LeetCode #3684 — EASY

Maximize Sum of At Most K Distinct Elements

Build confidence with an intuition-first walkthrough focused on array fundamentals.

The Problem

Problem Statement

You are given a positive integer array nums and an integer k.

Choose at most k elements from nums so that their sum is maximized. However, the chosen numbers must be distinct.

Return an array containing the chosen numbers in strictly descending order.

Example 1:

Input: nums = [84,93,100,77,90], k = 3

Output: [100,93,90]

Explanation:

The maximum sum is 283, which is attained by choosing 93, 100 and 90. We rearrange them in strictly descending order as [100, 93, 90].

Example 2:

Input: nums = [84,93,100,77,93], k = 3

Output: [100,93,84]

Explanation:

The maximum sum is 277, which is attained by choosing 84, 93 and 100. We rearrange them in strictly descending order as [100, 93, 84]. We cannot choose 93, 100 and 93 because the chosen numbers must be distinct.

Example 3:

Input: nums = [1,1,1,2,2,2], k = 6

Output: [2,1]

Explanation:

The maximum sum is 3, which is attained by choosing 1 and 2. We rearrange them in strictly descending order as [2, 1].

Constraints:

1 <= nums.length <= 100
1 <= nums[i] <= 10⁹
1 <= k <= nums.length

Step 01

Brute Force Baseline

Problem summary: You are given a positive integer array nums and an integer k. Choose at most k elements from nums so that their sum is maximized. However, the chosen numbers must be distinct. Return an array containing the chosen numbers in strictly descending order.

Baseline thinking

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

Pattern signal: Array · Hash Map · Greedy

Example 1

[84,93,100,77,90]
3

Example 2

[84,93,100,77,93]
3

Example 3

[1,1,1,2,2,2]
6

Step 02

Core Insight

What unlocks the optimal approach

Use a set to dedupe then sort descending and pick top <code>k</code>.

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 #3684: Maximize Sum of At Most K Distinct Elements
class Solution {
    public int[] maxKDistinct(int[] nums, int k) {
        Arrays.sort(nums);
        int n = nums.length;
        List<Integer> ans = new ArrayList<>();
        for (int i = n - 1; i >= 0; --i) {
            if (i + 1 < n && nums[i] == nums[i + 1]) {
                continue;
            }
            ans.add(nums[i]);
            if (--k == 0) {
                break;
            }
        }
        return ans.stream().mapToInt(x -> x).toArray();
    }
}

// Accepted solution for LeetCode #3684: Maximize Sum of At Most K Distinct Elements
func maxKDistinct(nums []int, k int) (ans []int) {
	slices.Sort(nums)
	n := len(nums)
	for i := n - 1; i >= 0; i-- {
		if i+1 < n && nums[i] == nums[i+1] {
			continue
		}
		ans = append(ans, nums[i])
		if k--; k == 0 {
			break
		}
	}
	return
}

# Accepted solution for LeetCode #3684: Maximize Sum of At Most K Distinct Elements
class Solution:
    def maxKDistinct(self, nums: List[int], k: int) -> List[int]:
        nums.sort()
        n = len(nums)
        ans = []
        for i in range(n - 1, -1, -1):
            if i + 1 < n and nums[i] == nums[i + 1]:
                continue
            ans.append(nums[i])
            k -= 1
            if k == 0:
                break
        return ans

// Accepted solution for LeetCode #3684: Maximize Sum of At Most K Distinct Elements
fn max_k_distinct(nums: Vec<i32>, k: i32) -> Vec<i32> {
    use std::collections::HashSet;

    let k = k as usize;
    let mut nums = nums;
    nums.sort_unstable();

    let mut ret = vec![];
    let mut s = HashSet::new();
    for n in nums.into_iter().rev() {
        if s.contains(&n) {
            continue;
        }

        ret.push(n);
        s.insert(n);

        if s.len() >= k {
            break;
        }
    }

    ret
}

fn main() {
    let nums = vec![84, 93, 100, 77, 93];
    let ret = max_k_distinct(nums, 3);
    println!("ret={ret:?}");
}

#[test]
fn test() {
    {
        let nums = vec![84, 93, 100, 77, 90];
        let ret = max_k_distinct(nums, 3);
        assert_eq!(ret, vec![100, 93, 90]);
    }
    {
        let nums = vec![84, 93, 100, 77, 93];
        let ret = max_k_distinct(nums, 3);
        assert_eq!(ret, vec![100, 93, 84]);
    }
    {
        let nums = vec![1, 1, 1, 2, 2, 2];
        let ret = max_k_distinct(nums, 6);
        assert_eq!(ret, vec![2, 1]);
    }
}

// Accepted solution for LeetCode #3684: Maximize Sum of At Most K Distinct Elements
function maxKDistinct(nums: number[], k: number): number[] {
    nums.sort((a, b) => a - b);
    const ans: number[] = [];
    const n = nums.length;
    for (let i = n - 1; ~i; --i) {
        if (i + 1 < n && nums[i] === nums[i + 1]) {
            continue;
        }
        ans.push(nums[i]);
        if (--k === 0) {
            break;
        }
    }
    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.