How do you solve Largest Local Values in a Matrix on LeetCode?

Largest Local Values in a Matrix (LeetCode #2373) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(n²) time complexity and O(1) space complexity. Key insight: Understanding how to efficiently navigate through a matrix is crucial.

What is the brute force solution for Largest Local Values in a Matrix?

The brute force approach for Largest Local Values in a Matrix is Brute Force, with O(n²) time complexity and O(1) space complexity. The brute force approach involves checking every possible 3x3 submatrix in the given grid and finding the maximum value within each of those submatrices. This is straightforward but can be inefficient for larger matrices. The optimal Optimal Solution approach improves this to O(n²).

What algorithm or data structure is used to solve Largest Local Values in a Matrix?

Largest Local Values in a Matrix uses the following concepts: Array, Matrix. The recommended patterns to study are: Sliding Window, Matrix Traversal.

What are the interview tips for Largest Local Values in a Matrix?

Always clarify the problem and constraints before jumping into coding. Think about edge cases, especially with matrix boundaries. Explain your thought process while coding to showcase your understanding.

What are common mistakes when solving Largest Local Values in a Matrix?

Failing to account for the boundaries of the matrix when accessing elements. Not initializing the output matrix correctly.

#2373

Largest Local Values in a Matrix

Easy

ArrayMatrixSliding WindowMatrix Traversal

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

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

💡

Intuition

Time O(n²)Space O(1)

While the brute force approach is simple, we can optimize it by using a sliding window technique to keep track of the maximum values more efficiently. This reduces the number of redundant calculations.

⚙️

Algorithm

3 steps

1Step 1: Create a new matrix `maxLocal` of size (n-2) x (n-2).
2Step 2: Loop through the grid starting from index 1 to n-2 for both rows and columns.
3Step 3: For each position, calculate the maximum of the 3x3 submatrix using a sliding window approach, updating the maximum as we move through the grid.

solution.py13 lines

1# Full working Python code
2
3def largestLocal(grid):
4    n = len(grid)
5    maxLocal = [[0] * (n - 2) for _ in range(n - 2)]
6    for i in range(1, n - 1):
7        for j in range(1, n - 1):
8            maxLocal[i - 1][j - 1] = max(
9                grid[i - 1][j - 1], grid[i - 1][j], grid[i - 1][j + 1],
10                grid[i][j - 1], grid[i][j], grid[i][j + 1],
11                grid[i + 1][j - 1], grid[i + 1][j], grid[i + 1][j + 1]
12            )
13    return maxLocal

ℹ

Complexity note: The time complexity remains O(n²) as we still iterate through the grid, but we reduce the number of comparisons by directly accessing the relevant elements instead of looping through them. The space complexity is O(1) since we only use the output matrix.

1Understanding how to efficiently navigate through a matrix is crucial.
2Recognizing patterns in submatrices can lead to optimized solutions.

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