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Formatted question description: https://leetcode.ca/all/2319.html

# 2319. Check if Matrix Is X-Matrix

• Difficulty: Easy.
• Related Topics: Array, Matrix.
• Similar Questions: Matrix Diagonal Sum.

## Problem

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

## Solution (Java, C++, Python)

• class Solution {
public boolean checkXMatrix(int[][] grid) {
for (int i = 0; i < grid.length; i++) {
for (int j = 0; j < grid[0].length; j++) {
if (i == j || i + j == grid.length - 1) {
if (grid[i][j] == 0) {
return false;
}
} else {
if (grid[i][j] != 0) {
return false;
}
}
}
}
return true;
}
}

############

class Solution {
public boolean checkXMatrix(int[][] grid) {
int n = grid.length;
for (int i = 0; i < n; ++i) {
for (int j = 0; j < n; ++j) {
if (i == j || i + j == n - 1) {
if (grid[i][j] == 0) {
return false;
}
} else if (grid[i][j] != 0) {
return false;
}
}
}
return true;
}
}

• class Solution:
def checkXMatrix(self, grid: List[List[int]]) -> bool:
for i, row in enumerate(grid):
for j, v in enumerate(row):
if i == j or i + j == len(grid) - 1:
if v == 0:
return False
elif v:
return False
return True

############

# 2319. Check if Matrix Is X-Matrix
# https://leetcode.com/problems/check-if-matrix-is-x-matrix

class Solution:
def checkXMatrix(self, grid: List[List[int]]) -> bool:
rows, cols = len(grid), len(grid[0])

for i in range(rows):
for j in range(cols):
if i == j or i + j == cols - 1:
if grid[i][j] == 0:
return False
elif grid[i][j] != 0:
return False

return True


• class Solution {
public:
bool checkXMatrix(vector<vector<int>>& grid) {
int n = grid.size();
for (int i = 0; i < n; ++i) {
for (int j = 0; j < n; ++j) {
if (i == j || i + j == n - 1) {
if (!grid[i][j]) {
return false;
}
} else if (grid[i][j]) {
return false;
}
}
}
return true;
}
};

• func checkXMatrix(grid [][]int) bool {
for i, row := range grid {
for j, v := range row {
if i == j || i+j == len(row)-1 {
if v == 0 {
return false
}
} else if v != 0 {
return false
}
}
}
return true
}

• function checkXMatrix(grid: number[][]): boolean {
const n = grid.length;
for (let i = 0; i < n; ++i) {
for (let j = 0; j < n; ++j) {
if (i == j || i + j == n - 1) {
if (!grid[i][j]) {
return false;
}
} else if (grid[i][j]) {
return false;
}
}
}
return true;
}


• public class Solution {
public bool CheckXMatrix(int[][] grid) {
int n = grid.Length;
for (int i = 0; i < n; ++i) {
for (int j = 0; j < n; ++j) {
if (i == j || i + j == n - 1) {
if (grid[i][j] == 0) {
return false;
}
} else if (grid[i][j] != 0) {
return false;
}
}
}
return true;
}
}

• impl Solution {
pub fn check_x_matrix(grid: Vec<Vec<i32>>) -> bool {
let n = grid.len();
for i in 0..n {
for j in 0..n {
if i == j || i + j == n - 1 {
if grid[i][j] == 0 {
return false;
}
} else if grid[i][j] != 0 {
return false;
}
}
}
true
}
}



Explain:

nope.

Complexity:

• Time complexity : O(n).
• Space complexity : O(n).