# Find the Number of Maximum Product Quadruples

Given an array of N positive elements find the number of quadruple, (i, j, k, m) such that i < j < k < m such that the product **a _{i}a_{j}a_{k}a_{m}** is the maximum possible

**Examples:**

Input : N = 7, arr = {1, 2, 3, 3, 3, 3, 5}Output : 4ExplanationThe maximum quadruple product possible is 135, which can be achieved by the following quadruples {i, j, k, m} such thata= 135: 1)_{i}a_{j}a_{k}a_{m}a2)_{3}a_{4}a_{5}a_{7}a3)_{3}a_{4}a_{6}a_{7}a4)_{4}a_{5}a_{6}a_{7}a_{3}a_{5}a_{6}a_{7}Input : N = 4, arr = {1, 5, 2, 1}Output : 1ExplanationThe maximum quadruple product possible is 10, which can be achieved by the following quadruple {1, 2, 3, 4} asa= 10_{1}a_{2}a_{3}a_{4}

**Brute Force: O(n ^{4})**

Generate all possible quadruples 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 maximum. So, the problem can now be reduced to finding the number of ways of selecting the four largest elements. To do so, maintain a frequency array which 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 this given a maximum product and the number of ways to do so are ^{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 are 4 – F_{X}. 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 remaining_choices = 0, it means the quadruples are selected, so we can stop the algorithm

## 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 Frequeny 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; ` `} ` |

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## 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 Frequeny 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 ` |

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**Output:**

1

**Time Complexity:** O(NlogN), where N is sizeof the array