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 Python (3.4.3) (judge verdict: Wrong Answer)

import sys

N, K = map(int, input().split())
S = input()
"""
Sに対して操作を行い1を何個連続で並べられるかという問題
SはN桁の2進数で表すことができる
0を潰していくほうが効率がいいので0を潰す
連続している数字の数を配列にする
3
1 00 1 0000 1111 0 1111 000 1 0 11
[1, 2, 1, 4, 4, 1, 4, 3, 1, 1, 2]
000 1 000 111 0000 1 0 1 000 11 0 1
[3, 1, 3, 3, 4, 1, 1, 1, 3, 2, 1, 1]

数え始める先頭(S_i)が1の場合連続で並ぶ1の数は得られた配列のS_i+2K  までの合計で与えられる
数え始める先頭(S_i)が0の場合連続で並ぶ1の数は得られた配列のS_i+2K-1までの合計で与えられる
"""
S_1 = S[0] # Sの先頭の数字

S_arr = []
checker = S_1 # 現在の並んでいる数字
num = 0
for i in range(N):
    if checker == S[i]:
        num += 1
    
    else:
        S_arr.append(num)
        num = 1
        checker = S[i]
S_arr.append(num)

S_1 = int(S_1)

l = len(S_arr)
if S_1 == 1 and l <= K * 2 + 1:
    print(N)
    sys.exit()

if S_1 == 0 and l <= K * 2:
    print(N)
    sys.exit()
"""
sum(S_arr[i:i+K*2+1]) # 数え始めが1
sum(S_arr[i:i+K*2])   # 数え始めが0
"""
S_ini = S_1
def f1(i): # i はS_arrのindex
    ini = S_1 ^ (i % 2)
   
    if i+K*2 + 1 * ini > l:
        return 0
    return sum(S_arr[i:i+K*2 + 1 * ini])
kazunoko = [0] * l
for i in range(l):
    kazunoko[i] = f1(i)
# print(S_arr)
# print(kazunoko)
print(max(kazunoko))