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

Sum vs XOR

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251 Discussions

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  • mohammadkh123
     + 3 comments

    long sumXor(long n) { int c = 0; c = popcount((~n)&large(n)); return (1ULL<<c); }

    also wrote the popcount function for 128 bit using parallel bit shifting and counting (spam checker isnt allowing long code pasting)

  • srivastavapakhi2
     + 0 comments

    long sumXor(long n) { long result=1; while(n){ if((n&1)==0)result*=2; n>>=1; } return result; }

  • quote_unquote
     + 0 comments
    /*
 * Complete the 'sumXor' function below.
 *
 * The function is expected to return a LONG_INTEGER.
 * The function accepts LONG_INTEGER n as parameter.
 */

long sumXor(long n) {
    long count = 0;
    auto s = bitset<64>(n).to_string();
    auto t = s.find_first_of("1");
    for (size_t i = t+1; i < 64; ++i) {
        if (s[i] == '0') { count++; }
    }    
    // cout << count << endl;
    return 1L << count;
}

  • skbenz
     + 0 comments

    sum is essentially xor except, there's the possibility of a carry with addition: 1 ^ 1 == 0, while 1 + 1 = 10. That's why you want to count the number of zeroes in the input. Once you have that, you need all the permutations of toggling those zeroes, which is 2 ** n_zeroes, or 1 << n_zeroes.

    If we choose 100, which is b'110 0100, there are four zeroes, so there are 16 ways to toggle the zeroes -> 0, 1, 10, 11, 1000, 1001, 1010, 1011, 10000, 10001, 10010, 10011, 11000, 11001, 11010, 11011.

  • patrickgao2020
     + 1 comment

    Here is my Python solution! A really clever trick is to realize that the amount is just 2 raised to the amount of 0s in the binary representation of the number.

    def sumXor(n):
    if n == 0:
        return 1
    return 2 ** (bin(n).count("0") - 1)