Number of sub arrays with negative product

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

Input: arr[] = {-1, 2, -2}
Output: 4
Subarray with negative product are {-1}, {-2}, {-1, 2} and {2, -2}.

Input: arr[] = {5, -4, -3, 2, -5}
Output: 8

Approach:

Replace the positive array elements with 1 and negative array elements with -1.
Create a prefix product array pre[] where pre[i] stores the product of all the elements from index arr[0] to arr[i].
Now, it can be noted that the sub-array arr[i…j] has a negative product only if pre[i] * pre[j] is negative.
Hence, the total count of sub-arrays with negative product will be the product of the count positive and negative elements in the prefix product array.

Below is the implementation of the above approach:

C++

// 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);
}
 
// Driver code
int main()
{
    int arr[] = { 5, -4, -3, 2, -5 };
    int n = sizeof(arr) / sizeof(arr[0]);
 
    cout << negProdSubArr(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); 
    } 
     
    // Driver code 
    public static void main (String[] args) 
    { 
        int arr[] = { 5, -4, -3, 2, -5 }; 
        int n = arr.length; 
     
        System.out.println(negProdSubArr(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)
 
# Driver code
arr = [5, -4, -3, 2, -5]
n = len(arr)
 
print(negProdSubArr(arr, n))
 
# This code is contributed by Mohit Kumar

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); 
    } 
     
    // Driver code 
    static public void Main ()
    {
        int []arr = { 5, -4, -3, 2, -5 }; 
        int n = arr.Length; 
     
        Console.Write(negProdSubArr(arr, n)); 
    } 
}
 
// This code is contributed by Sachin.

Javascript

<script>
 
// Javascript implementation of the approach
 
// Function to return the count of
// subarrays with negative product
function negProdSubArr(arr, n)
{
    let positive = 1, negative = 0;
    for (let 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);
}
 
// Driver code
    let arr = [ 5, -4, -3, 2, -5 ];
    let n = arr.length;
 
    document.write(negProdSubArr(arr, n));
 
</script>

Output:

Time Complexity: O(n)
Auxiliary Space: O(1), since no extra space has been taken.

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