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# Number of subarrays for which product and sum are equal

• Difficulty Level : Easy
• Last Updated : 29 Apr, 2021

Given a array of n numbers. We need to count the number of subarrays having the product and sum of elements are equal

Examples:

```Input  : arr[] = {1, 3, 2}
Output : 4
The subarrays are :
[0, 0] sum = 1, product = 1,
[1, 1] sum = 3, product = 3,
[2, 2] sum = 2, product = 2 and
[0, 2] sum = 1+3+2=6, product = 1*3*2 = 6

Input : arr[] = {4, 1, 2, 1}
Output : 5```

The idea is simple, we check for each subarray that if product and sum of its elements are equal or not. If it is then increase the counter variable by 1

## C++

 `// C++ program to count subarrays with``// same sum and product.``#include``using` `namespace` `std;` `// returns required number of subarrays``int` `numOfsubarrays(``int` `arr[] , ``int` `n)``{``    ``int` `count = 0; ``// Initialize result` `    ``// checking each subarray``    ``for` `(``int` `i=0; i

## Java

 `// Java program to count subarrays with``// same sum and product.` `class` `GFG``{``    ``// returns required number of subarrays``    ``static` `int` `numOfsubarrays(``int` `arr[] , ``int` `n)``    ``{``        ``int` `count = ``0``; ``// Initialize result``     ` `        ``// checking each subarray``        ``for` `(``int` `i=``0``; i

## Python3

 `# python program to``# count subarrays with``# same sum and product.` `# returns required``# number of subarrays``def` `numOfsubarrays(arr,n):` `    ``count ``=` `0` `# Initialize result`` ` `    ``# checking each subarray``    ``for` `i ``in` `range``(n):``    ` `        ``product ``=` `arr[i]``        ``sum` `=` `arr[i]``        ``for` `j ``in` `range``(i``+``1``,n):``        ` `            ``# checking if product is equal``            ``# to sum or not``            ``if` `(product``=``=``sum``):``                ``count``+``=``1`` ` `            ``product ``*``=` `arr[j]``            ``sum` `+``=` `arr[j]``        ` ` ` `        ``if` `(product``=``=``sum``):``            ``count``+``=``1``    ` `    ``return` `count` `# Driver code` `arr ``=` `[``1``,``3``,``2``]``n ``=``len``(arr)``print``(numOfsubarrays(arr , n))` `# This code is contributed``# by Anant Agarwal.`

## C#

 `// C# program to count subarrays``// with same sum and product.``using` `System;``class` `GFG {``    ` `    ``// returns required number``    ``// of subarrays``    ``static` `int` `numOfsubarrays(``int` `[]arr ,``                              ``int` `n)``    ``{``        ` `        ``// Initialize result``        ``int` `count = 0;``    ` `        ``// checking each subarray``        ``for` `(``int` `i = 0; i < n; i++)``        ``{``            ``int` `product = arr[i];``            ``int` `sum = arr[i];``            ``for` `(``int` `j = i + 1; j < n; j++)``            ``{``                ` `                ``// checking if product is``                ``// equal to sum or not``                ``if` `(product == sum)``                    ``count++;``    ` `                ``product *= arr[j];``                ``sum += arr[j];``            ``}``    ` `            ``if` `(product == sum)``                ``count++;``        ``}``        ``return` `count;``    ``}``    ` `    ``// Driver Code``    ``public` `static` `void` `Main()``    ``{``        ``int` `[]arr = {1,3,2};``        ``int` `n = arr.Length;``        ``Console.Write(numOfsubarrays(arr , n));``    ``}``}` `// This code is contributed by Nitin Mittal.`

## PHP

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## Javascript

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Output:

`4`

Time Complexity : O(n2)

This article is contributed by Ayush Jha. If you like GeeksforGeeks and would like to contribute, you can also write an article using write.geeksforgeeks.org or mail your article to contribute@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.
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