Find the Number of Maximum Product Quadruples
Last Updated :
12 Sep, 2022
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 aiajakam 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 FX, 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 FX 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 - FX. For the second element, say Y, the number of ways are: FX 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 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 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# 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
<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>
Complexity Analysis:
- Time Complexity: O(NlogN), where N is the size of the array.
- Auxiliary Space: O(N)
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