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

Sum vs XOR

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245 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;
}

  • 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

  • chiranjeevwork
     + 0 comments

    For every un-set bit , there are 2 posiiblities of a number that when added to the original number would give the same answer , if xored

    So , count the number of unset bits , and then raise it to the power of 2 , to get all combinations of numbers that when added to n will give the same result as to when xored by n

    def sumXor(n): # Write your code here ct = 0 while n: if n&1==0: ct+=1 n>>=1 return 2**ct

    `

  • 2k23_cscys231212
     + 0 comments
    long sumXor(long n) {
    long count = 0;
    long binary = n;
    while (binary > 0) {
        count += binary & 1 ? 0 : 1;
        binary >>= 1;
    }
    return 1L << count;
}

int main() {
    long n;
    scanf("%ld", &n);

    printf("%ld\n", sumXor(n));
		
		
#Pratham Gupta
		
		
		
		

    return 0;
}