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Number of pairs in an array having sum equal to product
  • Difficulty Level : Easy
  • Last Updated : 04 Mar, 2021

Given an array arr[], the task is to find the number of pairs (arr[i], arr[j]) in the array such that arr[i] + arr[j] = arr[i] * arr[j]
Examples: 

Input: arr[] = {2, 2, 3, 4, 6} 
Output:
(2, 2) is the only possible pair as (2 + 2) = (2 * 2) = 4.
Input: arr[] = {1, 2, 3, 4, 5} 
Output:

Approach: The only possible pairs of integers that will satisfy the given conditions are (0, 0) and (2, 2). So the task now is to count the number of 0s and 2s in the array and store them in cnt0 and cnt2 respectively and then the required count will be (cnt0 * (cnt0 – 1)) / 2 + (cnt2 * (cnt2 – 1)) / 2.
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 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




// 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

Python3




# 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#




// 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
Output: 
3

 

Time Complexity : O(N)

Auxiliary Space : O(1)

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