766. Toeplitz Matrix

Description

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?

Solutions

Solution 1: Single Traversal

According to the problem description, the characteristic of a Toeplitz matrix is that each element is equal to the element in its upper left corner. Therefore, we only need to iterate through each element in the matrix and check if it is equal to the element in its upper left corner.

The time complexity is \(O(m \times n)\), where \(m\) and \(n\) are the number of rows and columns of the matrix, respectively. The space complexity is \(O(1)\).

Python3JavaC++GoTypeScriptRustJavaScript

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

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

class Solution {
public:
    bool isToeplitzMatrix(vector<vector<int>>& matrix) {
        int m = matrix.size(), n = matrix[0].size();
        for (int i = 1; i < m; ++i) {
            for (int j = 1; j < n; ++j) {
                if (matrix[i][j] != matrix[i - 1][j - 1]) {
                    return false;
                }
            }
        }
        return true;
    }
};

func isToeplitzMatrix(matrix [][]int) bool {
    m, n := len(matrix), len(matrix[0])
    for i := 1; i < m; i++ {
        for j := 1; j < n; j++ {
            if matrix[i][j] != matrix[i-1][j-1] {
                return false
            }
        }
    }
    return true
}

function isToeplitzMatrix(matrix: number[][]): boolean {
    const [m, n] = [matrix.length, matrix[0].length];
    for (let i = 1; i < m; ++i) {
        for (let j = 1; j < n; ++j) {
            if (matrix[i][j] !== matrix[i - 1][j - 1]) {
                return false;
            }
        }
    }
    return true;
}

impl Solution {
    pub fn is_toeplitz_matrix(matrix: Vec<Vec<i32>>) -> bool {
        let (m, n) = (matrix.len(), matrix[0].len());
        for i in 1..m {
            for j in 1..n {
                if matrix[i][j] != matrix[i - 1][j - 1] {
                    return false;
                }
            }
        }
        true
    }
}

/**
 * @param {number[][]} matrix
 * @return {boolean}
 */
var isToeplitzMatrix = function (matrix) {
    const [m, n] = [matrix.length, matrix[0].length];
    for (let i = 1; i < m; ++i) {
        for (let j = 1; j < n; ++j) {
            if (matrix[i][j] !== matrix[i - 1][j - 1]) {
                return false;
            }
        }
    }
    return true;
};

766. Toeplitz Matrix

Description

Solutions

Solution 1: Single Traversal

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