Single Operation Part 2 CodeChef Solution

Problem – Single Operation Part 2 CodeChef Solution

Chef has the binary representation S of a number X with him. He can modify the number by applying the following operation exactly once:

• Make X:=X⊕⌊X/2^Y​⌋, where (1≤Y≤∣S∣) and ⊕ denotes the bitwise XOR operation.

Chef wants to minimize the value of X after performing the operation. Help Chef in determining the value of Y which will minimize the value of X after the operation.

Input Format

• The first line of input will contain a single integer T, denoting the number of test cases.
• Each test case consists of two lines of inputs:
  • The first line contains the length of the binary string S.
  • The second line contains the binary string S.

Output Format

For each test case, output on a new line, the value of Y which will minimize the value of X after the operation.

Constraints

• 1≤T≤5⋅10^4
• 1≤∣S∣≤10^5
• The sum of |S| over all test cases won’t exceed 5⋅10^5.
• S contains the characters 0 and 1 only.

Sample 1:

Input:
4
2
10
2
11
3
101
3
110
Output:
2
1
2
1

Explanation:

Test case 1: Since S=10 is the binary representation of 2, the current value of X=2. On choosing Y=2, X becomes 2⊕⌊2/2^2​⌋=2. We can show that this is the minimum value of X we can achieve after one operation.

Test case 2: Since S=11 is the binary representation of 3, the current value of X=3. On choosing Y=1, X becomes 3⊕⌊3/2^1​⌋=2. We can show that this is the minimum value of X we can achieve after one operation.

Test case 3: Since S=101 is the binary representation of 5, the current value of X=5. On choosing Y=2, X becomes 5⊕⌊5/2^2​⌋=4. We can show that this is the minimum value of X we can achieve after one operation.

Test case 4: Since S=110 is the binary representation of 6, the current value of X=6. On choosing Y=1, X becomes 6⊕⌊6/2^1​⌋=5. We can show that this is the minimum value of X we can achieve after one operation.

Single Operation Part 2 CodeChef Solution in C++17

#include<bits/stdc++.h>
int main(){
    int t;  std::cin >> t;
    while(t--){
        int n;  std::string s;
        std::cin >> n >> s;
        int res = n;
        for(int i = 1; i < n; ++i){
            if(s[i] == '1'){
                res = std::min(res, i);
                break;
            }
        }
        std::cout << res << '\n';
    }
}

Single Operation Part 2 CodeChef Solution in Pyth 3

msd = int(input())
for aj in range(msd):
    n = int(input())
    S = str(input())
    y = n
    for i in range(1,n):
        if S[i] == '1':
            y = i 
            break
    print(y)
        

    
    

Single Operation Part 2 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
	{
		// your code goes here
// 		Scanner sc = new Scanner(System.in);
		BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
		
		int t = Integer.parseInt(br.readLine());
		
		while(t-->0) {
		    int l = Integer.parseInt(br.readLine());
		    String s = br.readLine();
		    
		    int i=0, j=0;
		    while(i<l && s.charAt(i)!='1') i++;
		    j = i+1;
		    while(j<l && s.charAt(j)!='1') j++;
		    System.out.println(j-i);
		}
		
		
		
	}
}

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