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  1. Prepare
  2. Algorithms
  3. Bit Manipulation
  4. Flipping bits
  5. Discussions

Flipping bits

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779 Discussions

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  • kryoton_98
     + 0 comments

    def flippingBits(n): num = ~n return 4294967296 + num

  • alban_tyrex
     + 0 comments

    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;
}

  • patrickgao2020
     + 0 comments

    Here is my Python solution!

    def binary(num):
    remainders = []
    while num != 0 :
        remainders.insert(0, str(num % 2))
        num //= 2
    return "".join(remainders)

def decimal(num):
    newnum = 0
    for i in range(len(num)):
        newnum += int(num[-i - 1]) * 2 ** i
    return newnum

def flippingBits(n):
    num = binary(n)
    num = "0" * (32 - len(num)) + num
    for i in range(len(num)):
        if num[i] == "0":
            num = num[:i] + "1" + num[i + 1:]
        else:
            num = num[:i] + "0" + num[i + 1:]
    return decimal(num)

  • georgiougeorgex1
     + 0 comments

    long flippingBits(uint32_t n) { return ~n; }

  • somaditya
     + 0 comments

    2 line Java hackish solution:

    public static long flippingBits(long n) {
        long maxLong = 4294967295L;
        
        return -1 * (~n ^ maxLong) - 1;
    }