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Count distinct Triplets with negative product

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  • Last Updated : 26 Sep, 2022
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Given an array arr[] of N integers, the task is to find count of triplets having a product less than 0.

Examples:

Input: arr[] = {-1, -2, 3, 4, 5}
Output: 6
Explanation: Triplets are {-1, 3, 4}, {-1, 4, 5},  
{-1, 3, 5}, {-2, 3, 4}, {-2, 4, 5}, {-2, 3, 5}

Input: arr[] = {8, 7, 0}
Output:  0

Input: arr[] = {-1, 2, 2, 2}
Output: 3
Explanation: Three triplets can be formed {-1, 2, 2}, {-1, 2, 2} and {-1, 2, 2}
by choosing 2s from the following pair of indices {1, 2}, {2, 3}, {1, 3}. 

Approach: The problem can be solved based on the following observation.

The observation is we can form a triplets having negative product, if all three elements are negative or only one element is negative.

Follow the steps to solve this problem:

  • If N<3, return 0, as triplets can’t be formed.
  • Else, check the positive and negative count in arr[].
  • Let’s say the negative count is NegCount and the positive count is PosCount.
    • If NegCount = 0, no such triplet is possible.
    • Else, initialize x = 0 and y = 0.
    • If PosCount > 1, we can choose any combination of two positive elements.
    • If NegCount > 2, we can choose any combination of three negative elements.
    • For the above calculation, use the formula of combinatorics nCr = n! / r! *(n – r)!.
  • Return the summation of counts of the above two possible ways as the required answer.

Below is the implementation of the above approach.

C++

// C++ code to implement the above approach

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

// Function to find factorial of given number
int fact(int n)
{
    if (n == 0)
        return 1;
    return n * fact(n - 1);
}

// Function to calculate nCr
int nCr(int n, int r)
{
    return fact(n) / (fact(r) * fact(n - r));
}

// Function to return count of triplets
// having negative product
int findTriplets(int* arr, int N)
{

    if (N <= 2)
        return 0;

    int PosCount = 0;
    int NegCount = 0;

    for (int i = 0; i < N; i++) {
        if (arr[i] > 0)
            PosCount++;
        else if (arr[i] < 0)
            NegCount++;
    }

    if (NegCount == 0)
        return 0;

    int x = 0, y = 0;

    if (PosCount >= 2)
        x = NegCount * nCr(PosCount, 2);

    if (NegCount >= 3)
        y = nCr(NegCount, 3);

    return x + y;
}

// Driver Code
int main()
{
    int arr[] = { -1, -2, 3, 4, 5 };
    int N = sizeof(arr) / sizeof(arr[0]);

    // Function call
    cout << findTriplets(arr, N);

    return 0;
}

Java

// Java code to implement the above approach
import java.io.*;

class GFG {
    // Function to find factorial of given number
    public static int fact(int n)
    {
        if (n == 0)
            return 1;
        return n * fact(n - 1);
    }

    // Function to calculate nCr
    public static int nCr(int n, int r)
    {
        return fact(n) / (fact(r) * fact(n - r));
    }

    // Function to return count of triplets
    // having negative product
    public static int findTriplets(int arr[], int N)
    {

        if (N <= 2)
            return 0;

        int PosCount = 0;
        int NegCount = 0;

        for (int i = 0; i < N; i++) {
            if (arr[i] > 0)
                PosCount++;
            else if (arr[i] < 0)
                NegCount++;
        }

        if (NegCount == 0)
            return 0;

        int x = 0, y = 0;

        if (PosCount >= 2)
            x = NegCount * nCr(PosCount, 2);

        if (NegCount >= 3)
            y = nCr(NegCount, 3);

        return x + y;
    }

    // Driver Code
    public static void main(String[] args)
    {
        int arr[] = { -1, -2, 3, 4, 5 };
        int N = arr.length;

        // Function call
        System.out.print(findTriplets(arr, N));
    }
}

// This code is contributed by Rohit Pradhan

Python3

# Python3 code to implement the approach

# Function to find factorial of given number
def fact(n) :
    
    if (n == 0):
        return 1
    return n * fact(n - 1)

# Function to calculate nCr
def nCr(n, r) :
    
    return fact(n) // (fact(r) * fact(n - r))

# Function to return count of triplets
# having negative product
def findTriplets(arr, N) :
 
    if (N <= 2) :
        return 0
 
    PosCount = 0
    NegCount = 0
 
    for i in range(N):
        if (arr[i] > 0):
            PosCount += 1
        elif (arr[i] < 0):
            NegCount += 1
    
    if (NegCount == 0):
        return 0
 
    x = 0
    y = 0
 
    if (PosCount >= 2):
        x = NegCount * nCr(PosCount, 2)
 
    if (NegCount >= 3):
        y = nCr(NegCount, 3)
 
    return x + y

# Driver Code
if __name__ == "__main__":
    
    arr = [ -1, -2, 3, 4, 5 ]
    N = len(arr) 
 
    # Function call
    print(findTriplets(arr, N))

    # This code is contributed by sanjoy_62.

C#

// C# code to implement the approach
using System;

class GFG
{

  // Function to find factorial of given number
  public static int fact(int n)
  {
    if (n == 0)
      return 1;
    return n * fact(n - 1);
  }

  // Function to calculate nCr
  public static int nCr(int n, int r)
  {
    return fact(n) / (fact(r) * fact(n - r));
  }

  // Function to return count of triplets
  // having negative product
  public static int findTriplets(int[] arr, int N)
  {

    if (N <= 2)
      return 0;

    int PosCount = 0;
    int NegCount = 0;

    for (int i = 0; i < N; i++) {
      if (arr[i] > 0)
        PosCount++;
      else if (arr[i] < 0)
        NegCount++;
    }

    if (NegCount == 0)
      return 0;

    int x = 0, y = 0;

    if (PosCount >= 2)
      x = NegCount * nCr(PosCount, 2);

    if (NegCount >= 3)
      y = nCr(NegCount, 3);

    return x + y;
  }

  // Driver Code
  public static void Main()
  {
    int[] arr = { -1, -2, 3, 4, 5 };
    int N = arr.Length;

    // Function call
    Console.Write(findTriplets(arr, N));
  }
}

// This code is contributed by code_hunt.

Javascript

// javascript code to implement the approach
// Function to find factorial of given number
function fact(no){

    if (no == 0)
        return 1;
        
    return no * fact(no - 1);
}

// Function to calculate nCr
function nCr(n, r){

    return fact(n) / (fact(r) * fact(n - r));
}

// Function to return count of triplets
 // having negative product
function findTriplets(arr, N){
    
    if (N <= 2)
        return 0
        
    let PosCount = 0;
    let NegCount = 0;
    
    for (i = 0; i < N; i++){
    
        if (arr[i] > 0)
            PosCount++;
        else
            NegCount++;
    }        
    if (NegCount == 0)
        return 0;
        
    let x = 0; 
    let y = 0;
    
    if (PosCount >= 2){
        x = NegCount * nCr(PosCount, 2);
        }
        
    if (NegCount >= 3){
        y = nCr(NegCount, 3);
        }
        
    return x + y;
}
    
// Driver Code
    let arr = [-1, -2, 3, 4, 5 ];
    let N = arr.length;
    
    // Function call
    console.log(findTriplets(arr, N));
    
 // this code is contributed by ksam24000   
Output

6

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

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