Skip to content
Related Articles

Related Articles

Improve Article

Number of unique triplets whose XOR is zero

  • Difficulty Level : Medium
  • Last Updated : 21 May, 2021

Given N numbers with no duplicates, count the number of unique triplets (ai, aj, ak) such that their XOR is 0. A triplet is said to be unique if all of the three numbers in the triplet are unique. 

Examples: 

Input : a[] = {1, 3, 5, 10, 14, 15};
Output : 2 
Explanation : {1, 14, 15} and {5, 10, 15} are the 
              unique triplets whose XOR is 0. 
              {1, 14, 15} and all other combinations of 
              1, 14, 15 are considered as 1 only.

Input : a[] = {4, 7, 5, 8, 3, 9};
Output : 1
Explanation : {4, 7, 3} is the only triplet whose XOR is 0 

Naive Approach: A naive approach is to run three nested loops, the first runs from 0 to n, the second from i+1 to n, and the last one from j+1 to n to get the unique triplets. Calculate the XOR of ai, aj, ak, check if it equals 0. If so, then increase the count. 
Time Complexity: O(n3)

Efficient Approach: An efficient approach is to use one of the properties of XOR: the XOR of two of the same numbers gives 0. So we need to calculate the XOR of unique pairs only, and if the calculated XOR is one of the array elements, then we get the triplet whose XOR is 0. Given below are the steps for counting the number of unique triplets:
Below is the complete algorithm for this approach:  

  1. With the map, mark all the array elements.
  2. Run two nested loops, one from i-n-1, and the other from i+1-n to get all the pairs.
  3. Obtain the XOR of the pair.
  4. Check if the XOR is an array element and not one of ai or aj.
  5. Increase the count if the condition holds.
  6. Return count/3 as we only want unique triplets. Since i-n and j+1-n give us unique pairs but not triplets, we do a count/3 to remove the other two possible combinations.

Below is the implementation of the above idea:  



C++




// CPP program to count the number of
// unique triplets whose XOR is 0
#include <bits/stdc++.h>
using namespace std;
 
// function to count the number of
// unique triplets whose xor is 0
int countTriplets(int a[], int n)
{
    // To store values that are present
    unordered_set<int> s;
    for (int i = 0; i < n; i++)
        s.insert(a[i]);
     
    // stores the count of unique triplets
    int count = 0;
     
    // traverse for all i, j pairs such that j>i
    for (int i = 0; i < n-1; i++) {
        for (int j = i + 1; j < n; j++) {
 
          // xor of a[i] and a[j]
          int xr = a[i] ^ a[j];
     
          // if xr of two numbers is present,
          // then increase the count
          if (s.find(xr) != s.end() && xr != a[i] &&
                                       xr != a[j])
            count++;
        }
    }
     
    // returns answer
    return count / 3;
}
 
// Driver code to test above function
int main()
{
    int a[] = {1, 3, 5, 10, 14, 15};
    int n = sizeof(a) / sizeof(a[0]);  
    cout << countTriplets(a, n);   
    return 0;
}

Java




// Java program to count
// the number of unique
// triplets whose XOR is 0
import java.io.*;
import java.util.*;
 
class GFG
{
    // function to count the
    // number of unique triplets
    // whose xor is 0
    static int countTriplets(int []a,
                             int n)
    {
        // To store values
        // that are present
        ArrayList<Integer> s =
                  new ArrayList<Integer>();
        for (int i = 0; i < n; i++)
            s.add(a[i]);
         
        // stores the count
        // of unique triplets
        int count = 0;
         
        // traverse for all i,
        // j pairs such that j>i
        for (int i = 0; i < n; i++)
        {
            for (int j = i + 1;
                     j < n; j++)
            {
     
            // xor of a[i] and a[j]
            int xr = a[i] ^ a[j];
         
            // if xr of two numbers
            // is present, then
            // increase the count
            if (s.contains(xr) &&
                xr != a[i] && xr != a[j])
                count++;
            }
        }
         
