N Buttons 1 Bulb CodeChef Solution

Problem – N Buttons 1 Bulb CodeChef Solution

Kulyash stays in room that has a single bulb and N buttons. The bulb is initially on.

The initial states of the buttons are stored in a binary string S of length N — if Si​ is 0, the i-th button is off, and if Si​ is 1, the i-th button is on. If Kulyash toggles any single button then the state of the bulb reverses i.e. the bulb lights up if it was off and vice versa.

Kulyash has toggled some buttons and the final states of the buttons are stored in another binary string R of length N. He asks you to determine the final state of the bulb.

Input Format

• The first line of input will contain a single integer T, denoting the number of test cases.
• Each test case consists of three lines of input.
  • The first line of each test case contains an integer N — the number of buttons.
  • The second line of each test case contains a binary string S — the initial states of the buttons.
  • The third line of each test case contains a binary string R — the final states of the buttons.

Output Format

For each test case, output on a new line the final state of the bulb (0 for off and 1 for on).

Constraints

• 1≤T≤100
• 1≤N≤100
• S and R are binary strings, i.e, contain only the characters 0 and 1.

Sample 1:

Input:
2
3
000
001
2
00
11
Output:
0
1

Explanation:

Test case 1: All the buttons are initially off and finally the state of only one button has changed so the state of the bulb reverses once i.e. it turns off.

Test case 2: All the buttons are initially off and finally the states of two buttons have changed so the state of the bulb reverses twice i.e. it remains on.

N Buttons 1 Bulb CodeChef Solution in C++14

#include <iostream>
using namespace std;

int main() {
	int t; cin >> t;
	while(t--){
	    int n, count = 0; cin >> n;
	    string s,r; cin >> s >> r;
	    for(int i = 0; i < n; i++){
	        if(s[i] != r[i])
	            count++;
	    }
	    if(count%2 == 0){
	        cout << 1 << endl;
	    }
	    else{
	        cout << 0 << endl;
	    }
	}
	return 0;
}

N Buttons 1 Bulb CodeChef Solution in Java

/* package codechef; // don't place package name! */

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

/* Name of the class has to be "Main" only if the class is public. */
class Codechef
{
	public static void main (String[] args) throws java.lang.Exception
	{
		Scanner sc=new Scanner(System.in);
		int t=sc.nextInt();
		while(t-->0){
		    int n=sc.nextInt();
		    int count=0;
		    String s=sc.next();
		    String r=sc.next();
		    for(int i=0;i<n;i++){
		        if(s.charAt(i)!=r.charAt(i)){
		            count++;
		        }
		    }
		    if(count%2==1){
		        System.out.println("0");
		    }else{
		        System.out.println("1");
		    }
		}
		
		
		
		
	    
	}
}


// Scanner sc = new Scanner(System.in);
// 		int t = Integer.parseInt(sc.nextLine());
// 	    while(t>0){
// 	        int count = 0;
// 	        int n = Integer.parseInt(sc.nextLine());
// 	        String s = sc.nextLine();
// 	        String r = sc.nextLine();
// 	        for(int i=0 ; i<n ; i++){
// 	            if(s.charAt(i)!=r.charAt(i)){
// 	                count++;
// 	            }
// 	        }
// 	        if(count%2 == 1) System.out.println(0);
// 	        else System.out.println(1);
	        
// 	        t--;
// 	    }

N Buttons 1 Bulb CodeChef Solution in Pyth 3

# cook your dish here
for i in range(int(input())):
    a=int(input())
    b=input()
    c=input()
    count=0
    for j in range(len(b)):
        if c[j]!=b[j]:
            count+=1
    if(count%2==0):
        print(1)
    else:
        print(0)
    
            

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