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