# 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 ` `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); ` `    ``int` `limit = 2; ` `     `  `    ``cout << MaxSumSubset(values, labels, ` `                         ``n, limit); ` `    ``return` `0; ` `} `

Output:

```17
```

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

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