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# Maximum product from array such that frequency sum of all repeating elements in product is less than or equal to 2 * k

• Difficulty Level : Medium
• Last Updated : 08 Apr, 2019

Given an array arr[] and an integer k, the task is to find the maximum product from the array such that the frequency sum of all repeating elements in the product is ≤ 2 * k where frequency sum is the sum of frequencies of all the elements in the product that appear more than once. For example, if we choose a product 1 * 1 * 2 * 3 * 4 * 5 * 6 * 6 * 7 * 8 * 8 then the frequency sum of repeating element in this product is 6 as the repeating elements in the product are 1, 6 and 8 (frequency of 1 + frequency of 6 + frequency of 8) = (2 + 2 + 2) = 6

Examples:

Input: arr[] = {5, 6, 7, 8, 2, 5, 6, 8}, k = 2
Output: 161280
The products can be:
5 * 5 * 7 * 8 * 8 * 2 * 6 = 134400
5 * 8 * 8 * 7 * 6 * 6 * 2 = 161280
2 * 7 * 6 * 5 * 8 * 6 * 5 = 100800
Out of which 161280 is the maximum

Input: arr[] = {1, 5, 1, 5, 4, 3, 8}, k = 2
Output: 2400

Approach: First take the product of all the elements from the array (include only a single occurrence of the elements as a single occurrence will not affect the frequency sum). Now in order to maximize the product, sort the array and start taking all the remaining occurrences of the elements starting from the greatest element until the frequency sum doesn’t exceed 2 * k. Print the calculated product in the end.

Below is the implementation of the above approach:

## C++

 `// C++ implementation of the approach``#include ``using` `namespace` `std;``#define ll long long int`` ` `// Function to return the maximum product value``ll maxProd(``int` `arr[], ``int` `n, ``int` `k)``{`` ` `    ``// To store the product``    ``ll product = 1;``    ``unordered_map<``int``, ``int``> s;`` ` `    ``// Sort the array``    ``sort(arr, arr + n);`` ` `    ``for` `(``int` `i = 0; i < n; i++) {``        ``if` `(s[arr[i]] == 0) {`` ` `            ``// Efficiently finding product``            ``// including every element once``            ``product = product * arr[i];``        ``}`` ` `        ``// Storing values in hash map``        ``s[arr[i]] = s[arr[i]] + 1;``    ``}`` ` `    ``for` `(``int` `j = n - 1; j >= 0 && k > 0; j--) {``        ``if` `((k > (s[arr[j]] - 1)) && ((s[arr[j]] - 1) > 0)) {`` ` `            ``// Including the greater repeating values``            ``// so that product can be maximized``            ``product *= ``pow``(arr[j], s[arr[j]] - 1);``            ``k = k - s[arr[j]] + 1;``            ``s[arr[j]] = 0;``        ``}``        ``if` `(k <= (s[arr[j]] - 1) && ((s[arr[j]] - 1) > 0)) {``            ``product *= ``pow``(arr[j], k);``            ``break``;``        ``}``    ``}`` ` `    ``return` `product;``}`` ` `// Driver code``int` `main()``{``    ``int` `arr[] = { 5, 6, 7, 8, 2, 5, 6, 8 };``    ``int` `n = ``sizeof``(arr) / ``sizeof``(arr);``    ``int` `k = 2;``    ``cout << maxProd(arr, n, k);`` ` `    ``return` `0;``}`

## Java

 `// Java implementation of the approach``import` `java.util.*;`` ` `class` `GFG``{``    ``// Function to return the maximum product value``    ``static` `long` `maxProd(``int` `arr[], ``int` `n, ``int` `k)``    ``{``     ` `        ``// To store the product``        ``long` `product = ``1``;``        ``HashMap s = ``new` `HashMap(); ``         ` `        ``// Sort the array``        ``Arrays.sort(arr);``     ` `        ``for` `(``int` `i = ``0``; i < n; i++)``        ``{``            ``if` `(s.containsKey(arr[i]) == ``false``)``            ``{``     ` `                ``// Efficiently finding product``                ``// including every element once``                ``product = product * arr[i];``                 ` `                ``s.put(arr[i], ``1``);``     ` `            ``}``     ` `            ``// Storing values in hash map``            ``else``                ``s.put(arr[i],s.get(arr[i]) +``1``);``     ` `        ``}``     ` `        ``for` `(``int` `j = n - ``1``; j >= ``0` `&& k > ``0``; j--) ``        ``{``            ``if` `((k > (s.get(arr[j]) - ``1``)) && ``                    ``((s.get(arr[j]) - ``1``) > ``0``)) ``            ``{``     ` `                ``// Including the greater repeating values``                ``// so that product can be maximized``                ``product *= Math.pow(arr[j], s.get(arr[j]) - ``1``);``                ``k = k - s.get(arr[j]) + ``1``;``                ``s.put(arr[j], ``0``);``            ``}``            ``if` `(k <= (s.get(arr[j]) - ``1``) && ``                    ``((s.get(arr[j]) - ``1``) > ``0``)) ``            ``{``                ``product *= Math.pow(arr[j], k);``                ``break``;``            ``}``        ``}``     ` `        ``return` `product;``    ``}``     ` `    ``// Driver code``    ``public` `static` `void` `main (String[] args) ``    ``{``        ``int` `arr[] = { ``5``, ``6``, ``7``, ``8``, ``2``, ``5``, ``6``, ``8` `};``        ``int` `n = arr.length;``        ``int` `k = ``2``;``        ``System.out.println(maxProd(arr, n, k));``    ``}``}`` ` `// This code is contributed by ihritik`

