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C++ Program for Number of unique triplets whose XOR is zero

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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. 


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:  


// 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++)
    // 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])
    // 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;



Time Complexity: O(n2)

Auxiliary Space: O(n) for unordered_set

Please refer complete article on Number of unique triplets whose XOR is zero for more details!

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Last Updated : 27 Aug, 2022
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