'd' as dummy. Then a possible tournament could be as follows:

round 1: (3 vs 4) , (5 vs 6) , (7 vs d) --- winner --> 3, 5, d
round 2: (3 vs 5), (d vs d) -- winner --> 5, d
round 3: (5 vs d) -- winner --> 5

If d wins the last round, we repeat the experiment. This algorithm is just the same as earlier algorithm.

Finding Majority Element

Jargon: Majority Element is an element of an array that occurs more than half the number of times in the array. Also, [x] means smallest integer greater than x.

Problem 1: Assume that an integer array A[1..n] has a majority element and elements other than majority element are distinct. How to find majority element. What's the space and time complexity of the algorithm ?

Complexity: Time Complexity = n/2(comparisons), Space Complexity = O(1)

Solution: show

We will use pigeonhole principle to make our clever observations. There can be a maximum of [n/2]-1 elements that are distinct, remaining [n/2] element have to be same (as majority element exists in the array). We can consider that the [n/2]-1 distinct elements to be laid out and the space between two laid out elements to be a basket of infinite size. So we will have [n/2] baskets. If we try to distribute [n/2] majority element to these baskets, then in only one permutation will each basket get one element. Otherwise, some baskets would get more than one element and some baskets would remain empty.

Using this observation, we can claim that to find majority element we need to check adjacent elements for equality or two hop neighbor (adjacent to neighbor) for equality. That requires two comparison per element and the algorithm would look like this:

for i <- 1 to n-2
  if A[i] == A[i+1] or A[i] == A[i+2]
    return A[i]

A quick observation tells that this algorithm's for loop would stop when i=[3n/4]+1 even in worst case and the number of comparisons done will be ~ 3n/2. Now, we can make another clever observation that is, if we do not find two adjacent elements to be equal after complete scanning of array, then first element has to be majority element by pigeonhole principle.

for i <- 1 to n-1
  if A[i] == A[i+1]
    return A[i]
return A[1]

In Worst case this algorithms stops at i=n+1 and the number of comparisons done is = n-1. Even though this looks better than the first solution but it could be worse in practice. On a highly pipelined architecture with branch prediction, first one is much better than the second one, even though it does more comparisons per loop. So testing for large array sizes with random shuffling of elements will give clear insight into which algorithm is faster. Now an extremely clever solution which is based on observation that if we pair array elements and strike out a pair if its both the elements are distinct or else return the pair element as majority element. If all pair are striked out then what ever element is left has to be majority element.

i = 1
while i <= n 
  if A[i] == A[i+1]: return A[i]
  i+= 2

return A[n]

Worst case analysis tells that algorithms would stop when i=n-1 and the number of comparisons done is ~ n/2. This is indeed the best in worst case that we can do.

Problem 2: Assume that an integer array A[1..n] has a majority element and elements other than majority element need NOT be distinct. How to find majority element. What's the space and the time complexity of the algorithm?

Complexity: Time Complexity=n(comparison), Space Complexity=O(1)

Solution: show

This algorithm uses an observation that if we consider our array to be list of voters and array value is the candidate to whom they are voting, then majority element is the winning candidate. Additionally, if all other candidates are merged to form one candidate then still it will fail to defeat the majority element candidate (hence the name majority element). So we can think of it this way. Assuming a negative vote decrease the count of majority element then still the count of majority element would be atleast 1. This algorithm implements the above idea.

element = A[1]
votes = 1
for i <- 2 to n:
  if A[i] == element
    votes += 1
  else if votes > 0
    votes -= 1
  else:
    element = A[i]
    votes = 1

return element

This algorithm does n-1 comparisons. The comparisons such as votes > 0 don't count as they are integer variable comparison.

Problem 3: Now consider that there are k majority elements, i.e., each of the k majority elements appear in the array more than ceil(n/(k+1)) times. How do we find the k majority elements. (Problem 2 is a special case of this problem with k=1).  

