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: Runtime Error)

import sys
def I(): return int(sys.stdin.readline())
def LI(): return list(map(int,sys.stdin.readline().split()))
mod = 10**9 + 7
inf = float('inf')
ans = int(0)

N, K = LI()
S = str(input())
S = S + "2"
d1 = []
d0 = []
k = S[0]
if k=="0":
    d1.append(0)
cou = int(1)
for i in range(N):
    if S[i]!=S[i+1]:
        if S[i]=="0":
            d0.append(cou)
        if S[i]=="1":
            d1.append(cou)
        cou = int(1)
    else:
        cou +=1

#print(d1)
#print(d0)

ansC = []
t = 2*K+1
newN = len(d1)
if len(d1)==1 or d0==[] or K>=N//2+1:
    ansC.append(N)

if newN!=len(d0):
    d0.append(0)
    d0.append(0)
sumd1 = [0]*(newN+t)
sumd0 = [0]*(newN+t)
for i in range(newN):
    sumd1[i+1] = d1[i]
    sumd0[i+1] = d0[i]
for i in range(newN+1):
    sumd1[i+1] += sumd1[i]
    sumd0[i+1] += sumd0[i]

#for i in range(max(newN-K+1,K)):
if ansC==[]:
    for i in range(newN - K + 1):
        ans = (sumd1[i + K + 1] - sumd1[i]) + (sumd0[i + K] - sumd0[i])
        ansC.append(ans)

print(max(ansC))
#print(ansC)
#print(sumd1)
#print(sumd0)