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 (Clang 3.8.0) (judge verdict: Wrong Answer)

#include<stdio.h>
#include<string.h>
#include <math.h>
#include <stdlib.h>
int main()
{
    int n,k;
    scanf("%d%d",&n,&k);getchar();
    int i,b[100005],num1=0,num2=0,j=0;
    char a[100005];
    for(i=0;i<n;i++)
    {scanf("%c",&a[i]);
    if(a[i]=='1'&&a[i-1]=='0')
        b[j++]=num2;
    if(a[i]=='0'&&a[i-1]=='1')
        b[j++]=num1;

      if(a[i]=='1')
        {num1++;num2=0;}
      if(a[i]=='0')
        {num2++;num1=0;}
        if(i==n-1)
       {
        if(a[i]=='1')
            b[j++]=num1;
        else
            b[j++]=num2;
        }
    }
    int t=j-1,max=0;
    j=1;
    int c[100005]={0};
    for(i=0;i<=t;i++)
        {c[j]=c[j-1]+b[i];j++;}
    if(t<2*k)
        printf("%d\n",c[j-1]);
    else
    {if(a[0]=='1')
    {
        for(i=0;i<=t-2*k;i++)
            max=(c[i+2*k+1]-c[i])>max?(c[i+2*k+1]-c[i]):max;
    }
    else
        for(i=0;i<=t-2*k;i++)
            max=(c[i+2*k]-c[i])>max?(c[i+2*k]-c[i]):max;
    printf("%d\n",max);}
	return 0;
}