g0701_0800.s0766_toeplitz_matrix.Solution Maven / Gradle / Ivy

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package g0701_0800.s0766_toeplitz_matrix;

// #Easy #Array #Matrix #2022_03_26_Time_1_ms_(83.45%)_Space_45.8_MB_(49.31%)

/**
 * 766 - Toeplitz Matrix\.
 *
 * Easy
 *
 * Given an `m x n` `matrix`, return _`true` if the matrix is Toeplitz. Otherwise, return `false`._
 *
 * A matrix is **Toeplitz** if every diagonal from top-left to bottom-right has the same elements.
 *
 * **Example 1:**
 *
 * ![](https://assets.leetcode.com/uploads/2020/11/04/ex1.jpg)
 *
 * **Input:** matrix = \[\[1,2,3,4],[5,1,2,3],[9,5,1,2]]
 *
 * **Output:** true
 *
 * **Explanation:** In the above grid, the diagonals are: "[9]", "[5, 5]", "[1, 1, 1]", "[2, 2, 2]", "[3, 3]", "[4]". In each diagonal all elements are the same, so the answer is True.
 *
 * **Example 2:**
 *
 * ![](https://assets.leetcode.com/uploads/2020/11/04/ex2.jpg)
 *
 * **Input:** matrix = \[\[1,2],[2,2]]
 *
 * **Output:** false
 *
 * **Explanation:** The diagonal "[1, 2]" has different elements.
 *
 * **Constraints:**
 *
 * *   `m == matrix.length`
 * *   `n == matrix[i].length`
 * *   `1 <= m, n <= 20`
 * *   `0 <= matrix[i][j] <= 99`
 *
 * **Follow up:**
 *
 * *   What if the `matrix` is stored on disk, and the memory is limited such that you can only load at most one row of the matrix into the memory at once?
 * *   What if the `matrix` is so large that you can only load up a partial row into the memory at once?
**/
public class Solution {
    public boolean isToeplitzMatrix(int[][] matrix) {
        int m = matrix.length;
        int n = matrix[0].length;
        int i = 0;
        int j = 0;
        int sameVal = matrix[i][j];
        while (++i < m && ++j < n) {
            if (matrix[i][j] != sameVal) {
                return false;
            }
        }

        for (i = 1, j = 0; i < m; i++) {
            int tmpI = i;
            int tmpJ = j;
            sameVal = matrix[i][j];
            while (++tmpI < m && ++tmpJ < n) {
                if (matrix[tmpI][tmpJ] != sameVal) {
                    return false;
                }
            }
        }
        for (i = 0, j = 1; j < n; j++) {
            int tmpJ = j;
            int tmpI = i;
            sameVal = matrix[tmpI][tmpJ];
            while (++tmpI < m && ++tmpJ < n) {
                if (matrix[tmpI][tmpJ] != sameVal) {
                    return false;
                }
            }
        }
        return true;
    }
}