LeetCode #1090 — MEDIUM

Largest Values From Labels

Move from brute-force thinking to an efficient approach using array strategy.

The Problem

Problem Statement

You are given n item's value and label as two integer arrays values and labels. You are also given two integers numWanted and useLimit.

Your task is to find a subset of items with the maximum sum of their values such that:

The number of items is at most numWanted.
The number of items with the same label is at most useLimit.

Return the maximum sum.

Example 1:

Input: values = [5,4,3,2,1], labels = [1,1,2,2,3], numWanted = 3, useLimit = 1

Output: 9

Explanation:

The subset chosen is the first, third, and fifth items with the sum of values 5 + 3 + 1.

Example 2:

Input: values = [5,4,3,2,1], labels = [1,3,3,3,2], numWanted = 3, useLimit = 2

Output: 12

Explanation:

The subset chosen is the first, second, and third items with the sum of values 5 + 4 + 3.

Example 3:

Input: values = [9,8,8,7,6], labels = [0,0,0,1,1], numWanted = 3, useLimit = 1

Output: 16

Explanation:

The subset chosen is the first and fourth items with the sum of values 9 + 7.

Constraints:

n == values.length == labels.length
1 <= n <= 2 * 10⁴
0 <= values[i], labels[i] <= 2 * 10⁴
1 <= numWanted, useLimit <= n

Step 01

Brute Force Baseline

Problem summary: You are given n item's value and label as two integer arrays values and labels. You are also given two integers numWanted and useLimit. Your task is to find a subset of items with the maximum sum of their values such that: The number of items is at most numWanted. The number of items with the same label is at most useLimit. Return the maximum sum.

Baseline thinking

Start with the most direct exhaustive search. That gives a correctness anchor before optimizing.

Pattern signal: Array · Hash Map · Greedy

Example 1

[5,4,3,2,1]
[1,1,2,2,3]
3
1

Example 2

[5,4,3,2,1]
[1,3,3,3,2]
3
2

Example 3

[9,8,8,7,6]
[0,0,0,1,1]
3
1

Core Insight

What unlocks the optimal approach

Consider the items in order from largest to smallest value, and greedily take the items if they fall under the use_limit. We can keep track of how many items of each label are used by using a hash table.

Interview move: turn each hint into an invariant you can check after every iteration/recursion step.

Step 03

Algorithm Walkthrough

Iteration Checklist

Define state (indices, window, stack, map, DP cell, or recursion frame).
Apply one transition step and update the invariant.
Record answer candidate when condition is met.
Continue until all input is consumed.

Use the first example testcase as your mental trace to verify each transition.

Step 04

Edge Cases

Minimum Input

Single element / shortest valid input

Validate boundary behavior before entering the main loop or recursion.

Duplicates & Repeats

Repeated values / repeated states

Decide whether duplicates should be merged, skipped, or counted explicitly.

Extreme Constraints

Upper-end input sizes

Re-check complexity target against constraints to avoid time-limit issues.

Invalid / Corner Shape

Empty collections, zeros, or disconnected structures

Handle special-case structure before the core algorithm path.

Step 05

Full Annotated Code

Source-backed implementations are provided below for direct study and interview prep.

// Accepted solution for LeetCode #1090: Largest Values From Labels
class Solution {
    public int largestValsFromLabels(int[] values, int[] labels, int numWanted, int useLimit) {
        int n = values.length;
        int[][] pairs = new int[n][2];
        for (int i = 0; i < n; ++i) {
            pairs[i] = new int[] {values[i], labels[i]};
        }
        Arrays.sort(pairs, (a, b) -> b[0] - a[0]);
        Map<Integer, Integer> cnt = new HashMap<>();
        int ans = 0, num = 0;
        for (int i = 0; i < n && num < numWanted; ++i) {
            int v = pairs[i][0], l = pairs[i][1];
            if (cnt.getOrDefault(l, 0) < useLimit) {
                cnt.merge(l, 1, Integer::sum);
                num += 1;
                ans += v;
            }
        }
        return ans;
    }
}

// Accepted solution for LeetCode #1090: Largest Values From Labels
func largestValsFromLabels(values []int, labels []int, numWanted int, useLimit int) (ans int) {
	n := len(values)
	pairs := make([][2]int, n)
	for i := 0; i < n; i++ {
		pairs[i] = [2]int{values[i], labels[i]}
	}
	sort.Slice(pairs, func(i, j int) bool { return pairs[i][0] > pairs[j][0] })
	cnt := map[int]int{}
	for i, num := 0, 0; i < n && num < numWanted; i++ {
		v, l := pairs[i][0], pairs[i][1]
		if cnt[l] < useLimit {
			cnt[l]++
			num++
			ans += v
		}
	}
	return
}

# Accepted solution for LeetCode #1090: Largest Values From Labels
class Solution:
    def largestValsFromLabels(
        self, values: List[int], labels: List[int], numWanted: int, useLimit: int
    ) -> int:
        ans = num = 0
        cnt = Counter()
        for v, l in sorted(zip(values, labels), reverse=True):
            if cnt[l] < useLimit:
                cnt[l] += 1
                num += 1
                ans += v
                if num == numWanted:
                    break
        return ans

