Find the Number of Maximum Product Quadruples

Given an array of N positive elements, find the number of quadruples, (i, j, k, m) such that i < j < k < m such that the product a_ia_ja_ka_m is the maximum possible.

Examples:

Input : N = 7, arr = {1, 2, 3, 3, 3, 3, 5}  
Output : 4 
 Explanation 
The maximum quadruple product possible is 135, which can be 
achieved by the following quadruples {i, j, k, m} such that aiajakam = 135:
1) a3a4a5a7
2) a3a4a6a7
3) a4a5a6a7
4) a3a5a6a7

Input : N = 4, arr = {1, 5, 2, 1} 
Output : 1 
Explanation 
The maximum quadruple product possible is 10, which can be 
achieved by the following quadruple {1, 2, 3, 4} as a1a2a3a4 = 10

Brute Force: O(n⁴)

Generate all possible quadruple and count the quadruples, giving the maximum product.

Optimized Solution: It is easy to see that the product of the four largest numbers would be the maximum. So, the problem can now be reduced to finding the number of ways of selecting the four largest elements. To do so, we maintain a frequency array that stores the frequency of each element of the array. 

Suppose the largest element is X with frequency F_X, then if the frequency of this element is >= 4, it is best suited to select the four elements as X, X, X, as that, given a maximum product and the number of ways to do so is ^F_X C ₄

and if the frequency is less than 4, the number of ways to select this is 1 and now the required number of elements is 4 – F_X. For the second element, say Y, the number of ways are: ^F_X C _{remaining_choices}. Remaining choices denotes the number of additional elements we need to select after selecting the first element. If at any time, the remaining_choices = 0, it means the quadruples are selected, so we can stop the algorithm. 

Implementation:

C++

// CPP program to find the number of Quadruples
// having maximum product
#include <bits/stdc++.h>
using namespace std;

// Returns the number of ways to select r objects
// out of available n choices
int NCR(int n, int r)
{
    int numerator = 1;
    int denominator = 1;

    // ncr = (n * (n - 1) * (n - 2) * .....
    // ... (n - r + 1)) / (r * (r - 1) * ... * 1)
    while (r > 0) {
        numerator *= n;
        denominator *= r;
        n--;
        r--;
    }

    return (numerator / denominator);
}

// Returns the number of quadruples having maximum product
int findWays(int arr[], int n)
{
    // stores the frequency of each element
    map<int, int> count;

    if (n < 4)
        return 0;

    for (int i = 0; i < n; i++) {
        count[arr[i]]++;
    }

    // remaining_choices denotes the remaining
    // elements to select inorder to form quadruple
    int remaining_choices = 4;
    int ans = 1;

    // traverse the elements of the map in reverse order
    for (auto iter = count.rbegin(); iter != count.rend(); ++iter) {
        int number = iter->first;
        int frequency = iter->second;

        // If Frequency of element < remaining choices,
        // select all of these elements, else select only
        // the number of elements required
        int toSelect = min(remaining_choices, frequency);
        ans = ans * NCR(frequency, toSelect);

        // Decrement remaining_choices acc to the number
        // of the current elements selected
        remaining_choices -= toSelect;

        // if the quadruple is formed stop the algorithm
        if (!remaining_choices) {
            break;
        }
    }
    return ans;
}

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

    int maxQuadrupleWays = findWays(arr, n);
    cout << maxQuadrupleWays;

    return 0;
}

Java

// Java program to find the number of Quadruples
// having maximum product
import java.util.*;
class Solution
{
// Returns the number of ways to select r objects
// out of available n choices
static int NCR(int n, int r)
{
    int numerator = 1;
    int denominator = 1;

    // ncr = (n * (n - 1) * (n - 2) * .....
    // ... (n - r + 1)) / (r * (r - 1) * ... * 1)
    while (r > 0) {
        numerator *= n;
        denominator *= r;
        n--;
        r--;
    }

    return (numerator / denominator);
}

// Returns the number of quadruples having maximum product
static int findWays(int arr[], int n)
{
    // stores the frequency of each element
    HashMap<Integer,Integer> count= new HashMap<Integer,Integer>();

    if (n < 4)
        return 0;

    for (int i = 0; i < n; i++) {
        count.put(arr[i],(count.get(arr[i])==null?0🙁int)count.get(arr[i])));
    }

    // remaining_choices denotes the remaining
    // elements to select inorder to form quadruple
    int remaining_choices = 4;
    int ans = 1;

    // Getting an iterator
        Iterator hmIterator = count.entrySet().iterator();

        while (hmIterator.hasNext()) {
            Map.Entry mapElement = (Map.Entry)hmIterator.next();
            int number =(int) mapElement.getKey();
            int frequency =(int)mapElement.getValue();

              // If Frequency of element < remaining choices,
        // select all of these elements, else select only
        // the number of elements required
        int toSelect = Math.min(remaining_choices, frequency);
        ans = ans * NCR(frequency, toSelect);

        // Decrement remaining_choices acc to the number
        // of the current elements selected
        remaining_choices -= toSelect;

        // if the quadruple is formed stop the algorithm
        if (remaining_choices==0) {
            break;
        }
        }

    return ans;
}

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

    int maxQuadrupleWays = findWays(arr, n);
    System.out.print( maxQuadrupleWays);
}
}
//contributed by Arnab Kundu

