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Check if Matrix Is X-Matrix - Solution & Explanation

EasyArrayMatrix16 min readAsked at: Amazon, Google
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Problem Statement

A square matrix is said to be an X-Matrix if both of the following conditions hold:

  1. All the elements in the diagonals of the matrix are non-zero.
  2. 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

Approach Overview

Problem Overview: You are given an n x n matrix. The matrix is considered an X-Matrix if every element on the primary diagonal (i == j) and secondary diagonal (i + j == n - 1) is non‑zero, and every other element is exactly zero. The task is to verify these conditions for all cells in the matrix.

Approach 1: Iterative Matrix Traversal (O(n^2) time, O(1) space)

Traverse the entire matrix using two nested loops. For each cell (i, j), determine whether it lies on one of the two diagonals. If i == j or i + j == n - 1, the value must be non‑zero. If the cell is not on either diagonal, the value must be zero. The moment a violation appears, return false. This approach works because every cell is validated exactly once, and the diagonal condition is computed with constant-time checks.

This solution uses straightforward iteration over a matrix and simple conditional checks. Since no extra data structures are required, the space usage remains constant. The runtime is O(n^2) because every element in the matrix must be inspected.

Approach 2: Separate Diagonal and Non-Diagonal Checks (O(n^2) time, O(1) space)

Another way to structure the validation is to handle diagonal and non-diagonal cells separately. First iterate through indices i and check the primary diagonal grid[i][i] and secondary diagonal grid[i][n - i - 1]. Both values must be non-zero. After confirming the diagonals, run a second pass through the matrix to verify that every remaining cell is zero.

This separation makes the logic slightly clearer in interviews because the X-structure is validated in two explicit phases. It still relies on basic array traversal and index math. The overall time complexity remains O(n^2) since the matrix is scanned, and the space complexity is O(1).

Recommended for interviews: The single-pass iterative traversal is usually preferred. It validates diagonal and non-diagonal conditions in one loop, keeps the implementation concise, and avoids redundant scans. The two-phase version is still acceptable and sometimes easier to reason about, but interviewers typically expect the clean single-pass matrix check once you recognize the diagonal conditions.

Approach 1: Iterative Approach

This approach involves iterating over every element of the matrix and evaluating their positions. For elements on the main or anti-diagonal, ensure they are non-zero. For other elements, check if they are zero.

This C function iterates across the entire grid. For each element, it checks if the element is on the main diagonal (i == j) or the anti-diagonal (i + j == gridSize - 1). If so, it ensures the element is non-zero. Elements not on these diagonals are checked to be zero. If the conditions break at any point, it returns false.

Code

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Complexity

Time Complexity: O(n^2) where n is the number of rows (or columns) in the matrix. Space Complexity: O(1), only constant space is used.

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Approach 2: Separate Diagonal and Non-Diagonal Check

This approach explicitly checks nodes based on their positions. First, identify and validate values on the diagonals, then on the fly, verify if non-diagonal elements maintain zero-value constraints.

This optimized C method checks elements by evaluating if they reside on any diagonals according to its conditions then checks if invalid conditions are met early. This enables early termination of the check if any rule is breached.

Code

CC++JavaPythonC#JavaScript

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C++

Java

Python

C#

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Complexity

Time Complexity: O(n^2), efficiently checks all node conditions with a potential early stop. Space Complexity: O(1), only uses base iterative space without extra structures.

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Approach 3: Simulation

We can directly traverse the matrix and check if each element satisfies the conditions of an X matrix. If any element does not satisfy the conditions, return false immediately. If all elements satisfy the conditions after traversal, return true.

The time complexity is O(n^2), where n is the number of rows or columns of the matrix. The space complexity is O(1).

Code

PythonJavaC++GoTypeScriptRustC#C

Python

Java

C++

Go

TypeScript

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Try this approach in the editor →

Complexity Comparison

ApproachComplexity
Iterative Approach

Time Complexity: O(n^2) where n is the number of rows (or columns) in the matrix. Space Complexity: O(1), only constant space is used.

Separate Diagonal and Non-Diagonal Check

Time Complexity: O(n^2), efficiently checks all node conditions with a potential early stop. Space Complexity: O(1), only uses base iterative space without extra structures.

Simulation—

Detailed Complexity Analysis

ApproachTimeSpaceWhen to Use
Iterative Matrix TraversalO(n^2)O(1)Best general solution. Validates all constraints in a single pass.
Separate Diagonal and Non-Diagonal CheckO(n^2)O(1)Useful when you want clearer logic by validating diagonals first, then remaining cells.

Video Solution

Weekly Contest 299 | Leetcode 2319 Check if Matrix Is X-Matrix 🔥🔥 | Matrix | Easy Peasy • Coding Decoded • 903 views views

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Frequently Asked Questions

Is Check if Matrix Is X-Matrix easy or hard?
Check if Matrix Is X-Matrix is classified as an Easy problem. It focuses on basic matrix traversal and index conditions rather than complex algorithms. The main challenge is correctly identifying the two diagonals and applying the validation rules.
Check if Matrix Is X-Matrix Python/Java solution
In both Python and Java, the implementation uses nested loops over the matrix indices. The program checks diagonal conditions with i == j or i + j == n - 1 and validates whether the value should be non-zero or zero accordingly. The logic is identical across languages and runs in O(n^2) time.
How to solve Check if Matrix Is X-Matrix in O(n^2)?
Iterate through the matrix using two nested loops. For each cell (i, j), check if it lies on a diagonal using the conditions i == j or i + j == n - 1. If it is a diagonal cell, ensure the value is non-zero; otherwise ensure the value is zero. If any condition fails, return false; otherwise return true after the full traversal.
What is the best approach for Check if Matrix Is X-Matrix?
The most efficient approach is a single-pass matrix traversal. Iterate through every cell and check whether it belongs to the primary diagonal (i == j) or secondary diagonal (i + j == n - 1). Diagonal cells must be non-zero, and all other cells must be zero. This solution runs in O(n^2) time with O(1) extra space.
Is Check if Matrix Is X-Matrix asked at Google/Amazon/Meta?
Matrix validation and traversal problems appear frequently in coding interviews at companies like Amazon, Google, and Meta. While this exact problem may vary, recognizing diagonal conditions and performing structured matrix checks is a common interview pattern.
What data structure is used in Check if Matrix Is X-Matrix?
The problem uses a two-dimensional array (matrix). The algorithm relies on index arithmetic to identify the primary and secondary diagonals rather than additional data structures.
What is the time complexity of Check if Matrix Is X-Matrix?
The time complexity is O(n^2) because every element in the n x n matrix must be checked at least once. Each validation involves constant-time comparisons to determine whether the cell lies on a diagonal. The space complexity remains O(1) since no extra data structures are used.

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