Largest Values From Labels - Leetcode Solution

Problem statement:

We have a set of items: the i-th item has value values[i] and label labels[i].

Then, we choose a subset S of these items, such that:

|S| <= num_wanted. For every label L, the number of items in S with label L is <= use_limit. Return the largest possible sum of the subset S.

Example:

Input: values = [5,4,3,2,1] labels = [1,1,2,2,3] num_wanted = 3 use_limit = 1 Output: 9 Explanation: The subset chosen is the first, third, and fifth item.

Solution:

This problem can be solved in O(n*log(n)) time using the greedy approach. We will sort the items in decreasing order of value and then iterate over them, adding the items to our subset S if they satisfy the label constraints. We keep track of the number of items with each label in S and stop adding items to S for a particular label when we reach the use limit for that label. We stop when we have added num_wanted items to S or there are no more items left.

Step-by-step Algorithm:

1. Create a list called items of tuples where each tuple contains the value and label of an item.
2. Sort the items in descending order of value.
3. Create a dictionary called used_labels to keep track of the number of items we have added to S for each label.
4. Initialize the variable ans to 0.
5. Iterate over the items in the order defined.
6. If the number of items with the label of the current item in S is less than the use limit, add the current item's value to ans and increment the count of the label in used_labels.
7. Stop iterating when we have added num_wanted items to S or there are no more items left.
8. Return the value of ans.

Pseudo-code:

1. items = [(values[i], labels[i]) for i in range(len(values))]
2. items.sort(reverse=True)
3. used_labels = {}
4. ans = 0
5. for value, label in items:
6. if label in used_labels and used_labels[label] >= use_limit:

7. continue
   

8. ans += value
9. used_labels[label] = used_labels.get(label, 0) + 1
10. if len(used_labels) == num_wanted:
11. break
12. return ans

Time Complexity:

The time complexity of the above algorithm is O(nlog(n)), where n is the number of items. The sort function in Python has a time complexity of O(nlog(n)), and iterating over n items has a time complexity of O(n). Therefore, the overall time complexity of the algorithm is O(n*log(n)).

Space Complexity:

The space complexity of the algorithm is O(n) because we are using a dictionary to keep track of the number of items with each label. This dictionary has a maximum size of n, assuming there are n unique labels. We are also storing the items as tuples in a list, which has size n. Therefore, the overall space complexity is O(n).

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