Python3

# Python3 program to find
# the number of Quadruples
# having maximum product
from collections import defaultdict

# Returns the number of ways
# to select r objects out of
# available n choices
def NCR(n, r):

    numerator = 1
    denominator = 1

    # ncr = (n * (n - 1) *
    # (n - 2) * .....
    # ... (n - r + 1)) /
    # (r * (r - 1) * ... * 1)
    while (r > 0):
        numerator *= n
        denominator *= r
        n -= 1
        r -= 1

    return (numerator // denominator)

# Returns the number of
# quadruples having
# maximum product
def findWays(arr, n):

    # stores the frequency
    # of each element
    count = defaultdict (int)

    if (n < 4):
        return 0

    for i in range (n):
        count[arr[i]] += 1

    # remaining_choices denotes
    # the remaining elements to
    # select inorder to form quadruple
    remaining_choices = 4
    ans = 1

    # traverse the elements of
    # the map in reverse order
    for it in reversed(sorted(count.keys())):
        number = it
        frequency = count[it]

        # If Frequency of element <
        # remaining choices, select
        # all of these elements,
        # else select only the
        # number of elements required
        toSelect = min(remaining_choices,
                       frequency)
        ans = ans * NCR(frequency,
                        toSelect)

        # Decrement remaining_choices
        # acc to the number of the
        # current elements selected
        remaining_choices -= toSelect

        # if the quadruple is
        # formed stop the algorithm
        if (not remaining_choices):
            break
    return ans

# Driver Code
if __name__ == "__main__":

    arr = [1, 2, 3, 3, 3, 5]
    n = len(arr)
    maxQuadrupleWays = findWays(arr, n)
    print (maxQuadrupleWays)

# This code is contributed by Chitranayal

C#

// C# program to find the number of Quadruples
// having maximum product
using System;
using System.Collections.Generic;

class GFG
{

  // Returns the number of ways to select r objects
  // out of available n choices
  static int NCR(int n, int r)
  {
    int numerator = 1;
    int denominator = 1;

    // ncr = (n * (n - 1) * (n - 2) * .....
    // ... (n - r + 1)) / (r * (r - 1) * ... * 1)
    while (r > 0) {
      numerator *= n;
      denominator *= r;
      n--;
      r--;
    }

    return (numerator / denominator);
  }

  // Returns the number of quadruples having maximum
  // product
  static int findWays(int[] arr, int n)
  {

    // stores the frequency of each element
    Dictionary<int, int> count
      = new Dictionary<int, int>();

    if (n < 4)
      return 0;

    for (int i = 0; i < n; i++) {
      if (!count.ContainsKey(arr[i]))
        count[arr[i]] = 1;
      else
        count[arr[i]] += 1;
    }

    // remaining_choices denotes the remaining
    // elements to select inorder to form quadruple
    int remaining_choices = 4;
    int ans = 1;

    List<int> mapElements = new List<int>(count.Keys);
    mapElements.Sort();
    mapElements.Reverse();

    // Getting an iterator
    foreach(var number in mapElements)
    {
      int frequency = count[number];

      // If Frequency of element < remaining choices,
      // select all of these elements, else select
      // only the number of elements required
      int toSelect
        = Math.Min(remaining_choices, frequency);
      ans = ans * NCR(frequency, toSelect);

      // Decrement remaining_choices acc to the number
      // of the current elements selected
      remaining_choices -= toSelect;

      // if the quadruple is formed stop the algorithm
      if (remaining_choices == 0) {
        break;
      }
    }

    return ans;
  }

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

    int maxQuadrupleWays = findWays(arr, n);
    Console.WriteLine(maxQuadrupleWays);
  }
}

// This code is contributed by phasing17

Javascript

<script>
// Javascript program to find the number of Quadruples
// having maximum product

// Returns the number of ways to select r objects
// out of available n choices
function NCR(n,r)
{
    let numerator = 1;
    let denominator = 1;

    // ncr = (n * (n - 1) * (n - 2) * .....
    // ... (n - r + 1)) / (r * (r - 1) * ... * 1)
    while (r > 0) {
        numerator *= n;
        denominator *= r;
        n--;
        r--;
    }

    return Math.floor(numerator / denominator);
}

// Returns the number of quadruples having maximum product
function findWays(arr,n)
{
    // stores the frequency of each element
    let count= new Map();

    if (n < 4)
        return 0;

    for (let i = 0; i < n; i++) {
        count.set(arr[i],(count.get(arr[i])==
                  null?0:count.get(arr[i])+1));
    }

    // remaining_choices denotes the remaining
    // elements to select inorder to form quadruple
    let remaining_choices = 4;
    let ans = 1;

    // Getting an iterator


        for(let [key, value] of count.entries()) {
            let number =key;
            let frequency =value;

            // If Frequency of element < remaining choices,
            // select all of these elements, else select only
            // the number of elements required
            let toSelect = Math.min(remaining_choices, frequency);
            ans = ans * NCR(frequency, toSelect);

            // Decrement remaining_choices acc to the number
            // of the current elements selected
            remaining_choices -= toSelect;

            // if the quadruple is formed stop the algorithm
            if (remaining_choices==0) {
                break;
            }
        }

    return ans;
}

// Driver Code
let arr=[1, 2, 3, 3, 3, 5 ];
let n = arr.length;
let maxQuadrupleWays = findWays(arr, n);
document.write( maxQuadrupleWays);


// This code is contributed by rag2127
</script>

Output

1

Complexity Analysis:

• Time Complexity: O(NlogN), where N is the size of the array.
• Auxiliary Space: O(N)