Equal Sum and XOR of three Numbers

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+2



So (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:-

  1. Represent the fixed number in binary form.
  2. 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).
  3. The following above triplets of bits will never go for a carry so they are valid.
  4. So the answer will be 3^(number of zeros in binary representation).

Below is the implementation of the above approach:

C++

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// C++ implementation of the above approach
#include <bits/stdc++.h>
using namespace std;
  
// Defining ull to unsigned long long int
typedef unsigned long long int ull;
  
// Function to calculate power of 3
ull 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 Code
int main()
{
    ull n;
    n = 2;
  
    int count = unset_bit_count(n);
  
    ull ans = calculate(count);
  
    cout << ans << endl;
  
    return 0;
}

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Java

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// Java implementation of the approach
import java.util.*;
  
class GFG
{
  
// Function to calculate power of 3
static 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 Code
public 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 

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Python3

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# Python3 implementation of the approach
  
# Function to calculate power of 3
def 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 Code
if __name__ == '__main__':
  
    n = 2;
  
    count = unset_bit_count(n);
  
    ans = calculate(count);
  
    print(ans);
  
# This code contributed by Rajput-Ji

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C#

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// C# implementation of the approach
using System;
  
class GFG
{
  
// Function to calculate power of 3
static 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 Code
public 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

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Output:

3

Time Complexity: O(Number of unset_bits)



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