[LeetCode] 766. Toeplitz Matrix_Easy tag: array

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:

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:

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?

Ideas:

1. T: O(m * n);     S: O(m * n)    将r - c 作为key来存入到dictionary里面， 如果r - c 在dictionary，再比较是否相等

class Solution:
    def isToeplitzMatrix(self, matrix: List[List[int]]) -> bool:
        group = dict()
        for r, row in enumerate(matrix):
            for c, val in enumerate(row):
                if r - c not in group:
                    group[r - c] = val
                else:
                    if val != group[r - c]:
                        return False
        return True

 

2. T: O(m * n)       S: O(1), 如果r > 0 and c > 0 , 将它与左上角的元素进行比较

class Solution:
    def isToeplitzMatrix(self, matrix: List[List[int]]) -> bool:
        for r, row in enumerate(matrix):
            for c, val in enumerate(row):
                if r == 0 or c == 0 or matrix[r][c] == matrix[r - 1][c - 1]:
                    continue
                else:
                    return False
        return True