algorithm notes: leetcode#766 toeplitz matrix

Problem

A matrix is Toeplitz if every diagonal from top-left to bottom-right has the same element.

Now given an M x N matrix, return True if and only if the matrix is Toeplitz.

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.

Note:

1. matrix will be a 2D array of integers.
2. matrix will have a number of rows and columns in range [1, 20].
3. matrix[i][j] will be integers in range [0, 99].

Solution

Analysis

Iterate each element in the matrix and check whether it equals to its top-left neighbor. Because there is no top-left neighbor for elements in the first row and first column, just traverse elements from the second row and second column.

Python implementation

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class :
    def isToeplitzMatrix(self, matrix):
        """
        :type matrix: List[List[int]]
        :rtype: bool
        """
        for i in range(1, len(matrix)):
            for j in range(1, len(matrix[0])):
                if matrix[i][j] != matrix[i-1][j-1]:
                    return False
        return True

Java implementation

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

Time complexity

O(MN).

Space complexity

O(1).

Link

766. Toeplitz Matrix
(中文版) 算法笔记: 力扣#766 托普利茨矩阵