Cuente pares de elementos desiguales de la array dada

Dada una array arr[] de N elementos. La tarea es contar el número total de índices (i, j) tales que arr[i] != arr[j] e i < j .
Ejemplos: 
 

Entrada: arr[] = {1, 1, 2} 
Salida: 2 
(1, 2) y (1, 2) son los únicos pares válidos.
Entrada: arr[] = {1, 2, 3} 
Salida: 3
Entrada: arr[] = {1, 1, 1} 
Salida: 0 
 

Enfoque: Inicialice una variable de conteo cnt = 0 y ejecute dos bucles anidados para verificar cada par posible si el par actual es válido o no. Si es válido, incremente la variable de conteo. Finalmente, imprima el conteo de pares válidos.
A continuación se muestra la implementación del enfoque anterior: 
 

// C++ implementation of the approach
 
#include <bits/stdc++.h>
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 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 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# 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

<script>
 
// Javascript implementation of the approach
 
// Function to return the
// count of valid pairs
function countPairs(arr, n)
{
 
    // To store the required count
    var cnt = 0;
 
    // For each index pair (i, j)
    for (var i = 0; i < n; i++) {
        for (var 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
var arr = [ 1, 1, 2 ];
var n = arr.length;
document.write(countPairs(arr, n));
 
</script>

2

Complejidad del Tiempo: O(N ² )