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 Ruby (2.3.3) (judge verdict: Accepted)


inputs = readlines

input=[]
for i in 1..inputs.size do
    input.push(inputs[i-1].split)
end

n = input[0][0].to_i
k = input[0][1].to_i
s = input[1][0].split("")

stack = []
if s[0] == "0"
    stack.push(0)
end

sum = 0
max = 0

count = 0
tar = s[0]

for i in 1..n do 
    if tar != s[i-1] 
        tar = s[i-1]
        stack.push(count)
        sum += count
        count = 1
        if stack.size > 2*k +1
            sum -= stack.shift
        end
        if tar == "0" and sum > max 
            max = sum
        end
    else
        count += 1
    end
end

stack.push(count)
sum += count
if s[n-1] == "0"
    stack.push(0)
    sum -= stack.shift
end

if stack.size > 2*k +1
    sum -= stack.shift
elsif stack.size == 1
    sum = n
end

if sum > max 
    max = sum
end

p max