LeetCode 766: Toeplitz Matrix Solution

Problem Explanation

Explanation:

To solve this problem, we can iterate through the matrix and check if each element is equal to the element on its top-left diagonal. If we find any mismatch, we return false. If we reach the end of the matrix without finding any mismatches, we return true.

We start by checking each element in the first row and first column of the matrix.
Then, we move diagonally to check the elements on the top-left diagonal for each element in the matrix.
If we find any mismatch, we return false.
If we reach the end of the matrix without any mismatches, we return true.

Time complexity: O(m*n) where m is the number of rows and n is the number of columns in the matrix. Space complexity: O(1) as we are using constant extra space.

Solution Code

class Solution {
    public boolean isToeplitzMatrix(int[][] matrix) {
        for (int i = 0; i < matrix.length; i++) {
            for (int j = 0; j < matrix[0].length; j++) {
                if (i > 0 && j > 0 && matrix[i][j] != matrix[i - 1][j - 1]) {
                    return false;
                }
            }
        }
        return true;
    }
}

Frequently Asked Questions

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What is the time complexity of LeetCode 766 (Toeplitz Matrix)?

The time complexity for LeetCode 766 (Toeplitz Matrix) varies by solution. Check the detailed explanation section for specific complexities in Java, C++, and Python implementations.

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766. Toeplitz Matrix