Sum vs XOR


Problem Statement :


Given an integer , find each  such that:
0 <= x <= n
n + x = n+x
where + denotes the bitwise XOR operator. Return the number of x's satisfying the criteria.

Example
n = 4
There are four values that meet the criteria:
4 + 0 = 4 + 0 = 4
4 + 1 = 4 + 1 = 5
4 + 2 = 4 + 2 = 6
4 + 3 = 4 + 3 = 7
Return 4.

Function Description

Complete the sumXor function in the editor below.

sumXor has the following parameter(s):
- int n: an integer

Returns
- int: the number of values found

Input Format

A single integer, n.

Constraints

0 <= n <= 10^15


Solution :



title-img 


                            Solution in C :

In C++ :





#include <map>
#include <set>
#include <list>
#include <cmath>
#include <ctime>
#include <deque>
#include <queue>
#include <stack>
#include <string>
#include <bitset>
#include <cstdio>
#include <limits>
#include <vector>
#include <climits>
#include <cstring>
#include <cstdlib>
#include <fstream>
#include <numeric>
#include <sstream>
#include <iostream>
#include <algorithm>
#include <unordered_map>

using namespace std;


int main(){
    long long n;
    scanf("%lld", &n);
    long long res=1;
    while(n){
        if(n%2 == 0) res *= 2;
        n /= 2;
    }
    printf("%lld", res);
    return 0;
}








In Java :





import java.io.*;
import java.util.*;
import java.text.*;
import java.math.*;
import java.util.regex.*;

public class Solution {

    public static void main(String[] args) {
        Scanner in = new Scanner(System.in);
        long n = in.nextLong();
        int zeroCount = 0;
        while (n>0) {
            if (n%2==0)
                zeroCount++;
            n/=2;
        }
        System.out.println(1l<<zeroCount);
    }
}








In C :





#include <stdio.h>

int main(){
	long long int n,m=1;
	scanf("%lld",&n);
	while(n>0){
		if(n%2==0)m*=2;
		n/=2;
	}
	printf("%lld\n",m);
	return 0;
}









In Python3 :





#!/bin/python3

import sys


n = int(input().strip())
print(1 if n == 0 else 1 << (bin(n)[2:].count('0')))








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