Complexity: Time Complexity=k*n(comparison), Space Complexity=O(k)  

Solution: show

Algorithm uses the same intuition as that presented for problem 2. But instead of one counter, we keep k counters. Each counter is initialized with a unique element from the array and we maintain count of how many times that element has been seen so far. Now for a new element x, if it matches an element in the counter then its count is incremented by 1. If the counter of an element is 0 then the x replaces that element and its count is set to 1. If x doesnt match any element on the counter, and all elements have count > 0, then all counters are decremented by 1. Then we move to next element. Following code implements this logic.

for i = 1:n
   // update element count in majority array
   for j = 1:k
      if Maj[j] == A[i] 
         Count[j] += 1
         break
      end
   end
   
   // succeeded in above operation (dont go below)
   if k > j
      continue
   end
   
   // Came here, it means that element not in 
   // majority array (put it in an empty slot)
   for j = 1:k
      if Count[j] == 0
         Maj[j] = A[i] 
         Count[j] = 1
         break
      end
   end
   
   // succeeded in above operation (dont go below)
   if k > j
      continue
   end
   
   // Came here, it means that no empty slow 
   // (decrement counts of all majority elements)
   for j = 1:k
      Count[j] -= 1
   end
end   

The algorithm runs in O(nk) time and takes O(k) extra space. It is easy to see that algorithm finds all the k majority elements. This is becase a non-majority element can knock of all the K majority element once. Since there are less than n - k*ceil(n/(k+1)) (< n/(k+1)) non-majority elements, majority elements will survive in the Maj array.

Airplane Seating Problem

100 passengers are boarding an airplane with 100 seats. Everyone has a ticket with his seat number. These 100 passengers boards the airplane in order. However, the first passenger lost his ticket so he just take a random seat. For any subsequent passenger, he either sits on his own seat or, if the seat is taken, he takes a random empty seat. What's the probability that the last passenger would sit on his own seat?

Answer: 1/2

Solution:
When 100th passenger arrives, only seat 1 or 100 is vacant (rest all must be non-empty). To see this, let seat 50 is vacant. Since passenger 50 came before passenger 100, he would sit in his seat only. Hence seat 50 cannot be vacant. So all permutations of the seating arrangement would result in last person sitting in either seat 1 or seat 100. Both the options are equi-likely.

Tip: Its wiser to solve problems like these for small values and then try to generalize. For e.g. let there be three passengers p1, p2, p3. p1 can sit in seat 1 with prob=1/3. In this case p3 sites in seat 3. p1 can sit in seat 2 with probability = 1/3, p2 can sit in either seat 1 or 3 (with prob=1/2). so p3 gets to sit in his seat with probability = 1/3 * 1/2 = 1/6. p1 can sit in seat 3 with probability 1/3. In this case p3 can sit in seat 3 with probability = 0. So the total probability for p3 to sit in his seat = 1/3 + 1/6 + 0 = 1/2

Tic Tac Toe

Two players are playing a game and take alternating turns. There are 9 cards on the table with numbers from 1 to 9. On each turn, a player picks one card from the table. The first player to have 3 cards that total a sum of 15 wins. If no one can after all cards are distributed, then it's a draw. Can you tell who wins, assume both players are highly and equally intelligent and what is the winning strategy ?
Source: Petr Mitrichev's Blog

Answer: Draw

Solution:
If player 1 picks any card except for 5 then there are three ways he can pick second card in order to total 15. Player 2 will block one way in his following turn. Leaving two options for player 1. For any way that player 1 chooses, he has just one way now to total 15. Player 2 will block that way. Player 2 has blocked all ways of player 1 so far. But in the two moves so far player 2 has one way to reach 15. Player 1's next turn can block that. But, This analysis is hard to follow and unclear to see for all the possible combination.

This problem can be modeled in form of tic-tac-toe by constructing a 3 x 3 magic square, as follows:

4  9  2
3  5  7
8  1  6

Now all rows, diagonals, columns sum to 15. So the problem is reducible to playing a tic-tac-toe. Its easy to see that if two players are highly and equally intelligent then the game would result in a draw.

Math Magician

A magician has one hundred cards numbered 1 to 100. He puts them into three boxes, a red one, a white one and a blue one, such that each box contains atleast one card. A member of audience draws two cards from two different boxes and announces the sum of the number on those cards. Given this information magician locates the box from which no card has been drawn. How many ways are there to put the cards in the boxes so that the trick works.
Source: International Maths Olympiad 2000

Answer: 12

Solution:
We split the possible arrangements into two different cases.

Case 1: Assume that there exists an i such that i, i+1, i+2 go in different box (say wrb). Since, (i+1) + (i+2) = i + (i+3) => (i+3) should go in white box. Similarly i-1 should go in blue box. Since the pattern repeats without much choice so it boils down to assigning different boxes to first three cards. Total of 6 ways.

