# Number of subarrays for which product and sum are equal

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++ program to count subarrays with`
`// same sum and product.`
`#include<bits/stdc++.h>`
`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<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 function`
`int` `main()`
`{`
`    ``int` `arr[] = {1,3,2};`
`    ``int` `n = ``sizeof``(arr)/``sizeof``(arr[0]);`
`    ``cout << numOfsubarrays(arr , n);`
`    ``return` `0;`
`}`

Output:

```4
```

Time Complexity : O(n2)\

Disclaimer: This does not belong to TechCodeBit, its an article taken from the below
source and credits.
source and credits: http://www.geeksforgeeks.org
We have built the accelerating growth-oriented website for budding engineers and aspiring job holders of technology companies such as Google, Facebook, and Amazon
If you would like to study our free courses you can join us at