Problem:
Score : 400 points

Problem Statement

N people are arranged in a row from left to right.

You are given a string S of length N consisting of 0 and 1, and a positive integer K.

The i-th person from the left is standing on feet if the i-th character of S is 0, and standing on hands if that character is 1.

You will give the following direction at most K times (possibly zero):

Direction: Choose integers l and r satisfying 1 \leq l \leq r \leq N, and flip the l-th, (l+1)-th, ..., and r-th persons. That is, for each i = l, l+1, ..., r, the i-th person from the left now stands on hands if he/she was standing on feet, and stands on feet if he/she was standing on hands.

Find the maximum possible number of consecutive people standing on hands after at most K directions.

Constraints

N is an integer satisfying 1 \leq N \leq 10^5.

K is an integer satisfying 1 \leq K \leq 10^5.

The length of the string S is N.

Each character of the string S is 0 or 1.

Input

Input is given from Standard Input in the following format:

N K
S

Output

Print the maximum possible number of consecutive people standing on hands after at most K directions.

Sample Input 1

5 1
00010

Sample Output 1

4

We can have four consecutive people standing on hands, which is the maximum result, by giving the following direction:

Give the direction with l = 1, r = 3, which flips the first, second and third persons from the left.

Sample Input 2

14 2
11101010110011

Sample Output 2

8

Sample Input 3

1 1
1

Sample Output 3

1

No directions are necessary.

Solution in C (GCC 5.4.1) (judge verdict: Accepted)

#include<stdio.h>

int main() {
  int N, K;
  char S[100000];
  char S0;
  int cont[100000];
  int count = 1;
  int count_stand;
  int max = 1;
  int i, j;
  int set_n = 0;

  scanf("%d %d", &N, &K);
  scanf("%s", S);
  S0 = S[0];

  for(i = 1; i < N; i++) {
    if(S[i] == S0) count++;
    else {
      cont[set_n] = count;
      count = 1;
      set_n++;
      S0 = S[i];
    }
  }
  cont[set_n] = count;

  //for(i = 0; i < set_n + K * 2 + 1; i++) printf("%d\n", cont[i]);

  for(i = 0; i < set_n + 1; i++) {
    count_stand = 0;
    if(i == 0 && S[0] == '0') {
      for(j = i; (j < i + K * 2 && j < set_n + 1); j++) {
        count_stand += cont[j];
      }
    } else if((S[0] == '0' && i % 2 != 0) || (S[0] == '1' && i % 2 == 0)) {
      for(j = i; (j < i + K * 2 + 1 && j < set_n + 1); j++) {
        count_stand += cont[j];
      }
    }
    if(count_stand > max) max = count_stand;
  }

  printf("%d\n", max);

  return 0;
}