# 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
```

## Recommended: Please solve it on “PRACTICE” first, before moving on to the solution.

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

 ` `

Output:

```4
```

Time Complexity : O(n2)

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Improved By : Hariesh, nitin mittal, jit_t

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