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

Sum vs XOR

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247 Discussions

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  • patrickgao2020
     + 0 comments

    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)

  • 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;
}

  • ze_hkrk
     + 0 comments

    C solution

    long sumXor (long n) {
    long result = 0;
    while(n) {
        if(!(n & 0x1)) {
            result++;
        }
        n = n>>0x1;
    }
    return 0x1ULL << (result);
}

  • pakbungdesu
     + 0 comments

    More at https://github.com/pakbungdesu/hackerrank-problem-solving.git

    Java

    public static long sumXor(long n) {
    if (n == 0) {
        return 1;
    } else {
        String bin = Long.toBinaryString(n);
        int zero_bits = bin.length() - bin.replace("0", "").length();
        return (long)Math.pow(2, zero_bits);
    }
}

    Pyhon

    def sumXor(n):
    if n == 0:
        return 1
    else:
        zero_bits = bin(n)[2:].count('0')  # Ignore the '0b' prefix
        return 2**zero_bits

  • jmcparland
     + 0 comments

    Haskell

    module Main where

import Data.Bits

countZeroBits :: Integer -> Int
countZeroBits n = go n 0
  where
    go :: Integer -> Int -> Int
    go 0 acc = acc
    go n acc = go (n `shiftR` 1) (acc + 1 - fromIntegral (n .&. 1))

solve :: Integer -> Integer
solve n = 1 `shiftL` countZeroBits n

main :: IO ()
main = readLn >>= print . solve