java(8)

    int[] binary = new int[32];
    int count = 0;
    int idx = 31;

    while (count < 32) {

        binary[idx--] = (int) (n % 2);
        n = n / 2;
        count++;

    }

    for (int i = 0; i < binary.length; i++) {
        if (binary[i] == 0) {
            binary[i] = 1;
        } else {
            binary[i] = 0;
        }
    }
    long flifResult = 0;
    int p = 0;
    for (int i = 31; i >= 0; i--) {
        flifResult += binary[i] * Math.pow(2, p++);
    }

    return flifResult;

Here is my c++ solution, you can watch the video explanation here : https://youtu.be/eZ0lTIzOjhQ

V1

long flippingBits(long n) {
    long result = 0;
    for(int i = 31; i >= 0; i--) {
        if(n >= pow(2, i)){
            n -= pow(2,i);
        }
        else result += pow(2, i);
    }
    return result;
}

V2

long flippingBits(long n) {
    return 4294967295 ^ n;
}

No need for complex code. Flipping bits meaning is to just get the biggest number in that requested bit and then substract those with the number asked in the question. For ex. : the question wanted a 32 bit, so the max number would be 4294967295. Now to get what the flipping bits of lets say 10 in 32 bit and return it to decimal, just do 4294967295 - 10 which is 4294967285, and there you go the decimal version of its flipping bit.

Here is in Go

func flippingBits(n int64) int64 {
    // Write your code here
    return 4294967295 - n
}

**SIMPLE SOLUTION subtract input from 4294967295 and print result ****NO NEED TO CONVERT INTO BINARY**

Kotlin Solution. Note that the type was changed from Long to UInt

fun flippingBits(n: UInt): UInt {
    return n.inv()
}