# Number of Subarrays with positive product

Given an array arr[] of N integers, the task is to find the count of subarrays with positive product.

Examples:

Input: arr[] = {-1, 2, -2}
Output: 2
Subarrays with positive product are {2} and {-1, 2, -2}.

Input: arr[] = {5, -4, -3, 2, -5}
Output: 7

## Recommended: Please try your approach on {IDE} first, before moving on to the solution.

Approach: The approach to find the subarrays with negative product has been discussed in this article. If cntNeg is the count of negative product subarrays and total is the count of all possible subarrays of the given array then the count of positive product subarrays will be cntPos = total – cntNeg.

Below is the implementation of the above approach:

## C++

 `// C++ implementation of the approach ` `#include ` `using` `namespace` `std; ` ` `  `// Function to return the count of ` `// subarrays with negative product ` `int` `negProdSubArr(``int` `arr[], ``int` `n) ` `{ ` `    ``int` `positive = 1, negative = 0; ` `    ``for` `(``int` `i = 0; i < n; i++) { ` ` `  `        ``// Replace current element with 1 ` `        ``// if it is positive else replace ` `        ``// it with -1 instead ` `        ``if` `(arr[i] > 0) ` `            ``arr[i] = 1; ` `        ``else` `            ``arr[i] = -1; ` ` `  `        ``// Take product with previous element ` `        ``// to form the prefix product ` `        ``if` `(i > 0) ` `            ``arr[i] *= arr[i - 1]; ` ` `  `        ``// Count positive and negative elements ` `        ``// in the prefix product array ` `        ``if` `(arr[i] == 1) ` `            ``positive++; ` `        ``else` `            ``negative++; ` `    ``} ` ` `  `    ``// Return the required count of subarrays ` `    ``return` `(positive * negative); ` `} ` ` `  `// Function to return the count of ` `// subarrays with positive product ` `int` `posProdSubArr(``int` `arr[], ``int` `n) ` `{ ` ` `  `    ``// Total subarrays possible ` `    ``int` `total = (n * (n + 1)) / 2; ` ` `  `    ``// Count to subarrays with negative product ` `    ``int` `cntNeg = negProdSubArr(arr, n); ` ` `  `    ``// Return the count of subarrays ` `    ``// with positive product ` `    ``return` `(total - cntNeg); ` `} ` ` `  `// Driver code ` `int` `main() ` `{ ` `    ``int` `arr[] = { 5, -4, -3, 2, -5 }; ` `    ``int` `n = ``sizeof``(arr) / ``sizeof``(arr); ` ` `  `    ``cout << posProdSubArr(arr, n); ` ` `  `    ``return` `0; ` `} `

## Java

 `// Java implementation of the approach  ` `class` `GFG ` `{ ` `     `  `// Function to return the count of  ` `// subarrays with negative product  ` `static` `int` `negProdSubArr(``int` `arr[], ``int` `n)  ` `{  ` `    ``int` `positive = ``1``, negative = ``0``;  ` `    ``for` `(``int` `i = ``0``; i < n; i++) ` `    ``{  ` ` `  `        ``// Replace current element with 1  ` `        ``// if it is positive else replace  ` `        ``// it with -1 instead  ` `        ``if` `(arr[i] > ``0``)  ` `            ``arr[i] = ``1``;  ` `        ``else` `            ``arr[i] = -``1``;  ` ` `  `        ``// Take product with previous element  ` `        ``// to form the prefix product  ` `        ``if` `(i > ``0``)  ` `            ``arr[i] *= arr[i - ``1``];  ` ` `  `        ``// Count positive and negative elements  ` `        ``// in the prefix product array  ` `        ``if` `(arr[i] == ``1``)  ` `            ``positive++;  ` `        ``else` `            ``negative++;  ` `    ``}  ` ` `  `    ``// Return the required count of subarrays  ` `    ``return` `(positive * negative);  ` `}  ` ` `  `// Function to return the count of  ` `// subarrays with positive product  ` `static` `int` `posProdSubArr(``int` `arr[], ``int` `n)  ` `{  ` ` `  `    ``// Total subarrays possible  ` `    ``int` `total = (n * (n + ``1``)) / ``2``;  ` ` `  `    ``// Count to subarrays with negative product  ` `    ``int` `cntNeg = negProdSubArr(arr, n);  ` ` `  `    ``// Return the count of subarrays  ` `    ``// with positive product  ` `    ``return` `(total - cntNeg);  ` `}  ` ` `  `// Driver code  ` `public` `static` `void` `main (String[] args) ` `{  ` `    ``int` `arr[] = { ``5``, -``4``, -``3``, ``2``, -``5` `};  ` `    ``int` `n = arr.length;  ` ` `  `    ``System.out.println(posProdSubArr(arr, n));  ` `} ` `} ` ` `  `// This code is contributed by AnkitRai01 `

