[Solved] Check if Matrix Is X-Matrix LeetCode Contest Problem

Check if Matrix Is X-Matrix: 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

Solution

Time Complexity: O(m + n)
Space Complexity: O(1)

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

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:
    def checkXMatrix(self, grid: List[List[int]]) -> bool:
        globalSum, diagonalSum = 0, 0
        for i in range(len(grid)):
            for j in range(len(grid[0])):
                globalSum += grid[i][j]
            
            #principle diagonal 
            if grid[i][i] == 0 or grid[i][len(grid)-1-i] == 0:
                return False
        
            diagonalSum += grid[i][i] + grid[i][len(grid)-1-i]
        
        if len(grid) % 2 != 0:
            mid = len(grid) // 2
            diagonalSum -= grid[mid][mid]
        
        return True if diagonalSum == globalSum else False

Happy Learning – If you require any further information, feel free to contact me.