Number of unique triplets whose XOR is zero

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

Recommended: Please try your approach on {IDE} first, before moving on to the solution.

Naive Approach: A naive approach is to run three nested loops, the first runs from 0 to n, second from i+1 to n, and the last one from j+1 to n to get the unique triplets. Calculate the XOR of a_i, a_j, a_k, check if it equals to 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 that XOR of two same numbers gives 0. So we need to calculate XOR of unique pairs only, and if the calculated XOR is one of the array element, 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 of this approach:

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

Below is the implementation of 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; 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) 

Output:

Time Complexity : O(n²)

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My Personal Notes

Recommended: Please try your approach on {IDE} first, before moving on to the solution.

C++

Java

Python3

C#

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