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  1. Prepare
  2. Algorithms
  3. Bit Manipulation
  4. Sum vs XOR
  5. Discussions

Sum vs XOR

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240 Discussions

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  • alban_tyrex
     + 0 comments

    Here is my C++ solution, you can watch the explanation here : https://youtu.be/yj0yNv6BZa8

    long sumXor(long n) {
    long result = 1;
    while(n) {
        result *= (n % 2) ? 1 : 2;
        n /= 2;
    }
    return result;
}

  • claudelounou75
     + 0 comments

    C solution 

    long sumXor(long n) {
    long count = 0;

    while (n > 0) {
        if ((n & 1) == 0) {
            count++;
        }
        n >>= 1;
    }

    return 1L << count;
}

  • kumar_cse_576
     + 0 comments

    easy cpp

    long sumXor(long n) {
int no0=0;
while(n!=0){
    if(!(n&1))no0++;
    n=n>>1;
}
return pow(2,no0);
}

  • jarekk
     + 0 comments
    def sumXor(n):
    return 2**(bin(n).lstrip('0b').count('0'))

  • ihorfinance
     + 0 comments

    C++ (more at https://github.com/IhorVodko/Hackerrank_solutions , feel free to give a star :) )

    long int sumXor(long int _num) {
    size_t satisfies = 0;
    while(_num){
        satisfies += _num % 2 == 0 ? 1 : 0;
        _num /= 2;
    }
    return std::pow(2, satisfies);
}