How do you solve Check if Matrix Is X-Matrix on LeetCode?

Check if Matrix Is X-Matrix (LeetCode #2319) 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: Diagonal elements must be non-zero while non-diagonal elements must be zero.

What is the brute force solution for Check if Matrix Is X-Matrix?

The brute force approach for Check if Matrix Is X-Matrix is Brute Force, with O(n²) time complexity and O(1) space complexity. The brute force approach checks each element of the matrix to see if it meets the criteria for being an X-Matrix. We will iterate through all elements and validate the diagonal and non-diagonal conditions. The optimal Optimal Solution approach improves this to O(n²).

What algorithm or data structure is used to solve Check if Matrix Is X-Matrix?

Check if Matrix Is X-Matrix uses the following concepts: Array, Matrix. The recommended patterns to study are: Array, Matrix.

What are the interview tips for Check if Matrix Is X-Matrix?

Always clarify the problem statement and constraints before jumping into coding. Think about edge cases, such as the smallest or largest possible matrices. Explain your thought process while coding; it shows your understanding and helps the interviewer follow along.

What are common mistakes when solving Check if Matrix Is X-Matrix?

Confusing diagonal indices with non-diagonal indices. Not checking all elements properly or missing edge cases.

#2319

Check if Matrix Is X-Matrix

Easy

ArrayMatrixArrayMatrix

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)

The optimal solution leverages the same checks but is structured to minimize unnecessary checks by directly focusing on the diagonal and non-diagonal elements in a single pass.

⚙️

Algorithm

4 steps

1Step 1: Loop through each element in the matrix using a single loop.
2Step 2: For each element at position (i, j), check if it is a diagonal element (i == j or i == n - 1 - j).
3Step 3: If it is a diagonal element, ensure it is non-zero; if it is not a diagonal element, ensure it is zero.
4Step 4: If any condition fails, return false. If all checks pass, return true.

solution.py11 lines

1# Full working Python code
2
3def checkXMatrix(grid):
4    n = len(grid)
5    for i in range(n):
6        if grid[i][i] == 0 or grid[i][n - 1 - i] == 0:
7            return False
8        for j in range(n):
9            if (j != i and j != n - 1 - i) and grid[i][j] != 0:
10                return False
11    return True

ℹ

Complexity note: The time complexity remains O(n²) since we still need to check each element of the matrix. The space complexity is O(1) as we are not using any additional data structures.

1Diagonal elements must be non-zero while non-diagonal elements must be zero.
2Understanding the structure of the matrix helps in efficiently checking the conditions.

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