Count of triplets that satisfy the given equation

0 0 4 minutes read

Given an array arr[] of N non-negative integers. The task is to count the number of triplets (i, j, k) where 0 ? i < j ? k < N such that A[i] ^ A[i + 1] ^ … ^ A[j – 1] = A[j] ^ A[j + 1] ^ … ^ A[k] where ^ is the bitwise XOR.
Examples:

Input: arr[] = {2, 5, 6, 4, 2}
Output: 2
The valid triplets are (2, 3, 4) and (2, 4, 4).
Input: arr[] = {5, 2, 7}
Output: 2

Naive approach: Consider each and every triplet and check whether the xor of the required elements is equal or not.
Efficient approach: If arr[i] ^ arr[i + 1] ^ … ^ arr[j – 1] = arr[j] ^ arr[j + 1] ^ … ^ arr[k] then arr[i] ^ arr[i + 1] ^ … ^ arr[k] = 0 since X ^ X = 0. Now the problem gets reduced to finding the sub-arrays with XOR 0. But every such sub-array can have multiple such triplets i.e.

If arr[i] ^ arr[i + 1] ^ … ^ arr[k] = 0
then, (arr[i]) ^ (arr[i + 1] ^ … ^ arr[k]) = 0
and, arr[i] ^ (arr[i + 1]) ^ … ^ arr[k] = 0
arr[i] ^ arr[i + 1] ^ (arr[i + 2]) ^ … ^ arr[k] = 0
…
j can have any value from i + 1 to k without violating the required property.

Below is the implementation of the above approach:

C++

// C++ implementation of the approach
#include <bits/stdc++.h>
using namespace std;
 
// Function to return the count
// of required triplets
int CountTriplets(int* arr, int n)
{
    int ans = 0;
    for (int i = 0; i < n - 1; i++) {
 
        // First element of the
        // current sub-array
        int first = arr[i];
        for (int j = i + 1; j < n; j++) {
 
            // XOR every element of
            // the current sub-array
            first ^= arr[j];
 
            // If the XOR becomes 0 then
            // update the count of triplets
            if (first == 0)
                ans += (j - i);
        }
    }
    return ans;
}
 
// Driver code
int main()
{
    int arr[] = { 2, 5, 6, 4, 2 };
    int n = sizeof(arr) / sizeof(arr[0]);
 
    cout << CountTriplets(arr, n);
 
    return 0;
}

Java

// Java implementation of the approach
class GFG
{
 
// Function to return the count
// of required triplets
static int CountTriplets(int[] arr, int n)
{
    int ans = 0;
    for (int i = 0; i < n - 1; i++)
    {
 
        // First element of the
        // current sub-array
        int first = arr[i];
        for (int j = i + 1; j < n; j++) 
        {
 
            // XOR every element of
            // the current sub-array
            first ^= arr[j];
 
            // If the XOR becomes 0 then
            // update the count of triplets
            if (first == 0)
                ans += (j - i);
        }
    }
    return ans;
}
 
// Driver code
public static void main(String[] args)
{
    int arr[] = {2, 5, 6, 4, 2};
    int n = arr.length;
 
    System.out.println(CountTriplets(arr, n));
}
} 
 
// This code is contributed by Princi Singh

Python3

# Python3 implementation of the approach
 
# Function to return the count
# of required triplets
def CountTriplets(arr, n):
 
    ans = 0
    for i in range(n - 1):
 
        # First element of the
        # current sub-array
        first = arr[i]
        for j in range(i + 1, n):
 
            # XOR every element of
            # the current sub-array
            first ^= arr[j]
 
            # If the XOR becomes 0 then
            # update the count of triplets
            if (first == 0):
                ans += (j - i)
 
    return ans
 
# Driver code
arr = [2, 5, 6, 4, 2 ]
n = len(arr)
print(CountTriplets(arr, n))
 
# This code is contributed by Mohit Kumar

C#

// C# implementation of the approach
using System;
 
class GFG
{
 
    // Function to return the count
    // of required triplets
    static int CountTriplets(int[] arr, int n)
    {
        int ans = 0;
        for (int i = 0; i < n - 1; i++)
        {
     
            // First element of the
            // current sub-array
            int first = arr[i];
            for (int j = i + 1; j < n; j++) 
            {
     
                // XOR every element of
                // the current sub-array
                first ^= arr[j];
     
                // If the XOR becomes 0 then
                // update the count of triplets
                if (first == 0)
                    ans += (j - i);
            }
        }
        return ans;
    }
     
    // Driver code
    public static void Main()
    {
        int []arr = {2, 5, 6, 4, 2};
        int n = arr.Length;
     
        Console.WriteLine(CountTriplets(arr, n));
    }
}
 
// This code is contributed by AnkitRai01

Javascript

// Function to return the count
// of required triplets
function CountTriplets(arr, n) {
  let ans = 0;
  for (let i = 0; i < n - 1; i++) {
    // First element of the current sub-array
    let first = arr[i];
    for (let j = i + 1; j < n; j++) {
      // XOR every element of the current sub-array
      first ^= arr[j];
      // If the XOR becomes 0 then update the count of triplets
      if (first == 0) ans += j - i;
    }
  }
  return ans;
}
 
// Driver code
const arr = [2, 5, 6, 4, 2];
const n = arr.length;
 
console.log(CountTriplets(arr, n));

Output:

Time Complexity: O(n²).
Auxiliary Space: O(1).

Feeling lost in the world of random DSA topics, wasting time without progress? It’s time for a change! Join our DSA course, where we’ll guide you on an exciting journey to master DSA efficiently and on schedule.
Ready to dive in? Explore our Free Demo Content and join our DSA course, trusted by over 100,000 zambiatek!