Count of index pairs in array whose range product is a positive integer

Given an array A of non-zero integers, the task is to find the number of pairs (l, r) where (l <= r) such that A[l]*A[l+1]*A[l+2]….A[r] is positive. 
Examples:

Input: A = {5, -3, 3, -1, 1} 
Output: 7 
Explanation:
First pair, (1, 1) = 5 is positive 
Second pair, (3, 3) = 3 is positive 
Third pair, (1, 4) = 5 * -3 * 3 * -1 = 45 is positive 
Forth pair, (2, 4) = -3 * 3 * -1 = 9 is positive 
Fifth pair, (1, 5) = 5 * -3 * 3 * -1 * 1 = 45 is positive 
Sixth pair, (2, 5) = -3 * 3 * -1 * 1 = 9 is positive 
Seventh pair, (5, 5) = 1 is positive 
So, there are seven pairs with positive product.
Input: A = {4, 2, -4, 3, 1, 2, -4, 3, 2, 3} 
Output: 27 

Approach:
The idea is to check possible number pairs for every array element. 

• Iterate through an array, follow the below steps for every element in array.
• Keep a track of the number of elements having an even number of negative elements before them (as even_count) and number of elements having odd number of negative elements before them (as odd_count).
• Store the total number of negative elements till now (as total_count).
• If total_count is even then add even_count to the answer. Otherwise add odd_count.

Below is the implementation of the above approach:

C++

// C++ Program to find the
// count of index pairs
// in the array positive
// range product

#include <bits/stdc++.h>
using namespace std;

void positiveProduct(int arr[], int n)
{
    int even_count = 0;
    int odd_count = 0;
    int total_count = 0;
    int ans = 0;

    for (int i = 0; i < n; i++) {

        // Condition if number of
        // negative elements is even
        // then increase even_count
        if (total_count % 2 == 0)
            even_count++;

        // Otherwise increase odd_count
        else
            odd_count++;

        // Condition if current element
        // is negative
        if (arr[i] < 0)
            total_count++;

        // Condition if number of
        // negative elements is even
        // then add even_count
        // in answer
        if (total_count % 2 == 0)
            ans += even_count;

        // Otherwise add odd_count
        // in answer
        else
            ans += odd_count;
    }

    cout << ans << "\n";
}

// Driver Code
int main()
{
    int A[] = { 5, -3, 3, -1, 1 };

    int size = sizeof(A) / sizeof(A[0]);

    positiveProduct(A, size);

    return 0;
}

Java

// Java program to find the count of
// index pairs in the array positive
// range product
class GFG{

public static void positiveProduct(int arr[],
                                   int n)
{
    int even_count = 0;
    int odd_count = 0;
    int total_count = 0;
    int ans = 0;

    for(int i = 0; i < n; i++)
    {

       // Condition if number of
       // negative elements is even
       // then increase even_count
       if (total_count % 2 == 0)
       {
           even_count++;
       }

       // Otherwise increase odd_count
       else
       {
           odd_count++;
       }

       // Condition if current element
       // is negative
       if (arr[i] < 0)
       {
           total_count++;
       }

       // Condition if number of
       // negative elements is even
       // then add even_count
       // in answer
       if (total_count % 2 == 0)
           ans += even_count;

       // Otherwise add odd_count
       // in answer
       else
           ans += odd_count;
    }
    System.out.println(ans);

}

// Driver Code
public static void main(String[] args)
{
    int A[] = { 5, -3, 3, -1, 1 };
    int size = A.length;

    positiveProduct(A, size);
}
}

// This code is contributed by divyeshrabadiya07

Python3

# Python3 program to find the count
# of index pairs in the array
# positive range product
def positiveProduct(arr, n):

    even_count = 0
    odd_count = 0
    total_count = 0
    ans = 0

    for i in range(n):

        # Condition if number of
        # negative elements is even
        # then increase even_count
        if(total_count % 2 == 0):
            even_count += 1

        # Otherwise increase odd_count
        else:
            odd_count += 1

        # Condition if current element
        # is negative
        if(arr[i] < 0):
            total_count += 1

        # Condition if number of
        # negative elements is even
        # then add even_count
        # in answer
        if(total_count % 2 == 0):
            ans += even_count

        # Otherwise add odd_count
        # in answer
        else:
            ans += odd_count

    print(ans)

# Driver Code
if __name__ == '__main__':

    A = [ 5, -3, 3, -1, 1 ]
    size = len(A)

    positiveProduct(A, size)

# This code is contributed by Shivam Singh

C#

// C# program to find the count of
// index pairs in the array positive
// range product
using System;

class GFG{

public static void positiveProduct(int []arr,
                                   int n)
{
    int even_count = 0;
    int odd_count = 0;
    int total_count = 0;
    int ans = 0;

    for(int i = 0; i < n; i++)
    {

       // Condition if number of
       // negative elements is even
       // then increase even_count
       if (total_count % 2 == 0)
       {
           even_count++;
       }

       // Otherwise increase odd_count
       else
       {
           odd_count++;
       }

       // Condition if current element
       // is negative
       if (arr[i] < 0)
       {
           total_count++;
       }

       // Condition if number of
       // negative elements is even
       // then add even_count
       // in answer
       if (total_count % 2 == 0)
           ans += even_count;

       // Otherwise add odd_count
       // in answer
       else
           ans += odd_count;
    }
    Console.WriteLine(ans);
}

// Driver Code
public static void Main(String[] args)
{
    int []A = { 5, -3, 3, -1, 1 };
    int size = A.Length;

    positiveProduct(A, size);
}
}

// This code is contributed by 29AjayKumar

Javascript

<script>

// Javascript program to find the count of
// index pairs in the array positive
// range product
function positiveProduct(arr,n)
{
    let even_count = 0;
    let odd_count = 0;
    let total_count = 0;
    let ans = 0;

    for(let i = 0; i < n; i++)
    {

    // Condition if number of
    // negative elements is even
    // then increase even_count
    if (total_count % 2 == 0)
    {
        even_count++;
    }

    // Otherwise increase odd_count
    else
    {
        odd_count++;
    }

    // Condition if current element
    // is negative
    if (arr[i] < 0)
    {
        total_count++;
    }

    // Condition if number of
    // negative elements is even
    // then add even_count
    // in answer
    if (total_count % 2 == 0)
        ans += even_count;

    // Otherwise add odd_count
    // in answer
    else
        ans += odd_count;
    }
    document.write(ans);

}

// Driver Code

    let A = [5, -3, 3, -1, 1 ];
    let size = A.length;

    positiveProduct(A, size);


// This code is contributed by sravan kumar Gottumukkala

</script>

Output:

7

Time Complexity: O(N)
Space Complexity: O(1)