Flipping bits


Problem Statement :


You will be given a list of 32 bit unsigned integers. Flip all the bits (1->0 and 0->1) and return the result as an unsigned integer.

Example
n=9(10)
9(10)=1001(2). We're working with 32 bits, so:
000000000000000000000000000001001 = 9(10)
111111111111111111111111111110110 = 4294967286(10)
Return 4294967286.

Function Description

Complete the flippingBits function in the editor below.

flippingBits has the following parameter(s):

int n: an integer
Returns

int: the unsigned decimal integer result
Input Format

The first line of the input contains q, the number of queries.
Each of the next q lines contain an integer, n, to process.

Constraints

1 <= q <= 100
0 <= n < 2^32



Solution :



title-img


                            Solution in C :

In C++ :





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

ll val = 4294967295L;
int main() {
    int T;
    cin >> T;
    for(int t=0;t<T;t++){
        ll cur;
        cin >> cur;
        cout << (val-cur) << endl;
    }
    return 0;
}








In Java :





import java.util.Scanner;

public class Solution {

	public static void main(String[] args) {
		Scanner scanner = new Scanner(System.in);

		int T = scanner.nextInt();

		long mask = (1L << 32) - 1;
		
		for (int t = 0; t < T; t++) {
			long n = scanner.nextLong();
			System.out.println(n ^ mask);			
		}
		
		scanner.close();

	}

}








In C :





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

int main() {

    int t;
    unsigned int n;
    scanf("%d", &t);
    while(t-- > 0) {
        scanf("%u", &n);
        printf("%u\n", ~n);
    }
    return 0;
}









In Python3 :





T = int(input())
for _ in range(T) :
    N = int(input())
    print(0xffffffff & ~N)
                        








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