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;
}

Python Solution

1. 			MaxNumBin = ''.join('1' for _ in range(32))
2. 			MaxNumDec = int(MaxNumBin,2)
3. 			return MaxNumDec - n

Solution in JS:

function flippingBits(n) {

return (~n) >>> 0;

}

A simple attempt using C++ bitset

long flippingBits(long n) { std::bitset<32> bits(n); for(int i=0;i<32;i++){ bits.flip(i);
} return bits.to_ullong(); }

Very simple solution: First perform NOT operation then Mask the number with 32-bit 1's (0xFFFF_FFFF) def FlippingBits(n): a = ~n & 0xFFFFFFFF return a