How do you solve Maximum Sum With at Most K Elements on LeetCode?

Maximum Sum With at Most K Elements (LeetCode #3462) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(n * m log m + k log k) time complexity and O(n) space complexity. Key insight: Sorting helps in quickly accessing the largest elements.

What is the time complexity of Maximum Sum With at Most K Elements?

The optimal solution for Maximum Sum With at Most K Elements runs in O(n * m log m + k log k) time and uses O(n) space. Sorting each row takes O(m log m) and extracting k elements from the heap takes O(k log k).

What is the brute force solution for Maximum Sum With at Most K Elements?

The brute force approach for Maximum Sum With at Most K Elements is Brute Force, with O(n²) time complexity and O(1) space complexity. We can generate all possible combinations of elements from the matrix while respecting the limits. This will allow us to find the maximum sum directly, but it's inefficient. The optimal Optimal Solution approach improves this to O(n * m log m + k log k).

What algorithm or data structure is used to solve Maximum Sum With at Most K Elements?

Maximum Sum With at Most K Elements uses the following concepts: Array, Greedy, Sorting, Heap (Priority Queue), Matrix. The recommended patterns to study are: Greedy, Heap.

What are the interview tips for Maximum Sum With at Most K Elements?

Explain your thought process clearly. Discuss edge cases and constraints. Practice coding under time constraints.

What are common mistakes when solving Maximum Sum With at Most K Elements?

Not respecting the limits while selecting elements. Forgetting to sort the rows before selection.

#3462

Maximum Sum With at Most K Elements

Medium

ArrayGreedySortingHeap (Priority Queue)MatrixGreedyHeap

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

	Brute Force	Optimal Solution★
Time	O(n²)	O(n * m log m + k log k)
Space	O(1)	O(n)

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Intuition

Time O(n * m log m + k log k)Space O(n)

By sorting each row and using a max-heap, we can efficiently select the largest elements while respecting the limits and the total count k.

⚙️

Algorithm

3 steps

1Step 1: Sort each row of the grid in descending order.
2Step 2: Use a max-heap to collect the largest elements up to the limits specified.
3Step 3: Extract the top k elements from the heap to compute the maximum sum.

solution.py9 lines

1import heapq
2
3def maxSum(grid, limits, k):
4    max_heap = []
5    for row, limit in zip(grid, limits):
6        row.sort(reverse=True)
7        for num in row[:limit]:
8            heapq.heappush(max_heap, num)
9    return sum(heapq.nlargest(k, max_heap))

ℹ

Complexity note: Sorting each row takes O(m log m) and extracting k elements from the heap takes O(k log k).

1Sorting helps in quickly accessing the largest elements.
2Using a max-heap allows efficient retrieval of the top k elements.

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