Número de pares en una array que tienen una suma igual al producto

Dada una array arr[] , la tarea es encontrar el número de pares (arr[i], arr[j]) en la array tal que arr[i] + arr[j] = arr[i] * arr[j ]
Ejemplos: 

Entrada: arr[] = {2, 2, 3, 4, 6} 
Salida: 1 
(2, 2) es el único par posible como (2 + 2) = (2 * 2) = 4.
Entrada: arr[] = {1, 2, 3, 4, 5} 
Salida: 0 

Enfoque: Los únicos pares posibles de enteros que satisfarán las condiciones dadas son (0, 0) y (2, 2) . Entonces, la tarea ahora es contar el número de 0 y 2 en la array y almacenarlos en cnt0 y cnt2 respectivamente y luego el conteo requerido será (cnt0 * (cnt0 – 1)) / 2 + (cnt2 * (cnt2 – 1) )) / 2 .
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 the required pairs
int sumEqualProduct(int a[], int n)
{
    int zero = 0, two = 0;
 
    // Find the count of 0s
    // and 2s in the array
    for (int i = 0; i < n; i++) {
        if (a[i] == 0) {
            zero++;
        }
        if (a[i] == 2) {
            two++;
        }
    }
 
    // Find the count of required pairs
    int cnt = (zero * (zero - 1)) / 2
              + (two * (two - 1)) / 2;
 
    // Return the count
    return cnt;
}
 
// Driver code
int main()
{
    int a[] = { 2, 2, 3, 4, 2, 6 };
    int n = sizeof(a) / sizeof(a[0]);
 
    cout << sumEqualProduct(a, n);
 
    return 0;
}

// Java implementation of the approach
import java.util.*;
 
class GFG {
    // Function to return the count
    // of the required pairs
    static int sumEqualProduct(int a[], int n)
    {
        int zero = 0, two = 0;
 
        // Find the count of 0s
        // and 2s in the array
        for (int i = 0; i < n; i++) {
            if (a[i] == 0) {
                zero++;
            }
            if (a[i] == 2) {
                two++;
            }
        }
 
        // Find the count of required pairs
        int cnt = (zero * (zero - 1)) / 2
                  + (two * (two - 1)) / 2;
 
        // Return the count
        return cnt;
    }
 
    // Driver code
    public static void main(String[] args)
    {
        int a[] = { 2, 2, 3, 4, 2, 6 };
        int n = a.length;
 
        System.out.print(sumEqualProduct(a, n));
    }
}
 
// This code is contributed by Rajput-Ji

# Python 3 implementation of the approach
 
# Function to return the count
# of the required pairs
def sumEqualProduct(a, n):
    zero = 0
    two = 0
     
    # Find the count of 0s
    # and 2s in the array
    for i in range(n):
        if a[i] == 0:
            zero += 1
        if a[i] == 2:
            two += 1
             
    # Find the count of required pairs
    cnt = (zero * (zero - 1)) // 2 + \
            (two * (two - 1)) // 2
     
    # Return the count
    return cnt
     
# Driver code
a = [ 2, 2, 3, 4, 2, 6 ]
n = len(a)
 
print(sumEqualProduct(a, n))
 
# This code is contributed by Ankit kumar

// C# implementation of the approach
using System;
 
class GFG
{
 
// Function to return the count
// of the required pairs
static int sumEqualProduct(int []a, int n)
{
    int zero = 0, two = 0;
 
    // Find the count of 0s
    // and 2s in the array
    for (int i = 0; i < n; i++)
    {
        if (a[i] == 0)
        {
            zero++;
        }
        if (a[i] == 2)
        {
            two++;
        }
    }
 
    // Find the count of required pairs
    int cnt = (zero * (zero - 1)) / 2 +
               (two * (two - 1)) / 2;
 
    // Return the count
    return cnt;
}
 
// Driver code
public static void Main(String[] args)
{
    int []a = { 2, 2, 3, 4, 2, 6 };
    int n = a.Length;
 
    Console.Write(sumEqualProduct(a, n));
}
}
 
// This code is contributed by 29AjayKumar

<script>
 
// Javascript implementation of the approach
 
// Function to return the count
// of the required pairs
function sumEqualProduct(a, n)
{
    var zero = 0, two = 0;
 
    // Find the count of 0s
    // and 2s in the array
    for (var i = 0; i < n; i++) {
        if (a[i] == 0) {
            zero++;
        }
        if (a[i] == 2) {
            two++;
        }
    }
 
    // Find the count of required pairs
    var cnt = (zero * (zero - 1)) / 2
              + (two * (two - 1)) / 2;
 
    // Return the count
    return cnt;
}
 
// Driver code
var a = [2, 2, 3, 4, 2, 6];
var n = a.length;
document.write( sumEqualProduct(a, n));
 
// This code is contributed by importantly.
</script>

3

Complejidad de tiempo : O(N)

Espacio Auxiliar : O(1)