Flipping bits

You will be given a list of 32 bit unsigned integers. Flip all the bits ( and ) and print the result as an unsigned integer.

For example, your decimal input . We're working with 32 bits, so:

Function Description

Complete the flippingBits function in the editor below. It should return the unsigned decimal integer result.

flippingBits has the following parameter(s):

• n: an integer

Input Format

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

Constraints

Output Format

Output one line per element from the list with the decimal value of the resulting unsigned integer.

Sample Input 0

3
2147483647
1
0

Sample Output 0

2147483648
4294967294
4294967295

Explanation 0

Sample Input 1

2
4
123456

Sample Output 1

4294967291
4294843839

Explanation 1

Sample Input 2

3
0
802743475
35601423

Sample Output 2

4294967295
3492223820
4259365872

Explanation 2

Solution:-

import java.io.*;

import java.math.*;

import java.security.*;

import java.text.*;

import java.util.*;

import java.util.concurrent.*;

import java.util.regex.*;

public class Solution {

    // Complete the flippingBits function below.

    static long flippingBits(long n) {

        long num = 4294967295L;

        return num-n;

        }

    private static final Scanner scanner = new Scanner(System.in);

    public static void main(String[] args) throws IOException {

        BufferedWriter bufferedWriter = new BufferedWriter(new FileWriter(System.getenv("OUTPUT_PATH")));

        int q = scanner.nextInt();

        scanner.skip("(\r\n|[\n\r\u2028\u2029\u0085])?");

        for (int qItr = 0; qItr < q; qItr++) {

            long n = scanner.nextLong();

            scanner.skip("(\r\n|[\n\r\u2028\u2029\u0085])?");

            long result = flippingBits(n);

            bufferedWriter.write(String.valueOf(result));

            bufferedWriter.newLine();

        }

        bufferedWriter.close();

        scanner.close();

    }

}