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.


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
1   <=   n  <=  2^32



Sample Input 0


3
2147483647
1
0


Sample Output 0

2147483648
4294967294
4294967295


Solution :



title-img 


                            Solution in C :

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;
}
                        


                        Solution in C++ :

In    C++ :





#include <cstdio>
using namespace std;

int main() {
    int n;
    scanf("%d", &n);
    for (int i = 0; i < n; i ++) {
        unsigned int val;
        scanf("%u", &val);
        printf("%u\n", ~val);
    }
    return 0;
}
                    


                        Solution in Java :

In   Java  :




import java.io.*;
import java.util.*;
import java.text.*;
import java.math.*;
import java.util.regex.*;

public class Solution {

    public static void main(String[] args) {
        Scanner sc = new Scanner(System.in);
        int n = sc.nextInt();
        long mask = 0x00000000ffffffff;
            
        for(int i = 0; i < n; i++) {
            long l = sc.nextLong();
            System.out.println(~l&mask + 4294967296l);
        }
    }
}
                    


                        Solution in Python : 
                            
In   Python3  :





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


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