# Number of pairs in an array having sum equal to product

• Difficulty Level : Easy
• Last Updated : 27 May, 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 ``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`

## Javascript

 ``
Output:
`3`

Time Complexity : O(N)

Auxiliary Space : O(1)

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