Description
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].length3 <= n <= 1000 <= grid[i][j] <= 105
Solutions
Solution 1: 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(n2), where n is the number of rows or columns of the matrix. The space complexity is O(1).
PythonJavaC++GoTypeScriptRustC#C
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(code-box)
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;
}
}(code-box)
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;
}
};(code-box)
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
}(code-box)
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;
}(code-box)
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
}
}(code-box)
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;
}
}(code-box)
bool checkXMatrix(int** grid, int gridSize, int* gridColSize) {
for (int i = 0; i < gridSize; i++) {
for (int j = 0; j < gridSize; j++) {
if (i == j || i + j == gridSize - 1) {
if (grid[i][j] == 0) {
return false;
}
} else if (grid[i][j] != 0) {
return false;
}
}
}
return true;
}(code-box)
