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 Java8 (OpenJDK 1.8.0) (judge verdict: Time Limit Exceeded)


import java.io.IOException;
import java.io.InputStream;
import java.io.PrintWriter;
import java.util.ArrayList;
import java.util.List;
import java.util.NoSuchElementException;

public class Main {

	public static void main(String[] args) throws IOException {
		new Main().solve();
	}

	private void solve() throws IOException {
		try {
			solveA();
			//			 solveB();
			//			 solveC();
			//			 solveD();
			//			 solveE();
			//			 solveF();
		} finally {
			if (in != null) {
				in.close();
			}
			if (out != null) {
				out.flush();
				out.close();
			}
		}

	}

	private void solveA() {
		int n = nextInt();
		int k = nextInt();
		String wk = next();

		int cnt = 0;
		int now = 1;
		List<Integer> oneZeroList = new ArrayList<Integer>();
		for (char i = 0; i < wk.length(); i++) {
			if (wk.charAt(i) == (char) ('0' + now)) {
				cnt++;
			} else {
				oneZeroList.add(cnt);
				cnt++;
				now = 1 - now;
			}
		}
		if (cnt != 0) {
			oneZeroList.add(cnt);
		}
		if (oneZeroList.size() % 2 == 0) {
			oneZeroList.add(0);
		}

		int[] resList = new int[oneZeroList.size() + 1];

		/*
		 * 累積和用配列に詰めていく
		 * テストケース1の
		 * 14 2
		 * 11101010110011
		 * を例に取ると、1,0の配列は以下になる。
		 * 3 1 1 1 1 1 2 2 2
		 * 1 0 1 0 1 0 1 0 1
		 * 上記の様な1-0リストだとしたら
		 * 累積和用配列には以下の様に詰められる
		 * resListのindex  0 1 2 3 4 5 6  7  8  9
		 * resListのvalue  0 3 4 5 6 7 8 10 12 14
		 * 0->1に変換可能な回数は2回
		 * なので、5この幅を取りたい
		 */
		for (int j = 0; j < oneZeroList.size(); j++) {
			resList[j] = oneZeroList.get(j);
		}

		int range = 2 * k;

		long res = 0;
		int maxSize = oneZeroList.size();
		int rightMax = maxSize - 1;
		for (int i = 0; i < maxSize; i += 2) {
			int left = i;
			int right = Math.min(i + range, rightMax);
			long leftNum = i != 0 ? resList[left - 1] : 0;
			long temp = resList[right] - leftNum;
			res = Math.max(res, temp);
		}

		out.println(res);
	}

	private void solveB() {
		int n = nextInt();

		out.println("");
	}

	private void solveC() {
		int n = nextInt();

		out.println("");
	}

	private void solveD() {
		int n = nextInt();

		out.println("");
	}

	private void solveE() {
		int n = nextInt();

		out.println("");
	}

	private void solveF() {
		int n = nextInt();

		out.println("");
	}

	private final PrintWriter out = new PrintWriter(System.out);
	private final InputStream in = System.in;
	private final byte[] buffer = new byte[1024];
	private int ptr = 0;
	private int buflen = 0;

	private boolean hasNextByte() {
		if (ptr < buflen) {
			return true;
		} else {
			ptr = 0;
			try {
				buflen = in.read(buffer);
			} catch (IOException e) {
				e.printStackTrace();
			}
			if (buflen <= 0) {
				return false;
			}
		}
		return true;
	}

	private int readByte() {
		if (hasNextByte())
			return buffer[ptr++];
		else
			return -1;
	}

	private static boolean isPrintableChar(int c) {
		return 33 <= c && c <= 126;
	}

	private void skipUnprintable() {
		while (hasNextByte() && !isPrintableChar(buffer[ptr]))
			ptr++;
	}

	public boolean hasNext() {
		skipUnprintable();
		return hasNextByte();
	}

	public int nextInt() {
		return Integer.parseInt(next());
	}

	public String next() {
		if (!hasNext())
			throw new NoSuchElementException();
		StringBuilder sb = new StringBuilder();
		int b = readByte();
		while (isPrintableChar(b)) {
			sb.appendCodePoint(b);
			b = readByte();
		}
		return sb.toString();
	}

	public long nextLong() {
		if (!hasNext())
			throw new NoSuchElementException();
		long n = 0;
		boolean minus = false;
		int b = readByte();
		if (b == '-') {
			minus = true;
			b = readByte();
		}
		if (b < '0' || '9' < b) {
			throw new NumberFormatException();
		}
		while (true) {
			if ('0' <= b && b <= '9') {
				n *= 10;
				n += b - '0';
			} else if (b == -1 || !isPrintableChar(b)) {
				return minus ? -n : n;
			} else {
				throw new NumberFormatException();
			}
			b = readByte();
		}
	}
}