How do you solve Sum in a Matrix on LeetCode?

Sum in a Matrix (LeetCode #2679) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(n log n) time complexity and O(1) space complexity. Key insight: Sorting each row allows for efficient access to maximum values.

What is the brute force solution for Sum in a Matrix?

The brute force approach for Sum in a Matrix is Brute Force, with O(n²) time complexity and O(1) space complexity. In the brute-force approach, we will repeatedly find the maximum number from each row, remove it, and keep track of the highest number removed in each iteration. This method is straightforward but inefficient for larger matrices. The optimal Optimal Solution approach improves this to O(n log n).

What algorithm or data structure is used to solve Sum in a Matrix?

Sum in a Matrix uses the following concepts: Array, Sorting, Heap (Priority Queue), Matrix, Simulation. The recommended patterns to study are: Sorting, Greedy Algorithm.

What are the interview tips for Sum in a Matrix?

Clearly explain your thought process while coding. Consider edge cases and constraints upfront. Use comments to clarify complex logic.

What are common mistakes when solving Sum in a Matrix?

Not considering edge cases like empty rows. Failing to sort rows before accessing maximum values.

#2679

Sum in a Matrix

Medium

ArraySortingHeap (Priority Queue)MatrixSimulationSortingGreedy Algorithm

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

	Brute Force	Optimal Solution★
Time	O(n²)	O(n log n)
Space	O(1)	O(1)

💡

Intuition

Time O(n log n)Space O(1)

The optimal approach involves sorting each row in descending order first. This allows us to easily access the maximum elements in each iteration without needing to search for them, leading to a more efficient solution.

⚙️

Algorithm

4 steps

1Step 1: Sort each row of the matrix in descending order.
2Step 2: Initialize a score variable to 0.
3Step 3: For each column index, find the maximum element from the first elements of each row and add it to the score.
4Step 4: Move to the next column and repeat until all columns are processed.

solution.py10 lines

1# Full working Python code
2class Solution:
3    def getMatrixSum(self, nums):
4        for row in nums:
5            row.sort(reverse=True)
6        score = 0
7        for col in range(len(nums[0])):
8            max_value = max(row[col] for row in nums)
9            score += max_value
10        return score

ℹ

Complexity note: The time complexity is O(n log n) due to the sorting of each row, and the space complexity is O(1) since we are not using any additional data structures that grow with input size.

1Sorting each row allows for efficient access to maximum values.
2Removing elements can be avoided by simply tracking column indices.

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