# Count unequal element pairs from the given Array

Given an array arr[] of N elements. The task is to count the total number of indices (i, j) such that arr[i] != arr[j] and i < j.

Examples:

Input: arr[] = {1, 1, 2}
Output: 2
(1, 2) and (1, 2) are the only valid pairs.

Input: arr[] = {1, 2, 3}
Output: 3

Input: arr[] = {1, 1, 1}
Output: 0

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

Approach: Initialise a count variable cnt = 0 and run two nested loops to check every possible pair whether the the current pair is valid or not. If it is valid, then increment the count variable. Finally, print the count of valid pairs.

Below is the implementation of the above approach:

## C++

 // C++ implementation of the approach    #include using namespace std;    // Function to return the // count of valid pairs int countPairs(int arr[], int n) {        // To store the required count     int cnt = 0;        // For each index pair (i, j)     for (int i = 0; i < n; i++) {         for (int j = i + 1; j < n; j++) {                // If current pair is valid             // then increment the count             if (arr[i] != arr[j])                 cnt++;         }     }     return cnt; }    // Driven code int main() {     int arr[] = { 1, 1, 2 };     int n = sizeof(arr) / sizeof(int);        cout << countPairs(arr, n);        return 0; }

## Java

 // Java implementation of the approach  class GFG  {            // Function to return the      // count of valid pairs      static int countPairs(int arr[], int n)      {                 // To store the required count          int cnt = 0;                 // For each index pair (i, j)          for (int i = 0; i < n; i++)          {              for (int j = i + 1; j < n; j++)              {                         // If current pair is valid                  // then increment the count                  if (arr[i] != arr[j])                      cnt++;              }          }          return cnt;      }             // Driven code      public static void main (String[] args)     {          int arr[] = { 1, 1, 2 };          int n = arr.length;                 System.out.println(countPairs(arr, n));      }  }    // This code is contributed by AnkitRai01

## Python3

 # Python3 implementation of the approach     # Function to return the # count of valid pairs def countPairs(arr, n):        # To store the required count     cnt = 0;        # For each index pair (i, j)     for i in range(n):         for j in range(i + 1, n):                # If current pair is valid             # then increment the count             if (arr[i] != arr[j]):                 cnt += 1;        return cnt;    # Driver code if __name__ == '__main__':     arr = [ 1, 1, 2 ];     n = len(arr);        print(countPairs(arr, n));        # This code is contributed by 29AjayKumar

## C#

 // C# implementation of the approach  using System;    class GFG  {            // Function to return the      // count of valid pairs      static int countPairs(int []arr, int n)      {                 // To store the required count          int cnt = 0;                 // For each index pair (i, j)          for (int i = 0; i < n; i++)          {              for (int j = i + 1; j < n; j++)              {                         // If current pair is valid                  // then increment the count                  if (arr[i] != arr[j])                      cnt++;              }          }          return cnt;      }             // Driven code      public static void Main()     {          int []arr = { 1, 1, 2 };          int n = arr.Length;                 Console.WriteLine(countPairs(arr, n));      }  }    // This code is contributed by AnkitRai01

Output:

2

Time Complexity: O(N2)

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Improved By : AnkitRai01, 29AjayKumar