Problem Statement :

Jack and Daniel are friends. They want to encrypt their conversations so that they can save themselves from interception by a detective agency so they invent a new cipher.

Every message is encoded to its binary representation. Then it is written down  times, shifted by  bits. Each of the columns is XORed together to get the final encoded string.

If  and  it looks like so:

1001011     shift 0 
01001011    shift 1
001001011   shift 2
0001001011  shift 3
1110101001  <- XORed/encoded string s
Now we have to decode the message. We know that . The first digit in  so our output string is going to start with . The next two digits are also , so they must have been XORed with . We know the first digit of our  shifted string is a  as well. Since the  digit of  is , we XOR that with our  and now know there is a  in the  position of the original string. Continue with that logic until the end.

Then the encoded message  and the key  are sent to Daniel.

Jack is using this encoding algorithm and asks Daniel to implement a decoding algorithm. Can you help Daniel implement this?

Function Description

Complete the cipher function in the editor below. It should return the decoded string.

cipher has the following parameter(s):

k: an integer that represents the number of times the string is shifted
s: an encoded string of binary digits
Input Format

The first line contains two integers  and , the length of the original decoded string and the number of shifts.
The second line contains the encoded string  consisting of  ones and zeros.

Output Format

Return the decoded message of length , consisting of ones and zeros.

Solution :


                            Solution in C :

In  C :

#include <stdio.h>
#include <string.h>
#include <math.h>
#include <stdlib.h>

int main() {

   int n, k, i, j, pre, cur, a[1000000];
	char input[2000000];

	scanf("%d", &n);
	scanf("%d", &k);
	scanf("%s", input);

	if (input[0] == '1')
		pre = 1;
	  	pre = 0;

	printf("%d", pre);
	a[0] = pre;
	for (i = 1; i < n; i++) {
		if (input[i] == '1') {
			printf("%d", (pre ^ 1) & 1);
			a[i] = (pre ^ 1) & 1;
		} else {
			printf("%d", (pre ^ 0) & 1);
			a[i] = (pre ^ 0) & 1;
		pre = pre ^ a[i];
		if ((i - k + 2) > 0)
			pre = pre ^ a[i - k + 1];
    return 0;

                        Solution in C++ :

In  C++  :

#include <cmath>
#include <cstdio>
#include <vector>
#include <iostream>
#include <algorithm>
using namespace std;

int main() {
    int N,K;
    char a[N+K-1];
    int arr[N],sum=0,i;
        if(a[i]=='0') arr[i]=sum%2;
        else arr[i]=(sum+1)%2;
//    arr[N-1]=a[N+K-2]-48;
    return 0;

                        Solution in Java :

In  Java :


public class Solution {

    public static void solve(Input in, PrintWriter out) throws IOException {
        int n = in.nextInt();
        int k = in.nextInt();
        char[] c =;
        int x = 0;
        char[] ans = new char[n];
        for (int i = 0; i < n; ++i) {
            ans[i] = (char) (((c[i] - '0') ^ x) + '0');
            x ^= ans[i] - '0';
            if (i >= k - 1) {
                x ^= ans[i - k + 1] - '0';

    public static void main(String[] args) throws IOException {
        PrintWriter out = new PrintWriter(System.out);
        solve(new Input(new BufferedReader(new InputStreamReader(, out);

    static class Input {
        BufferedReader in;
        StringBuilder sb = new StringBuilder();

        public Input(BufferedReader in) {
   = in;

        public Input(String s) {
   = new BufferedReader(new StringReader(s));

        public String next() throws IOException {
            while (true) {
                int c =;
                if (c == -1) {
                    return null;
                if (" \n\r\t".indexOf(c) == -1) {
            while (true) {
                int c =;
                if (c == -1 || " \n\r\t".indexOf(c) != -1) {
            return sb.toString();

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

        public long nextLong() throws IOException {
            return Long.parseLong(next());

        public double nextDouble() throws IOException {
            return Double.parseDouble(next());

                        Solution in Python : 
In  Python3 :

n, k = map(int, input().split())
s = input()
ans = []
pref = 0
for i in range(n):
    val = pref ^ (ord(s[i]) - ord('0'))
    pref ^= val
    if i >= k - 1:
        pref ^= ans[i - k + 1]
print(''.join(map(str, ans)))

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