Check if Matrix Is X-Matrix

Check if Matrix Is X-Matrix - Problem

Array Matrix Easy

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.

Input & Output

Example 1 — Valid X-Matrix

$ Input: grid = [[2,0,0,1],[0,3,1,0],[0,5,2,0],[4,0,0,2]]

› Output: true

💡 Note: All diagonal elements (2,3,1,2,4,1,2,2) are non-zero, and all non-diagonal elements are 0. This forms a valid X-Matrix pattern.

Example 2 — Invalid X-Matrix

$ Input: grid = [[5,7,0],[0,3,1],[0,5,0]]

› Output: false

💡 Note: Element at position (0,1) has value 7 which is non-zero, but it's not on any diagonal. Non-diagonal elements must be 0 in an X-Matrix.

Example 3 — Zero on Diagonal

$ Input: grid = [[5,0,5],[0,0,1],[5,1,5]]

› Output: false

💡 Note: The center element at (1,1) is 0, but it's on the main diagonal. Diagonal elements must be non-zero in an X-Matrix.

Constraints

n == grid.length == grid[i].length
3 ≤ n ≤ 100
-10⁵ ≤ grid[i][j] ≤ 10⁵

Visualization

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Asked in

a Amazon 15 M Microsoft 8

The key insight is identifying diagonal positions: main diagonal where i==j and anti-diagonal where i+j==n-1. Check that diagonal cells are non-zero and non-diagonal cells are zero. Best approach uses single pass with early return. Time: O(n²), Space: O(1)

Common Approaches

✓ Brute Force - Check Every Cell

⏱️ Time: O(n²) Space: O(1)

Check each cell to see if it's on a diagonal (should be non-zero) or not on a diagonal (should be zero). A cell at (i,j) is on the main diagonal if i==j, and on the anti-diagonal if i+j==n-1.

Single Pass with Early Return

⏱️ Time: O(n²) Space: O(1)

Same logic as brute force but optimized to return false as soon as we find a violation. This provides better average case performance when the matrix is not an X-Matrix.

Brute Force - Check Every Cell — Algorithm Steps

Iterate through all cells in the matrix
For each cell, determine if it's on a diagonal
Check if diagonal cells are non-zero and non-diagonal cells are zero

Visualization

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Step-by-Step Walkthrough

Identify Diagonals

Main diagonal (i==j) and anti-diagonal (i+j==n-1)

Check Each Cell

Diagonal cells must be non-zero, others must be zero

Result

Return true if all cells follow the rules

Code -

solution.c — C

#include <stdio.h>
#include <stdlib.h>
#include <string.h>

int solution(char* objType, char* classFunction) {
    if (strcmp(objType, "null") == 0 || classFunction == NULL || strcmp(classFunction, "null") == 0) {
        return 0;
    }
    
    // JavaScript-style type checking
    if (strcmp(objType, "object") == 0 && strcmp(classFunction, "Object") == 0) {
        return 1;
    }
    
    if (strcmp(objType, "number") == 0 && strcmp(classFunction, "Number") == 0) {
        return 1;
    }
    
    if (strcmp(objType, "string") == 0 && strcmp(classFunction, "String") == 0) {
        return 1;
    }
    
    return 0;
}

int main() {
    char objStr[1000], classStr[1000];
    
    fgets(objStr, sizeof(objStr), stdin);
    objStr[strcspn(objStr, "\n")] = 0;
    
    fgets(classStr, sizeof(classStr), stdin);
    classStr[strcspn(classStr, "\n")] = 0;
    
    char objType[100] = "null";
    
    // Determine object type
    if (strcmp(objStr, "5") == 0) {
        strcpy(objType, "number");
    } else if (strcmp(objStr, "{}") == 0) {
        strcpy(objType, "object");
    } else if (strcmp(objStr, "null") != 0) {
        strcpy(objType, "string");
    }
    
    int result = solution(objType, classStr);
    printf(result ? "true\n" : "false\n");
    
    return 0;
}

Time & Space Complexity

Time Complexity

⏱️

O(n²)

We iterate through all n×n cells in the matrix once

⚠ Quadratic Growth

Space Complexity

O(1)

Only using a few variables, no extra data structures

✓ Linear Space

18.5K Views

Medium Frequency

~15 min Avg. Time

542 Likes

Ln 1, Col 1

Smart Actions

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Input & Output

Constraints

Visualization

Related Problems

Common Approaches

Brute Force - Check Every Cell — Algorithm Steps

Visualization

Code -

Time & Space Complexity

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