How do you solve Largest Values From Labels on LeetCode?

Largest Values From Labels (LeetCode #1090) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(n log n) time complexity and O(n) space complexity. Key insight: Sorting items by value allows for a greedy selection of the highest values.

What is the brute force solution for Largest Values From Labels?

The brute force approach for Largest Values From Labels is Brute Force, with O(n²) time complexity and O(1) space complexity. The brute-force approach involves generating all possible subsets of items and checking which ones meet the criteria of maximum sum, numWanted, and useLimit. This is straightforward but inefficient for larger inputs. The optimal Optimal Solution approach improves this to O(n log n).

What algorithm or data structure is used to solve Largest Values From Labels?

Largest Values From Labels uses the following concepts: Array, Hash Table, Greedy, Sorting, Counting. The recommended patterns to study are: Hash Map, Array.

What are the interview tips for Largest Values From Labels?

Always clarify constraints and edge cases with the interviewer. Discuss your thought process and reasoning as you work through the problem. Consider both time and space complexity when proposing solutions.

What are common mistakes when solving Largest Values From Labels?

Failing to sort the items before selection. Not properly managing the count of items per label.

#1090

Largest Values From Labels

Medium

ArrayHash TableGreedySortingCountingHash MapArray

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

	Brute Force	Optimal Solution★
Time	O(n²)	O(n log n)
Space	O(1)	O(n)

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Intuition

Time O(n log n)Space O(n)

The optimal solution sorts the items by value and uses a greedy approach to select the highest values while respecting the label limits. This is efficient and leverages a hash map to track how many items from each label are selected.

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Algorithm

3 steps

1Step 1: Pair each value with its corresponding label and sort the pairs by value in descending order.
2Step 2: Initialize a hash map to count the number of items selected for each label.
3Step 3: Iterate through the sorted pairs, adding values to the sum while respecting numWanted and useLimit constraints.

solution.py12 lines

1def largestValuesFromLabels(values, labels, numWanted, useLimit):
2    items = sorted(zip(values, labels), key=lambda x: -x[0])
3    label_count = {}
4    total_sum = 0
5    count = 0
6    for value, label in items:
7        if count < numWanted:
8            if label_count.get(label, 0) < useLimit:
9                label_count[label] = label_count.get(label, 0) + 1
10                total_sum += value
11                count += 1
12    return total_sum

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Complexity note: The time complexity is O(n log n) due to the sorting step, which is efficient for the input size. The space complexity is O(n) because we store the items in a list.

1Sorting items by value allows for a greedy selection of the highest values.
2Using a hash map to track label counts helps enforce the useLimit constraint.

Solutions and explanations are original Tejav content. Problem titles © LeetCode — use the LeetCode button above for the full problem statement.