Approach 1: Prefix Sums


For say a 5 digit string, the answer is either '00000', '00001', '00011', '00111', '01111', or '11111'. Let's try to calculate the cost of switching to that answer. The answer has two halves, a left (zero) half, and a right (one) half.

Evidently, it comes down to a question of knowing, for each candidate half: how many ones are in the left half, and how many zeros are in the right half.

We can use prefix sums. Say P[i+1] = A[0] + A[1] + ... + A[i], where A[i] = 1 if S[i] == '1', else A[i] = 0. We can calculate P in linear time.

Then if we want x zeros followed by N-x ones, there are P[x] ones in the start that must be flipped, plus (N-x) - (P[N] - P[x]) zeros that must be flipped. The last calculation comes from the fact that there are P[N] - P[x] ones in the later segment of length N-x, but we want the number of zeros.


For example, with S = "010110": we have P = [0, 0, 1, 1, 2, 3, 3]. Now say we want to evaluate having x=3 zeros.

There are P[3] = 1 ones in the first 3 characters, and P[6] - P[3] = 2 ones in the later N-x = 3 characters.

So, there is (N-x) - (P[N] - P[x]) = 1 zero in the later N-x characters.

We take the minimum among all candidate answers to arrive at the final answer.

Complexity Analysis

  • Time Complexity: , where is the length of S.

  • Space Complexity: .

Analysis written by: @awice.