Count ways to form minimum product triplets

Given an array of positive integers. We need to find how many triples of indices (i, j, k) (i < j < k), such that a[i] * a[j] * a[k] is minimum possible.

Examples:
Input : 5
        1 3 2 3 4
Output : 2
The triplets are (1, 3, 2)
and (1, 2, 3)

Input : 5
        2 2 2 2 2
Output : 5
In this example we choose three 2s 
out of five, and the number of ways 
to choose them is 5C3.

Input : 6
        1 3 3 1 3 2
Output : 1
There is only one way (1, 1, 2).

Following cases arise in this problem.



  1. All three minimum elements are same. For example {1, 1, 1, 1, 2, 3, 4}. The solution for such cases is nC3.
  2. Two elements are same. For example {1, 2, 2, 2, 3} or {1, 1, 2, 2}. In this case, count of occurrences of first (or minimum element) cannot be more than 2. If minimuum element appears two times, then answer is count of second element (We get to choose only 1 from all occurrences of second element. If minimum element appears once, the count is nC2.
  3. All three elements are distinct. For example {1, 2, 3, 3, 5}. In this case, answer is count of occurrences of third element (or nC1).

We first sort the array in increasing order. Then count the frequency of 3 element of 3rd element from starting. Let the frequency be ‘count’. Following cases arise.

  • If 3rd element is equal to the first element, no. of triples will be (count-2)*(count-1)*(count)/6, where count is the frequency of 3rd element.
  • If 3rd element is equal to 2nd element, no. of triples will be (count-1)*(count)/2. Otherwise no. of triples will be value of count.

C++

filter_none

edit
close

play_arrow

link
brightness_4
code

// CPP program to count number of ways we can
// form triplets with minimum product.
#include <bits/stdc++.h>
using namespace std;
  
// function to calculate number of triples
long long noOfTriples(long long arr[], int n)
{
    // Sort the array
    sort(arr, arr + n);
  
    // Count occurrences of third element
    long long count = 0;
    for (long long i = 0; i < n; i++) 
        if (arr[i] == arr[2])
            count++;
      
    // If all three elements are same (minimum
    // element appears at least 3 times). Answer
    // is nC3.
    if (arr[0] == arr[2])
        return (count - 2) * (count - 1) * (count) / 6;
  
    // If minimum element appears once.  
    // Answer is nC2.
    else if (arr[1] == arr[2])
        return (count - 1) * (count) / 2;
   
    // Minimum two elements are distinct.
    // Answer is nC1.
    return count;
}
  
// Driver code
int main()
{
    long long arr[] = { 1, 3, 3, 4 };
    int n = sizeof(arr) / sizeof(arr[0]);
    cout << noOfTriples(arr, n);
    return 0;
}

chevron_right


Java

filter_none

edit
close

play_arrow

link
brightness_4
code

// Java program to count number of ways we can
// form triplets with minimum product.
import java.util.Arrays;
  
class GFG {
          
    // function to calculate number of triples
    static long noOfTriples(long arr[], int n)
    {
          
        // Sort the array
        Arrays.sort(arr);
      
        // Count occurrences of third element
        long count = 0;
        for (int i = 0; i < n; i++) 
            if (arr[i] == arr[2])
                count++;
          
        // If all three elements are same (minimum
        // element appears at least 3 times). Answer
        // is nC3.
        if (arr[0] == arr[2])
            return (count - 2) * (count - 1) * 
                                      (count) / 6;
      
        // If minimum element appears once. 
        // Answer is nC2.
        else if (arr[1] == arr[2])
            return (count - 1) * (count) / 2;
      
        // Minimum two elements are distinct.
        // Answer is nC1.
        return count;
    }
      
    //driver code
    public static void main(String arg[])
    {
          
        long arr[] = { 1, 3, 3, 4 };
        int n = arr.length;
          
        System.out.print(noOfTriples(arr, n));
    }
}
  
// This code is contributed by Anant Agarwal.

