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 <stdint.h>

#define STR_MAX    (100000)

#define FAST_TRIAL  (1)

uint64_t serialNum[STR_MAX+2];

int main(void)
{
    uint64_t n;
    uint64_t k;
    char s[STR_MAX+1];

    scanf("%ld %ld", &n, &k);
    scanf("%s", s);

    uint64_t ptr = 0;
    char curC;
    char nxtC;

    uint64_t indx   = 0;
    uint64_t head1  = 0;
    uint64_t max    = 0;
    uint64_t cntK   = 0;
    uint64_t curSum = 0;

    curC = s[ptr++];
    /* serialNumは 1 の数 -> 0の数としたいので0ならindxを送る */
    if ( curC == '0' ) {
        indx++;
    }

#ifdef FAST_TRIAL
    while ( (curC != '\0') && (cntK < k) ) {
        /* nxtCを得る */
        nxtC = s[ptr++];

        serialNum[indx]++;

        /* どうやら次は別の文字のようだ */
        if ( curC !=  nxtC ) {
            curSum += serialNum[indx++];

            if ( curC == '0' ) {
                /* 逆立ちできる人たちが増える */
                cntK++;
            }

            /* ここまでで今回のserialNumによるcurSumが決定するので最大値更新 */
            max = ( curSum >= max ) ? curSum : max;
        }

        curC = nxtC;
    }
#endif

    while ( curC != '\0' ) {
        /* nxtCを得る */
        nxtC = s[ptr++];

        serialNum[indx]++;

//        uint64_t dbgIndx = indx;

        /* どうやら次は別の文字のようだ */
        if ( curC !=  nxtC ) {
            curSum += serialNum[indx++];

            /* 0の連続が終わったときは昔の0を捨てる場合がある */
            if ( curC == '0' ) {
#ifdef FAST_TRIAL
                curSum -= (serialNum[head1] + serialNum[head1+1]);
                head1 += 2;
#else
                /* 逆立ちできる人たちが増える */
                cntK++;

                /* 逆立ちできる人たちがK人より多いとき */
                if ( cntK > k ) {
                    /* 
                     * 1Grp -> 0Grp -> 1Grp -> 0Grp ... 1Grp -> NEW 0Grp
                     * となるので先頭の0Grpと1Grp分を引いて, 0Grpのheadを移動させる
                     */
                    curSum -= (serialNum[head1] + serialNum[head1+1]);
                    head1 += 2;
                }
#endif
            }

            /* ここまでで今回のserialNumによるcurSumが決定するので最大値更新 */
            max = ( curSum >= max ) ? curSum : max;
        }

//        printf("%c : cntK=%ld, curSum=%ld, serialNum=%ld, head1=%ld, max=%ld\n", curC, cntK, curSum, serialNum[dbgIndx], head1, max);

        curC = nxtC;
    }

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

    return 0;
}