Number of sub arrays with negative product

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:



Below is the implementation of the above approach:

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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);
}
  
// Driver code
int main()
{
    int arr[] = { 5, -4, -3, 2, -5 };
    int n = sizeof(arr) / sizeof(arr[0]);
  
    cout << negProdSubArr(arr, n);
  
    return (0);
}
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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); 
    
      
    // 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
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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)
  
# Driver code
arr = [5, -4, -3, 2, -5]
n = len(arr)
  
print(negProdSubArr(arr, n))
  
# This code is contributed by Mohit Kumar
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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); 
    
      
    // 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.
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Output:
8

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