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>
#include <math.h>

typedef struct {
  int left_count_1;
  int right_count_1;
  int count_0;
} Count_1;

int main(void) {
  
  int N, K;
  char S[100001];
  
  int n = 0;
  Count_1 count_1[100001];
  
  scanf("%d %d", &N, &K);
  scanf("%s", &S);
  
  int current_count_1 = 0, current_count_0 = 0;
  int find_0 = 0, last_index_0;
  for(int i = 0; i < N; ++i) {
    
    if(S[i] == '0') {
      
      if(!find_0) {
        find_0 = 1;
        count_1[n].left_count_1 = current_count_1;
        current_count_1 = 0;
      } else {
        if(i - last_index_0 > 1) {
          count_1[n].count_0 = current_count_0;
          count_1[n].right_count_1 = current_count_1;
          ++n;
          current_count_0 = 0;
          count_1[n].left_count_1 = current_count_1;
          current_count_1 = 0;
        }
      }
      
      current_count_0++;
      last_index_0 = i;

    } else current_count_1++;
    
  }
  
  if(current_count_0 > 0 || current_count_1 > 0) {
    count_1[n].count_0 = current_count_0;
    count_1[n].right_count_1 = current_count_1;
  }
  
  int max_count_1 = 0;
  for(int i = 0; i <= n; ++i) {
    
    int total_count_1 = count_1[i].left_count_1;
    for(int j = 0; j < K && j + i <= n; ++j) {
      total_count_1 += count_1[i + j].count_0 + count_1[i + j].right_count_1;
    }
    
    if(max_count_1 < total_count_1) max_count_1 = total_count_1;
    
  }
  
  printf("%d\n", max_count_1);
  
  return 0;
  
}