chevron_right


Python3

filter_none

edit
close

play_arrow

link
brightness_4
code

# Python3 program to count number
# of ways we can form triplets 
# with minimum product.
  
# function to calculate number of triples
def noOfTriples (arr, n):
  
    # Sort the array
    arr.sort()
      
    # Count occurrences of third element
    count = 0
    for i in range(n):
        if arr[i] == arr[2]:
            count+=1
      
    # If all three elements are same 
    # (minimum element appears at l
    # east 3 times). Answer is nC3.
    if arr[0] == arr[2]:
        return (count - 2) * (count - 1) * (count) / 6
      
    # If minimum element appears once.
    # Answer is nC2.
    elif arr[1] == arr[2]:
        return (count - 1) * (count) / 2
      
    # Minimum two elements are distinct.
    # Answer is nC1.
    return count
      
# Driver code
arr = [1, 3, 3, 4]
n = len(arr)
print (noOfTriples(arr, n))
  
# This code is contributed by "Abhishek Sharma 44"

chevron_right


C#

filter_none

edit
close

play_arrow

link
brightness_4
code

// C# program to count number of ways
// we can form triplets with minimum product.
using System;
  
class GFG {
      
// function to calculate number of triples
static long noOfTriples(long []arr, int n)
{
    // Sort the array
    Array.Sort(arr);
   
    // Count occurrences of third element
    long count = 0;
    for (int i = 0; i < n; i++) 
        if (arr[i] == arr[2])
            count++;
       
    // If all three elements are same (minimum
    // element appears at least 3 times). Answer
    // is nC3.
    if (arr[0] == arr[2])
        return (count - 2) * (count - 1) * (count) / 6;
   
    // If minimum element appears once.  
    // Answer is nC2.
    else if (arr[1] == arr[2])
        return (count - 1) * (count) / 2;
    
    // Minimum two elements are distinct.
    // Answer is nC1.
    return count;
}
  
//driver code
public static void Main()
{
    long []arr = { 1, 3, 3, 4 };
    int n = arr.Length;
    Console.Write(noOfTriples(arr, n));
}
}
  
//This code is contributed by Anant Agarwal.

chevron_right


PHP

filter_none

edit
close

play_arrow

link
brightness_4
code

<?php
// PHP program to count number of ways 
// we can form triplets with minimum
// product.
  
// function to calculate number of
// triples
function noOfTriples( $arr, $n)
{
    // Sort the array
    sort($arr);
  
    // Count occurrences of third element
    $count = 0;
    for ( $i = 0; $i < $n; $i++) 
        if ($arr[$i] == $arr[2])
            $count++;
      
    // If all three elements are same 
    // (minimum element appears at least 
    // 3 times). Answer is nC3.
    if ($arr[0] == $arr[2])
        return ($count - 2) * ($count - 1)  
                           * ($count) / 6;
  
    // If minimum element appears once. 
    // Answer is nC2.
    else if ($arr[1] == $arr[2])
        return ($count - 1) * ($count) / 2;
  
    // Minimum two elements are distinct.
    // Answer is nC1.
    return $count;
}
  
// Driver code
    $arr = array( 1, 3, 3, 4 );
    $n = count($arr);
    echo noOfTriples($arr, $n);
  
// This code is contributed by anuj_67.
?>

chevron_right



Output:

1

Time Complexity: O(n Log n)

The solution can be optimized by first finding minimum element and its frequency and if frequency is less than 3, then finding second minimum and its frequency. If overall frequency is less than 3, then finding third minimum and its frequency. Time complexity of this optimized solution would be O(n)

This article is contributed by Sagar Shukla. If you like GeeksforGeeks and would like to contribute, you can also write an article using contribute.geeksforgeeks.org or mail your article to contribute@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.

Please write comments if you find anything incorrect, or you want to share more information about the topic discussed above.



My Personal Notes arrow_drop_up

Improved By : vt_m, BharathVarma1