How do you solve Max Sum of Rectangle No Larger Than K on LeetCode?

Max Sum of Rectangle No Larger Than K (LeetCode #363) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(m * n * log(n)) time complexity and O(n) space complexity. Key insight: Understanding how to efficiently calculate subarray sums is crucial.

What is the brute force solution for Max Sum of Rectangle No Larger Than K?

The brute force approach for Max Sum of Rectangle No Larger Than K is Brute Force, with O(m^2 * n^2) time complexity and O(1) space complexity. The brute-force approach involves checking every possible rectangle in the matrix and calculating its sum. This is straightforward but inefficient, especially for larger matrices. The optimal Optimal Solution approach improves this to O(m * n * log(n)).

What algorithm or data structure is used to solve Max Sum of Rectangle No Larger Than K?

Max Sum of Rectangle No Larger Than K uses the following concepts: Array, Binary Search, Matrix, Prefix Sum, Ordered Set. The recommended patterns to study are: Hash Map, Array.

What are the interview tips for Max Sum of Rectangle No Larger Than K?

Always start with a brute-force solution to understand the problem better. Think about how you can reduce the search space or use data structures to improve efficiency. Practice explaining your thought process clearly, as communication is key in interviews.

What are common mistakes when solving Max Sum of Rectangle No Larger Than K?

Not considering all possible rectangles, leading to missed solutions. Failing to optimize the search for maximum sums, resulting in timeouts.

#363

Max Sum of Rectangle No Larger Than K

Hard

ArrayBinary SearchMatrixPrefix SumOrdered SetHash MapArray

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

	Brute Force	Optimal Solution★
Time	O(m^2 * n^2)	O(m * n * log(n))
Space	O(1)	O(n)

💡

Intuition

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

The optimal solution uses a combination of prefix sums and a data structure to efficiently find the maximum sum rectangle that is less than or equal to k. This reduces the time complexity significantly.

⚙️

Algorithm

3 steps

1Step 1: Iterate over all pairs of rows to define the top and bottom of the rectangle.
2Step 2: For each pair of rows, create an array that represents the sum of columns between these two rows.
3Step 3: Use a sorted data structure to find the maximum sum of subarrays that is less than or equal to k using binary search.

solution.py19 lines

1import bisect
2
3def maxSumSubmatrix(matrix, k):
4    m, n = len(matrix), len(matrix[0])
5    max_sum = float('-inf')
6    for r1 in range(m):
7        sums = [0] * n
8        for r2 in range(r1, m):
9            for c in range(n):
10                sums[c] += matrix[r2][c]
11            sorted_sums = [0]
12            curr_sum = 0
13            for sum_val in sums:
14                curr_sum += sum_val
15                idx = bisect.bisect_left(sorted_sums, curr_sum - k)
16                if idx < len(sorted_sums):
17                    max_sum = max(max_sum, curr_sum - sorted_sums[idx])
18                bisect.insort(sorted_sums, curr_sum)
19    return max_sum

ℹ

Complexity note: The time complexity is O(m * n * log(n)) because we iterate through pairs of rows and use a sorted set to find the maximum sum efficiently. The space complexity is O(n) due to the array used for column sums.

1Understanding how to efficiently calculate subarray sums is crucial.
2Using sorted data structures can help in quickly finding the maximum sum that meets constraints.

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