// Accepted solution for LeetCode #1090: Largest Values From Labels
struct Solution;
use std::collections::HashMap;
type Pair = (i32, i32);

impl Solution {
    fn largest_vals_from_labels(
        values: Vec<i32>,
        labels: Vec<i32>,
        num_wanted: i32,
        use_limit: i32,
    ) -> i32 {
        let n = values.len();
        let use_limit = use_limit as usize;
        let mut num_wanted = num_wanted as usize;
        let mut pairs: Vec<Pair> = values.into_iter().zip(labels.into_iter()).collect();
        pairs.sort_unstable();
        let mut hm: HashMap<i32, usize> = HashMap::new();
        let mut res = 0;
        for i in (0..n).rev() {
            let count = hm.entry(pairs[i].1).or_default();
            if *count < use_limit {
                *count += 1;
                res += pairs[i].0;
                num_wanted -= 1;
            }
            if num_wanted == 0 {
                break;
            }
        }
        res
    }
}

#[test]
fn test() {
    let values = vec![5, 4, 3, 2, 1];
    let labels = vec![1, 1, 2, 2, 3];
    let num_wanted = 3;
    let use_limit = 1;
    let res = 9;
    assert_eq!(
        Solution::largest_vals_from_labels(values, labels, num_wanted, use_limit),
        res
    );
    let values = vec![5, 4, 3, 2, 1];
    let labels = vec![1, 3, 3, 3, 2];
    let num_wanted = 3;
    let use_limit = 2;
    let res = 12;
    assert_eq!(
        Solution::largest_vals_from_labels(values, labels, num_wanted, use_limit),
        res
    );
    let values = vec![9, 8, 8, 7, 6];
    let labels = vec![0, 0, 0, 1, 1];
    let num_wanted = 3;
    let use_limit = 1;
    let res = 16;
    assert_eq!(
        Solution::largest_vals_from_labels(values, labels, num_wanted, use_limit),
        res
    );
    let values = vec![9, 8, 8, 7, 6];
    let labels = vec![0, 0, 0, 1, 1];
    let num_wanted = 3;
    let use_limit = 2;
    let res = 24;
    assert_eq!(
        Solution::largest_vals_from_labels(values, labels, num_wanted, use_limit),
        res
    );
}

// Accepted solution for LeetCode #1090: Largest Values From Labels
function largestValsFromLabels(
    values: number[],
    labels: number[],
    numWanted: number,
    useLimit: number,
): number {
    const n = values.length;
    const pairs = new Array(n);
    for (let i = 0; i < n; ++i) {
        pairs[i] = [values[i], labels[i]];
    }
    pairs.sort((a, b) => b[0] - a[0]);
    const cnt: Map<number, number> = new Map();
    let ans = 0;
    for (let i = 0, num = 0; i < n && num < numWanted; ++i) {
        const [v, l] = pairs[i];
        if ((cnt.get(l) || 0) < useLimit) {
            cnt.set(l, (cnt.get(l) || 0) + 1);
            ++num;
            ans += v;
        }
    }
    return ans;
}

Step 06

Interactive Study Demo

Use this to step through a reusable interview workflow for this problem.

Custom checkpoints (one per line)

Press Step or Run All to begin.

Step 07

Complexity Analysis

Time

O(n log n)

Space

O(1)

Approach Breakdown

EXHAUSTIVE

O(2ⁿ) time

O(n) space

Try every possible combination of choices. With n items each having two states (include/exclude), the search space is 2ⁿ. Evaluating each combination takes O(n), giving O(n × 2ⁿ). The recursion stack or subset storage uses O(n) space.

GREEDY

O(n log n) time

O(1) space

Greedy algorithms typically sort the input (O(n log n)) then make a single pass (O(n)). The sort dominates. If the input is already sorted or the greedy choice can be computed without sorting, time drops to O(n). Proving greedy correctness (exchange argument) is harder than the implementation.

Shortcut: Sort + single pass → O(n log n). If no sort needed → O(n). The hard part is proving it works.

Coach Notes

Common Mistakes

Review these before coding to avoid predictable interview regressions.

Off-by-one on range boundaries

Wrong move: Loop endpoints miss first/last candidate.

Usually fails on: Fails on minimal arrays and exact-boundary answers.

Fix: Re-derive loops from inclusive/exclusive ranges before coding.

Mutating counts without cleanup

Wrong move: Zero-count keys stay in map and break distinct/count constraints.

Usually fails on: Window/map size checks are consistently off by one.

Fix: Delete keys when count reaches zero.

Using greedy without proof

Wrong move: Locally optimal choices may fail globally.

Usually fails on: Counterexamples appear on crafted input orderings.

Fix: Verify with exchange argument or monotonic objective before committing.

Largest Values From Labels

Problem Statement

Roadmap

Brute Force Baseline

Baseline thinking

Example 1

Example 2

Example 3

Related Problems

Core Insight

What unlocks the optimal approach

Algorithm Walkthrough

Iteration Checklist

Edge Cases

Full Annotated Code

Interactive Study Demo

Complexity Analysis

Approach Breakdown

Common Mistakes

Off-by-one on range boundaries

Mutating counts without cleanup

Using greedy without proof