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

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++

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// C++ implementation of the approach
#include <bits/stdc++.h>
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;
}

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Java














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// 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 



















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Python3














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# 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 



















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C#














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// 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 



















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Output:


7






          
          
          

          

    

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