Find subset with maximum sum under given condition

Given values[] and labels[] of n items and a positive integer limit, We need to choose a subset of these items in such a way that the number of the same type of label in the subset should be <= limit and sum of values are maximum among all possible subset choices.

Examples:

Input: values[] = [5, 3, 7, 1, 2],
       labels[] = [5, 7, 7, 7, 6],
       limit = 2
Output: 17
Explanation:
You can select first, second, third 
and Fifth values.
So, there is 1 value of the label 5 -> {5},
    2 value of the label 7 -> {3, 7} ,
    1 value of the label 6 -> {2}.
Final subset = {5, 3, 7, 2}
Sum  = 5 + 3 + 7 + 2 = 17.

Input: values[] = [9, 8, 7, 6, 5],
       labels[] = [5, 7, 7, 7, 6],
       limit = 2
Output: 29

Recommended: Please try your approach on {IDE} first, before moving on to the solution.

Approach: The idea is to use a Multimap and Hashmap to solve this problem.

We will store all the values and the respective labels as a pair in the Multimap.
In Multimap the {values, labels} pairs are sorted in increasing order. So, we will traverse the Multimap in reverse to get the pairs in decreasing order.
Now, we will add the values in our answer and store the occurrence of each label in Hashmap to check if the number of occurrence is <= limit.

Below is the implementation of the above approach:

// C++ program to Find subset with 
// maximum sum under given condition. 
#include <bits/stdc++.h> 
using namespace std; 
  
// Function to return the maximum 
// sum of the subset 
int MaxSumSubset(vector<int>& values,  
                 vector<int>& labels, 
                 int n, int limit) 
{ 
    int res = 0; 
    multimap<int, int> s; 
    unordered_map<int, int> map; 
  
    if (n == 0) 
    { 
        return 0; 
    } 
  
    // Pushing the pair into  
    // the multimap 
    for (int i = 0; i < n; i++) 
    { 
        s.insert({ values[i],  
                  labels[i] }); 
    } 
  
    // Traversing the multimap  
    // in reverse 
    for (auto it = s.rbegin(); 
         it != s.rend() && n > 0; it++) 
    { 
  
        if (++map[it->second] <= limit) 
        { 
            res += it->first; 
            n--; 
        } 
    } 
    return res; 
} 
// Driver code 
int main() 
{ 
    vector<int> values = { 5, 3, 7, 1, 2 }; 
    vector<int> labels = { 5, 7, 7, 7, 6 }; 
      
    int n = sizeof(values) / sizeof(values[0]); 
    int limit = 2; 
      
    cout << MaxSumSubset(values, labels, 
                         n, limit); 
    return 0; 
} 

Output:

Time Complexity: O(N), where N is the length of the array.

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My Personal Notes

Recommended: Please try your approach on {IDE} first, before moving on to the solution.

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