# 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[0]);        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

My Personal Notes arrow_drop_up

Check out this Author's contributed articles.

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

Please Improve this article if you find anything incorrect by clicking on the "Improve Article" button below.