Solving Leetcode Interviews in Seconds with AI: Check if Matrix Is X-Matrix
Introduction
In this blog post, we will explore how to solve the LeetCode problem "2319" using AI. LeetCode is a popular platform for preparing for coding interviews, and with the help of AI tools like Chatmagic, we can generate solutions quickly and efficiently - helping you pass the interviews and get the job offer without having to study for months.
Problem Statement
A square matrix is said to be an X-Matrix if both of the following conditions hold: All the elements in the diagonals of the matrix are non-zero. All other elements are 0. Given a 2D integer array grid of size n x n representing a square matrix, return true if grid is an X-Matrix. Otherwise, return false. Example 1: Input: grid = [[2,0,0,1],[0,3,1,0],[0,5,2,0],[4,0,0,2]] Output: true Explanation: Refer to the diagram above. An X-Matrix should have the green elements (diagonals) be non-zero and the red elements be 0. Thus, grid is an X-Matrix. Example 2: Input: grid = [[5,7,0],[0,3,1],[0,5,0]] Output: false Explanation: Refer to the diagram above. An X-Matrix should have the green elements (diagonals) be non-zero and the red elements be 0. Thus, grid is not an X-Matrix. Constraints: n == grid.length == grid[i].length 3 <= n <= 100 0 <= grid[i][j] <= 105
Explanation
Here's the solution to determine if a square matrix is an X-Matrix, following the requested format:
High-Level Approach:
- Iterate through each element of the grid.
- Check if diagonal elements are non-zero and non-diagonal elements are zero.
- Return
Falseimmediately if any condition is violated; otherwise, returnTrueafter checking all elements.
Complexity Analysis:
- Runtime Complexity: O(n^2), where n is the size of the grid.
- Storage Complexity: O(1).
Code
def isXMatrix(grid):
n = len(grid)
for i in range(n):
for j in range(n):
if i == j or i + j == n - 1:
if grid[i][j] == 0:
return False
else:
if grid[i][j] != 0:
return False
return True