Equal Sum and XOR of three Numbers - GeeksforGeeks

17 Aug, 2021
Given an integer N. The task is to count the numbers of pairs of integers A and B such that A + B + N = A ^ B ^ N and A and B are less than N.Examples:  
Input: N = 2 Output: 3 Explanation:- For N = 2 2 XOR 0 XOR 0 = 2+0+0 2 XOR 0 XOR 1 = 2+0+1 2 XOR 0 XOR 2 != 2+0+2 2 XOR 1 XOR 0 = 2+1+0 2 XOR 1 XOR 1 != 2+1+1 2 XOR 1 XOR 2 != 2+1+2 2 XOR 2 XOR 0 != 2+2+0 2 XOR 2 XOR 1 != 2+2+1 2 XOR 2 XOR 2 != 2+2+2So (0, 0), (0, 1) and (1, 0) are the required pairs. So the output is 3.Input: N = 4 Output: 9  
 
Approach : To make the sum of three numbers equal to the xor of three number with one of the number given we can do following:-  
Represent the fixed number in binary form.Traverse the binary expansion of the fixed number. If you find a 1 there is only one condition i.e. you take the other two number’s binary bits as 0 and 0.If you find a 0 there will be three conditions i.e. either you can have binary bits as (0, 0), (1, 0) or (0, 1).The following triplets of bits will never go for a carry so they are valid.So the answer will be 3^(number of zeros in binary representation).
Represent the fixed number in binary form.
Traverse the binary expansion of the fixed number. If you find a 1 there is only one condition i.e. you take the other two number’s binary bits as 0 and 0.If you find a 0 there will be three conditions i.e. either you can have binary bits as (0, 0), (1, 0) or (0, 1).
If you find a 1 there is only one condition i.e. you take the other two number’s binary bits as 0 and 0.
If you find a 0 there will be three conditions i.e. either you can have binary bits as (0, 0), (1, 0) or (0, 1).
The following triplets of bits will never go for a carry so they are valid.
So the answer will be 3^(number of zeros in binary representation).
Below is the implementation of the above approach:  
C++
Java
Python3
C#
Javascript
// C++ implementation of the above approach#include <bits/stdc++.h>using namespace std; // Defining ull to unsigned long long inttypedef unsigned long long int ull; // Function to calculate power of 3ull calculate(int bit_cnt){    ull res = 1;    while (bit_cnt--) {        res = res * 3;    }     return res;} // Function to return the count of the// unset bit ( zeros )int unset_bit_count(ull n){    int count = 0;    while (n) {         // Check the bit is 0 or not        if ((n & 1) == 0)            count++;        // Right shifting ( dividing by 2 )        n = n >> 1;    }    return count;} // Driver Codeint main(){    ull n;    n = 2;     int count = unset_bit_count(n);     ull ans = calculate(count);     cout << ans << endl;     return 0;}
// Java implementation of the approachimport java.util.*; class GFG{ // Function to calculate power of 3static long calculate(int bit_cnt){    long res = 1;    while (bit_cnt-- > 0)    {        res = res * 3;    }     return res;} // Function to return the count of the// unset bit ( zeros )static int unset_bit_count(long n){    int count = 0;    while (n > 0)    {         // Check the bit is 0 or not        if ((n & 1) == 0)            count++;                     // Right shifting ( dividing by 2 )        n = n >> 1;    }    return count;} // Driver Codepublic static void main(String[] args){    long n;    n = 2;     int count = unset_bit_count(n);     long ans = calculate(count);     System.out.println(ans);}} // This code is contributed by PrinciRaj1992
# Python3 implementation of the approach # Function to calculate power of 3def calculate(bit_cnt):     res = 1;    while (bit_cnt > 0):        bit_cnt -= 1;        res = res * 3;    return res; # Function to return the count of the# unset bit ( zeros )def unset_bit_count(n):     count = 0;    while (n > 0):                 # Check the bit is 0 or not        if ((n & 1) == 0):            count += 1;                     # Right shifting ( dividing by 2 )        n = n >> 1;         return count; # Driver Codeif __name__ == '__main__':     n = 2;     count = unset_bit_count(n);     ans = calculate(count);     print(ans); # This code contributed by Rajput-Ji
// C# implementation of the approachusing System; class GFG{ // Function to calculate power of 3static long calculate(int bit_cnt){    long res = 1;    while (bit_cnt-- > 0)    {        res = res * 3;    }     return res;} // Function to return the count of the// unset bit (zeros)static int unset_bit_count(long n){    int count = 0;    while (n > 0)    {         // Check the bit is 0 or not        if ((n & 1) == 0)            count++;                     // Right shifting ( dividing by 2 )        n = n >> 1;    }    return count;} // Driver Codepublic static void Main(String[] args){    long n;    n = 2;     int count = unset_bit_count(n);     long ans = calculate(count);     Console.WriteLine(ans);}} // This code is contributed by 29AjayKumar
<script> // Javascript implementation of// the above approach // Function to calculate power of 3function calculate(bit_cnt){    let res = 1;    while (bit_cnt--) {        res = res * 3;    }     return res;} // Function to return the count of the// unset bit ( zeros )function unset_bit_count(n){    let count = 0;    while (n) {         // Check the bit is 0 or not        if ((n & 1) == 0)            count++;        // Right shifting ( dividing by 2 )        n = n >> 1;    }    return count;} // Driver Code    let n;    n = 2;     let count = unset_bit_count(n);     let ans = calculate(count);     document.write(ans); </script>
3
 
Time Complexity: O(Number of unset_bits). Auxiliary Space: O(1).  
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