Count ways to obtain triplets with positive product consisting of at most one negative element

Given an array arr[] of size N (1 ? N ? 10⁵), the task is to find the number of ways to select triplet i, j, and k such that i < j < k and the product arr[i] * arr[j] * arr[k] is positive. 
Note: Each triplet can consist of at most one negative element.

Examples:

Input: arr[] = {2, 5, -9, -3, 6}
Output: 1
Explanation: The total number ways to obtain a triplet i, j and k to satisfy given conditions is 1 {0, 1, 4}.

Input : arr[] = {2, 5, 6, -2, 5}
Output : 4
Explanation: The total number ways to obtain a triplet i, j and k to satisfy given conditions are 4 {0, 1, 2}, {0, 1, 4}, {1, 2, 4} and {0, 2, 4}.

Approach: All possible combinations of a triplet are as follows:

• # negative elements or 2 negative elements and 1 positive element. Both these combinations cannot be considered as the maximum allowed negative elements in a triplet is 1.
• 2 negative (-ve) elements and 1 positive (+ve) element. Since the product of the triplet will be negative, the triplet cannot be considered.
• 3 positive elements.

Follow the steps below to solve the problem: 

• Traverse the array and count frequency of positive array elements, say freq. 
• Count of ways to select a valid triplet from freq number of array elements using the formula PnC = ^NC₃= (N * (N – 1) * (N – 2)) / 6. Add the count obtained to the answer. 
• Print the count obtained.

Below is the implementation of the above approach:

C++

// C++ Program to implement
// the above approach

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

// Function to calculate
// possible number of triplets
long long int possibleTriplets(int arr[], int N)
{
    int freq = 0;
    // counting frequency of positive numbers
    // in array

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

        // If current array
        // element is positive
        if (arr[i] > 0) {

            // Increment frequency
            freq++;
        }
    }

    // Select a triplet from freq
    // elements such that i < j < k.
    return (freq * 1LL * (freq - 1)
            * (freq - 2))
           / 6;
}

// Driver Code
int main()
{
    int arr[] = { 2, 5, -9, -3, 6 };
    int N = sizeof(arr) / sizeof(arr[0]);

    cout << possibleTriplets(arr, N);

    return 0;
}

Java

// Java Program to implement
// the above approach
import java.util.*;
class GFG
{

// Function to calculate
// possible number of triplets
static int possibleTriplets(int arr[], int N)
{
    int freq = 0;

    // counting frequency of positive numbers
    // in array
    for (int i = 0; i < N; i++)
    {

        // If current array
        // element is positive
        if (arr[i] > 0)
        {

            // Increment frequency
            freq++;
        }
    }

    // Select a triplet from freq
    // elements such that i < j < k.
    return (int) ((freq * 1L * (freq - 1)
            * (freq - 2))
           / 6);
}

// Driver Code
public static void main(String[] args)
{
    int arr[] = { 2, 5, -9, -3, 6 };
    int N = arr.length;
    System.out.print(possibleTriplets(arr, N));
}
}

// This code is contributed by 29AjayKumar

Python3

# Python3 Program to implement
# the above approach

# Function to calculate
# possible number of triplets
def possibleTriplets(arr, N):
    freq = 0

    # counting frequency of positive numbers
    # in array
    for i in range(N):

        # If current array
        # element is positive
        if (arr[i] > 0):

            # Increment frequency
            freq += 1

    # Select a triplet from freq
    # elements such that i < j < k.
    return (freq * (freq - 1) * (freq - 2)) // 6

# Driver Code
if __name__ == '__main__':
    arr = [2, 5, -9, -3, 6]
    N = len(arr)

    print(possibleTriplets(arr, N))

    # This code is contributed by mohit kumar 29

C#

// C# Program to implement
// the above approach
using System;

public class GFG
{

// Function to calculate
// possible number of triplets
static int possibleTriplets(int []arr, int N)
{
    int freq = 0;

    // counting frequency of positive numbers
    // in array
    for (int i = 0; i < N; i++)
    {

        // If current array
        // element is positive
        if (arr[i] > 0)
        {

            // Increment frequency
            freq++;
        }
    }

    // Select a triplet from freq
    // elements such that i < j < k.
    return (int) ((freq * 1L * (freq - 1)
            * (freq - 2)) / 6);
}

// Driver Code
public static void Main(String[] args)
{
    int []arr = { 2, 5, -9, -3, 6 };
    int N = arr.Length;
    Console.Write(possibleTriplets(arr, N));
}
}

// This code is contributed by 29AjayKumar

Javascript

<script>

// Javascript Program to implement
// the above approach

// Function to calculate
// possible number of triplets
function possibleTriplets(arr, N)
{
    var freq = 0;
    // counting frequency of positive numbers
    // in array

    for (var i = 0; i < N; i++) {

        // If current array
        // element is positive
        if (arr[i] > 0) {

            // Increment frequency
            freq++;
        }
    }

    // Select a triplet from freq
    // elements such that i < j < k.
    return (freq * 1 * (freq - 1)
            * (freq - 2))
        / 6;
}

// Driver Code
var arr = [ 2, 5, -9, -3, 6 ];
var N = arr.length;
document.write( possibleTriplets(arr, N));

</script>

Output:

1

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