Header Ad

HackerRank Cipher problem solution

In this HackerRank Cipher problem, solution 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 K times, shifted by 0,1,..., K - 1 bits. Each of the columns is XORed together to get the final encoded string.

HackerRank Cipher problem solution


Problem solution in Python.

def function(s,k,n):
	result=[]
	sum=0
	appendence=0

	for i in range(n):
		appendence = (sum+int(s[i]))&1
		result.append(appendence)
		prev=i-k+1
		if prev <0:
		    remain = 0
		else:
		    remain = result[prev]
		sum=sum + appendence - remain
	return ''.join(map(str,result))

n, k = map(int, input().split(' '))
s = input()
print(function(s,k,n))


Problem solution in Java.

import java.io.*;
import java.util.*;

public class Solution {

    public static void main(String[] args) {
        Scanner scanner = new Scanner(System.in);
        String[] line = scanner.nextLine().split(" ");
        int n = Integer.valueOf(line[0]);
        int k = Integer.valueOf(line[1]);
        String s = scanner.nextLine();
        
        StringBuilder builder = new StringBuilder(n);
        int previous = 0;
        for(int position = 0; position<n; position++) {
            if(position>=k) {
                previous ^= builder.charAt(position-k)-48;
            }
            
            int sgap = s.charAt(s.length()-1-position)-48;
            if((previous^sgap)==1) {
                builder.append('1');
            } else {
                builder.append('0');
            }
            previous = sgap;
        }
        
        System.out.println(builder.reverse().toString());
    }
}


Problem solution in C++.

#include <iostream>
#include <vector>
#include <queue>
#include <algorithm>
#include <map>
#include <string>
#include <climits>

using namespace std;

int main(){
	int N,K;
	cin >> N >> K;
	string s;
	cin >> s;
	
	vector<int> v(s.size());
	for(int i=0; i<s.size(); i++){
		v[i] = s[i] - '0';
	}

	vector<int> x = v;
	for(int i=1; i<s.size(); i++){
		x[i] = v[i] ^ v[i-1];
		if(i-K >= 0) x[i] ^= x[i-K];
	}
	for(int i=0; i<N; i++){
		cout << x[i];
	}
	cout << endl;
	return 0;
}


Problem solution in C.

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

int main() {

    int k, n, x, i;
    char *t;
    int *r;
    scanf("%d %d", &n, &k);
    t = (char*) malloc(sizeof(int)*(n + k - 1));
    r = (int*) malloc(sizeof(int)*n);
    scanf("%c", t);
    for (i = 0; i < n + k - 1; i++) scanf("%c", t + i);
    r[0] = t[0] - 48;
    for (i = 1; i < k; i++) {
        r[i] = (t[i - 1] - 48) ^ (t[i] - 48);
    }
    for (i; i < n - k; i++)
        r[i] = r[i - k] ^ (t[i - 1] - 48) ^ (t[i] - 48);
    r[n - 1] = (t[n + k - 2] - 48);
    for (i = 1; i < k; i++) {
        r[n - 1 - i] = (t[n + k - i - 1] - 48) ^ (t[n + k - 2 - i] - 48);
    }
    for (i = 0; i < n; i++) printf("%d", r[i]);
    free(t);
    free(r);
    return 0;
}


Post a Comment

0 Comments