## Python3

 `# Python3 implementation of the approach `` ` `# Function to return the maximum ``# product value ``def` `maxProd(arr, n, k) :`` ` `    ``# To store the product ``    ``product ``=` `1``; ``    ``s ``=` `dict``.fromkeys(arr, ``0``); `` ` `    ``# Sort the array ``    ``arr.sort(); `` ` `    ``for` `i ``in` `range``(n) :``        ``if` `(s[arr[i]] ``=``=` `0``) :`` ` `            ``# Efficiently finding product ``            ``# including every element once ``            ``product ``=` `product ``*` `arr[i]; `` ` `        ``# Storing values in hash map ``        ``s[arr[i]] ``=` `s[arr[i]] ``+` `1``; `` ` `    ``j ``=` `n ``-` `1``;``    ``while` `(j >``=` `0` `and` `k > ``0``) :``         ` `        ``if` `((k > (s[arr[j]] ``-` `1``)) ``and` `           ``((s[arr[j]] ``-` `1``) > ``0``)) :`` ` `            ``# Including the greater repeating values ``            ``# so that product can be maximized ``            ``product ``*``=` `pow``(arr[j], s[arr[j]] ``-` `1``); ``            ``k ``=` `k ``-` `s[arr[j]] ``+` `1``; ``            ``s[arr[j]] ``=` `0``; ``         ` `        ``if` `(k <``=` `(s[arr[j]] ``-` `1``) ``and` `                ``((s[arr[j]] ``-` `1``) > ``0``)) : ``            ``product ``*``=` `pow``(arr[j], k); ``            ``break``;``        ``j ``-``=` `1``         ` `    ``return` `product; `` ` `# Driver code ``if` `__name__ ``=``=` `"__main__"` `: `` ` `    ``arr ``=` `[ ``5``, ``6``, ``7``, ``8``, ``2``, ``5``, ``6``, ``8` `]; ``    ``n ``=` `len``(arr) ;``    ``k ``=` `2``; ``     ` `    ``print``(maxProd(arr, n, k)); ``     ` `# This code is contributed by Ryuga`

## C#

 `// C# implementation of the approach``using` `System; ``using` `System.Collections.Generic; `` ` `class` `GFG``{``    ``// Function to return the maximum product value``    ``static` `long` `maxProd(``int` `[]arr, ``int` `n, ``int` `k)``    ``{``     ` `        ``// To store the product``        ``long` `product = 1;``        ``Dictionary<``int``, ``int``> s = ``new` `Dictionary<``int``, ``int``>(); ``        ``// Sort the array``        ``Array.Sort(arr);``     ` `        ``for` `(``int` `i = 0; i < n; i++) ``        ``{``            ``if` `(!s.ContainsKey(arr[i]))``            ``{``     ` `                ``// Efficiently finding product``                ``// including every element once``                ``product = product * arr[i];``     ` `                ``s[arr[i]] = 1;``            ``}``     ` `            ``// Storing values in hash map``                ``else``            ``s[arr[i]]++;``     ` `        ``}``     ` `        ``for` `(``int` `j = n - 1; j >= 0 && k > 0; j--)``        ``{``            ``if` `((k > (s[arr[j]] - 1)) &&``                    ``((s[arr[j]] - 1) > 0))``            ``{``     ` `                ``// Including the greater repeating values``                ``// so that product can be maximized``                ``product *= (``long``)Math.Pow(arr[j], s[arr[j]] - 1);``                ``k = k - s[arr[j]] + 1;``                ``s[arr[j]] = 0;``            ``}``            ``if` `(k <= (s[arr[j]] - 1) && ((s[arr[j]] - 1) > 0)) ``            ``{``                ``product *= (``long``)Math.Pow(arr[j], k);``                ``break``;``            ``}``        ``}``     ` `        ``return` `product;``    ``}``     ` `    ``// Driver code``    ``public` `static` `void` `Main () ``    ``{``        ``int` `[]arr = { 5, 6, 7, 8, 2, 5, 6, 8 };``        ``int` `n = arr.Length;``        ``int` `k = 2;``        ``Console.WriteLine(maxProd(arr, n, k));``    ``}``}`` ` `// This code is contributed by ihritik`
Output:
```161280
```

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