        // returns answer
        return count / 3;
    }
     
    // Driver code
    public static void main(String srgs[])
    {
        int []a = {1, 3, 5,
                   10, 14, 15};
        int n = a.length;
        System.out.print(countTriplets(a, n));
    }
}
 
// This code is contributed by
// Manish Shaw(manishshaw1)

Python3




# Python 3 program to count the number of
# unique triplets whose XOR is 0
 
# function to count the number of
# unique triplets whose xor is 0
def countTriplets(a, n):
     
    # To store values that are present
    s = set()
    for i in range(n):
        s.add(a[i])
     
    # stores the count of unique triplets
    count = 0
     
    # traverse for all i, j pairs such that j>i
    for i in range(n):
        for j in range(i + 1, n, 1):
             
            # xor of a[i] and a[j]
            xr = a[i] ^ a[j]
             
            # if xr of two numbers is present,
            # then increase the count
            if (xr in s and xr != a[i] and
                            xr != a[j]):
                count += 1;
         
    # returns answer
    return int(count / 3)
 
# Driver code
if __name__ == '__main__':
    a = [1, 3, 5, 10, 14, 15]
    n = len(a)
    print(countTriplets(a, n))
     
# This code is contributed by
# Surendra_Gangwar

C#




// C# program to count
// the number of unique
// triplets whose XOR is 0
using System;
using System.Collections.Generic;
 
class GFG
{
    // function to count the
    // number of unique triplets
    // whose xor is 0
    static int countTriplets(int []a,
                             int n)
    {
        // To store values
        // that are present
        List<int> s = new List<int>();
        for (int i = 0; i < n; i++)
            s.Add(a[i]);
         
        // stores the count
        // of unique triplets
        int count = 0;
         
        // traverse for all i,
        // j pairs such that j>i
        for (int i = 0; i < n; i++)
        {
            for (int j = i + 1;
                     j < n; j++)
            {
     
            // xor of a[i] and a[j]
            int xr = a[i] ^ a[j];
         
            // if xr of two numbers
            // is present, then
            // increase the count
            if (s.Exists(item => item == xr) &&
                   xr != a[i] && xr != a[j])
                count++;
            }
        }
         
        // returns answer
        return count / 3;
    }
     
    // Driver code
    static void Main()
    {
        int []a = new int[]{1, 3, 5,
                            10, 14, 15};
        int n = a.Length;
        Console.Write(countTriplets(a, n));
    }
}
 
// This code is contributed by
// Manish Shaw(manishshaw1)

Javascript




<script>
    // javascript program to count
    // the number of unique
    // triplets whose XOR is 0
 
    // function to count the
    // number of unique triplets
    // whose xor is 0
    function countTriplets(a , n) {
        // To store values
        // that are present
        var s = [];
        for (i = 0; i < n; i++)
            s.push(a[i]);
 
        // stores the count
        // of unique triplets
        var count = 0;
 
        // traverse for all i,
        // j pairs such that j>i
        for (i = 0; i < n; i++) {
            for (j = i + 1; j < n; j++) {
 
                // xor of a[i] and a[j]
                var xr = a[i] ^ a[j];
 
                // if xr of two numbers
                // is present, then
                // increase the count
                if (s.includes(xr) && xr != a[i] && xr != a[j])
                    count++;
            }
        }
 
        // returns answer
        return count / 3;
    }
 
    // Driver code
    var a = [ 1, 3, 5, 10, 14, 15 ];
    var n = a.length;
    document.write(countTriplets(a, n));
 
// This code contributed by Rajput-Ji
</script>

Output: 

2

Time Complexity: O(n2)
 

Attention reader! Don’t stop learning now. Get hold of all the important DSA concepts with the DSA Self Paced Course at a student-friendly price and become industry ready.  To complete your preparation from learning a language to DS Algo and many more,  please refer Complete Interview Preparation Course.

In case you wish to attend live classes with experts, please refer DSA Live Classes for Working Professionals and Competitive Programming Live for Students.




My Personal Notes arrow_drop_up
Recommended Articles
Page :