Spiral Matrix

#leetcode #dsa #theabbie

Given an m x n matrix, return all elements of the matrix in spiral order.

Example 1:

Input: matrix = [[1,2,3],[4,5,6],[7,8,9]]
Output: [1,2,3,6,9,8,7,4,5]

Example 2:

Input: matrix = [[1,2,3,4],[5,6,7,8],[9,10,11,12]]
Output: [1,2,3,4,8,12,11,10,9,5,6,7]

Constraints:

m == matrix.length
n == matrix[i].length
1 <= m, n <= 10
-100 <= matrix[i][j] <= 100

SOLUTION:

class Solution:
    def spiralOrder(self, matrix: List[List[int]]) -> List[int]:
        m = len(matrix)
        n = len(matrix[0])
        op = []
        mark = "X"
        ctr = 0
        direction = 0
        x, y = 0, 0
        while ctr < m * n:
            if matrix[x][y] != mark:
                ctr += 1
                op.append(matrix[x][y])
            matrix[x][y] = mark
            switch = False
            if direction == 0:
                if (y + 1) < n and matrix[x][y + 1] != mark:
                    y += 1
                else:
                    switch = True
            elif direction == 1:
                if (x + 1) < m and matrix[x + 1][y] != mark:
                    x += 1
                else:
                    switch = True
            elif direction == 2:
                if (y - 1) >= 0 and matrix[x][y - 1] != mark:
                    y -= 1
                else:
                    switch = True
            else:
                if (x + 1) >= 0 and matrix[x - 1][y] != mark:
                    x -= 1
                else:
                    switch = True
            if switch:
                direction = (direction + 1) % 4
        return op