LeetCode #3462 — MEDIUM

Maximum Sum With at Most K Elements

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

The Problem

Problem Statement

You are given a 2D integer matrix grid of size n x m, an integer array limits of length n, and an integer k. The task is to find the maximum sum of at most k elements from the matrix grid such that:

The number of elements taken from the i^th row of grid does not exceed limits[i].

Return the maximum sum.

Example 1:

Input: grid = [[1,2],[3,4]], limits = [1,2], k = 2

Output: 7

Explanation:

From the second row, we can take at most 2 elements. The elements taken are 4 and 3.
The maximum possible sum of at most 2 selected elements is 4 + 3 = 7.

Example 2:

Input: grid = [[5,3,7],[8,2,6]], limits = [2,2], k = 3

Output: 21

Explanation:

From the first row, we can take at most 2 elements. The element taken is 7.
From the second row, we can take at most 2 elements. The elements taken are 8 and 6.
The maximum possible sum of at most 3 selected elements is 7 + 8 + 6 = 21.

Constraints:

n == grid.length == limits.length
m == grid[i].length
1 <= n, m <= 500
0 <= grid[i][j] <= 10⁵
0 <= limits[i] <= m
0 <= k <= min(n * m, sum(limits))

Step 01

Brute Force Baseline

Problem summary: You are given a 2D integer matrix grid of size n x m, an integer array limits of length n, and an integer k. The task is to find the maximum sum of at most k elements from the matrix grid such that: The number of elements taken from the ith row of grid does not exceed limits[i]. Return the maximum sum.

Baseline thinking

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

Pattern signal: Array · Greedy

Example 1

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

Example 2

[[5,3,7],[8,2,6]]
[2,2]
3

Step 02

Core Insight

What unlocks the optimal approach

Sort each row in descending order and extract the top <code>limits[i]</code> elements.
Use a max-heap to efficiently pick the largest <code>k</code> elements across all rows.

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 #3462: Maximum Sum With at Most K Elements
class Solution {
    public long maxSum(int[][] grid, int[] limits, int k) {
        PriorityQueue<Integer> pq = new PriorityQueue<>();
        int n = grid.length;
        for (int i = 0; i < n; ++i) {
            int[] nums = grid[i];
            int limit = limits[i];
            Arrays.sort(nums);
            for (int j = 0; j < limit; ++j) {
                pq.offer(nums[nums.length - j - 1]);
                if (pq.size() > k) {
                    pq.poll();
                }
            }
        }
        long ans = 0;
        for (int x : pq) {
            ans += x;
        }
        return ans;
    }
}

// Accepted solution for LeetCode #3462: Maximum Sum With at Most K Elements
type MinHeap []int

func (h MinHeap) Len() int           { return len(h) }
func (h MinHeap) Less(i, j int) bool { return h[i] < h[j] }
func (h MinHeap) Swap(i, j int)      { h[i], h[j] = h[j], h[i] }
func (h *MinHeap) Push(x interface{}) {
	*h = append(*h, x.(int))
}
func (h *MinHeap) Pop() interface{} {
	old := *h
	n := len(old)
	x := old[n-1]
	*h = old[0 : n-1]
	return x
}

func maxSum(grid [][]int, limits []int, k int) int64 {
	pq := &MinHeap{}
	heap.Init(pq)
	n := len(grid)

	for i := 0; i < n; i++ {
		nums := make([]int, len(grid[i]))
		copy(nums, grid[i])
		limit := limits[i]
		sort.Ints(nums)

		for j := 0; j < limit; j++ {
			heap.Push(pq, nums[len(nums)-j-1])
			if pq.Len() > k {
				heap.Pop(pq)
			}
		}
	}

	var ans int64 = 0
	for pq.Len() > 0 {
		ans += int64(heap.Pop(pq).(int))
	}

	return ans
}

# Accepted solution for LeetCode #3462: Maximum Sum With at Most K Elements
class Solution:
    def maxSum(self, grid: List[List[int]], limits: List[int], k: int) -> int:
        pq = []
        for nums, limit in zip(grid, limits):
            nums.sort()
            for _ in range(limit):
                heappush(pq, nums.pop())
                if len(pq) > k:
                    heappop(pq)
        return sum(pq)

// Accepted solution for LeetCode #3462: Maximum Sum With at Most K Elements
// Rust example auto-generated from java reference.
// Replace the signature and local types with the exact LeetCode harness for this problem.
impl Solution {
    pub fn rust_example() {
        // Port the logic from the reference block below.
    }
}

// Reference (java):
// // Accepted solution for LeetCode #3462: Maximum Sum With at Most K Elements
// class Solution {
//     public long maxSum(int[][] grid, int[] limits, int k) {
//         PriorityQueue<Integer> pq = new PriorityQueue<>();
//         int n = grid.length;
//         for (int i = 0; i < n; ++i) {
//             int[] nums = grid[i];
//             int limit = limits[i];
//             Arrays.sort(nums);
//             for (int j = 0; j < limit; ++j) {
//                 pq.offer(nums[nums.length - j - 1]);
//                 if (pq.size() > k) {
//                     pq.poll();
//                 }
//             }
//         }
//         long ans = 0;
//         for (int x : pq) {
//             ans += x;
//         }
//         return ans;
//     }
// }

// Accepted solution for LeetCode #3462: Maximum Sum With at Most K Elements
function maxSum(grid: number[][], limits: number[], k: number): number {
    const pq = new MinPriorityQueue<number>();
    const n = grid.length;
    for (let i = 0; i < n; i++) {
        const nums = grid[i];
        const limit = limits[i];
        nums.sort((a, b) => a - b);
        for (let j = 0; j < limit; j++) {
            pq.enqueue(nums[nums.length - j - 1]);
            if (pq.size() > k) {
                pq.dequeue();
            }
        }
    }
    let ans = 0;
    while (!pq.isEmpty()) {
        ans += pq.dequeue();
    }
    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 × m × (log m + log k)

Space

O(k)

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.

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.