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

Last Updated : 22 Mar, 2023

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 elements 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 <bits/stdc++.h> 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[0]);     int k = 2;     cout << maxProd(arr, n, k);      return 0; }

Python

# 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

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<Integer,Integer> s = new HashMap<Integer,Integer>();                  // 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

// 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

JavaScript

// JavaScript implementation of the approach function maxProd(arr, n, k) { // To store the product let product = 1; let s = {};  // Sort the array arr.sort((a, b) => a - b);  for (let i = 0; i < n; i++) { if (!s[arr[i]]) { // Efficiently finding product // including every element once product *= arr[i]; } // Storing values in hash map s[arr[i]] = (s[arr[i]] || 0) + 1; }  for (let 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 *= Math.pow(arr[j], s[arr[j]] - 1); k -= s[arr[j]] - 1; s[arr[j]] = 0; } if (k <= s[arr[j]] - 1 && s[arr[j]] - 1 > 0) { product *= Math.pow(arr[j], k); break; } }  return product; }  // Driver code const arr = [5, 6, 7, 8, 2, 5, 6, 8]; const n = arr.length; const k = 2; console.log(maxProd(arr, n, k));

Output:

Time Complexity: O(n logn)
Auxiliary Space: O(n)

Largest possible Subset from an Array such that no element is K times any other element in the Subset

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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

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