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Count ways to obtain triplets with positive product consisting of at most one negative element
  • Last Updated : 03 Feb, 2021

Given an array arr[] of size N (1 ≤ N ≤ 105), the task is to find the number of ways to select a 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 atmost 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 elements. 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: 
 



 

  • 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
Output: 
1

 

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

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