Case 2: Assume that no three neighbors are put in different boxes. Let card 1 be in white box. Let i be the smallest card to go in red box (such that i > 2). Let j be the smallest card to go in blue box (such that j < 100). Let j > i.

==> i-1 is in white box.
==> i + j = (i-1) + (j+1)
==> j+1 is in white box.

==> i + (j+1) = (i+1) + j
==> i+1 is in blue box
==> j = i+1

This violates the main assumption from both ends i-1, i, i+1 are in different boxes and also i, i+1, i+2, hence j = 100. so that j+1 doesn't exist and both 99 and 98 belong to same box other than blue.

also (i-1) + 100 = i + 99 => card 99 is in red box => card 98 is in red box

but 2 + 99 = 1 + 100 => 2 is red. in general assume that there exists a card c (>1) and it belongs to white box and also c-1 to white box. => c + 99 = (c-1) + 100. which leads to failure of the trick. Assuming cards surrounding c are red. c + 99 = c-1 + 100 => failure of trick. Hence no such c exists and c=1 is the only card in white box.

hence the combination looks like wrrrr.........rrrrrb i.e. one card (value 1) in white box, one card (value 100) in blue box and rest all cards in red box. There are six ways of such arrangements.

Unbiased Coin Tossing

Given a biased coin, with probability of Heads equal to x. How to do unbiased coin tossing?

show

Lets define an event, E as tossing the biased coin twice. The possible outcomes with probabilities is as follows

P(h,h) = x^2
P(h,t) = x(1-x)
P(t,h) = x(1-x)
P(t,t) = (1-x)^2

The event h,t or t,h are equi-likely, without any bias we can call that if Event h,t occurs it means head, t,h means tails but if h,h or t,t occurs we repeat the experiment.

Try to find the expected number of coin toss that would be required to call heads or tails?

show

Let expected coin toss be e.

Probability that we get outcome in 1st event is 2x(1-x)
Total number of coin toss would be 2
Probability that we get no outcome in 1st event is [1 - 2x(1-x)]
Total number of coin toss would be 2 + e (we wasted 2 coin toss and still we expect e)

==> e = 2 * 2x(1-x) + (2+e) * [1 - 2x(1-x)]
==> e = 4x(1-x) + 2 - 4x(1-x) + e - 2ex(1-x)
==> e = 1/ [x(1-x)]

so for x = 1/2, e = 4
for x=2/3, e = 4.5
for x=1, e = INF (as expected, because all we get is chain of h h h h h)

What is the probability that there wont be any outcome in e coin tosses (expected outcomes)?

show

p = prob of e heads + prob of e tail
==> p = x^e + (1-x)^e

To find a bound on this p in polynomial terms can be done by using binomial expansion and Newtonian series is out of the scope of this blog. But some number crunching is.

For x = 1/2, e = 4, p = 0.125
For x = 2/3, e = 4.5, p = 0.168
For x = 3/4, e = 5.33, p = 0.216
For x = 0.99, e = 101, p = 0.362

This tells us that probability that outcome comes is quite good even for high coin biases.

How can we improve on the expected number of coin tosses?

show

In above method, we say that HT or TH terminates experiment. and continue the experiment a fresh when the outcome is HH or TT.

We can further combine outcome of two such events to increase the probability of outcome e.g. say HH TT => heads and TT HH means tails.

How much expected coin tosses are we doing in this case?

show

try

Cards Shuffling Problem

A common algorithm problem is Cards Shuffling Problem, which states "how to shuffle a deck of cards randomly" or in more general "how to randomize an array of elements". A naive algorithm for this problem can be:

for i = 0 to N
  r = random (0, N)
  exchange C[i] and C[r]

Assume N=2. We see 4 possible combination of (i,r) namely (0,0), (0,1), (1,0), (1,1), leading to 2^2 outputs (N^N in general). However, card shuffling should only lead to N! permutations. So even though this algorithm looks correct, it is logically incorrect. A small change in the above code makes it correct:

for i = 0 to N
  r = random (i, N)
  exchange C[i] and C[r]

Generating r randomly from between i and N ensures that once a card is swapped to a given i, it's position is fixed. Above algorithm, with iteration on cards in reverse order is Knuth-Fisher-Yates shuffle algorithm.