## Python3

 `# Python3 implementation of the approach  ` ` `  `# Function to return the count of  ` `# subarrays with negative product  ` `def` `negProdSubArr(arr, n):  ` ` `  `    ``positive ``=` `1` ` `  `    ``negative ``=` `0` ` `  `    ``for` `i ``in` `range``(n):  ` ` `  `        ``# Replace current element with 1  ` `        ``# if it is positive else replace  ` `        ``# it with -1 instead  ` `        ``if` `(arr[i] > ``0``):  ` ` `  `            ``arr[i] ``=` `1` ` `  `        ``else``:  ` ` `  `            ``arr[i] ``=` `-``1` ` `  `        ``# Take product with previous element  ` `        ``# to form the prefix product  ` `        ``if` `(i > ``0``):  ` ` `  `            ``arr[i] ``*``=` `arr[i ``-` `1``]  ` ` `  `        ``# Count positive and negative elements  ` `        ``# in the prefix product array  ` `        ``if` `(arr[i] ``=``=` `1``):  ` ` `  `            ``positive ``+``=` `1` ` `  `        ``else``:  ` ` `  `            ``negative ``+``=` `1` ` `  `    ``# Return the required count of subarrays  ` `    ``return` `(positive ``*` `negative)  ` ` `  `# Function to return the count of ` `# subarrays with positive product ` `def` `posProdSubArr(arr, n): ` ` `  `    ``total ``=` `(n ``*` `(n ``+` `1``)) ``/` `2``; ` ` `  `    ``# Count to subarrays with negative product ` `    ``cntNeg ``=` `negProdSubArr(arr, n); ` ` `  `    ``# Return the count of subarrays ` `    ``# with positive product ` `    ``return` `(total ``-` `cntNeg); ` ` `  `# Driver code  ` `arr ``=` `[``5``, ``-``4``, ``-``3``, ``2``, ``-``5``]  ` `n ``=` `len``(arr)  ` `print``(posProdSubArr(arr, n))  ` ` `  `# This code is contributed by Mehul Bhutalia `

## C#

 `// C# implementation of the approach ` `using` `System; ` ` `  `class` `GFG ` `{ ` `     `  `// Function to return the count of  ` `// subarrays with negative product  ` `static` `int` `negProdSubArr(``int` `[]arr, ``int` `n)  ` `{  ` `    ``int` `positive = 1, negative = 0;  ` `    ``for` `(``int` `i = 0; i < n; i++) ` `    ``{  ` ` `  `        ``// Replace current element with 1  ` `        ``// if it is positive else replace  ` `        ``// it with -1 instead  ` `        ``if` `(arr[i] > 0)  ` `            ``arr[i] = 1;  ` `        ``else` `            ``arr[i] = -1;  ` ` `  `        ``// Take product with previous element  ` `        ``// to form the prefix product  ` `        ``if` `(i > 0)  ` `            ``arr[i] *= arr[i - 1];  ` ` `  `        ``// Count positive and negative elements  ` `        ``// in the prefix product array  ` `        ``if` `(arr[i] == 1)  ` `            ``positive++;  ` `        ``else` `            ``negative++;  ` `    ``}  ` ` `  `    ``// Return the required count of subarrays  ` `    ``return` `(positive * negative);  ` `}  ` ` `  `// Function to return the count of  ` `// subarrays with positive product  ` `static` `int` `posProdSubArr(``int` `[]arr, ``int` `n)  ` `{  ` ` `  `    ``// Total subarrays possible  ` `    ``int` `total = (n * (n + 1)) / 2;  ` ` `  `    ``// Count to subarrays with negative product  ` `    ``int` `cntNeg = negProdSubArr(arr, n);  ` ` `  `    ``// Return the count of subarrays  ` `    ``// with positive product  ` `    ``return` `(total - cntNeg);  ` `}  ` ` `  `// Driver code  ` `public` `static` `void` `Main (String[] args) ` `{  ` `    ``int` `[]arr = { 5, -4, -3, 2, -5 };  ` `    ``int` `n = arr.Length;  ` ` `  `    ``Console.WriteLine(posProdSubArr(arr, n));  ` `} ` `} ` ` `  `// This code is contributed by 29AjayKumar `

Output:

```7
```

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