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2319. Check if Matrix Is X-Matrix

Description

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

Solutions

  • 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 {
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;
    }
};

  • 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


  • 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
    }
}

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