readme.md

54. Spiral Matrix

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]:
        res = []
        left, right = 0, len(matrix[0])
        top, bottom = 0, len(matrix)

        while left < right and top < bottom:
            # get every value in top row
            for i in range(left, right):
                res.append(matrix[top][i])
            top += 1

            # get every i in the right col
            for i in range(top, bottom):
                res.append(matrix[i][right - 1])
            right -= 1

            if not (left < right and top < bottom):
                break
            # get every i in the bottom row
            for i in range(right - 1, left - 1, -1):
                res.append(matrix[bottom - 1][i])
            bottom -= 1

            # get every i in the left col
            for i in range(bottom - 1, top - 1, -1):
                res.append(matrix[i][left])
            left += 1

        return res

Thoughts

Neetcodes explaination is the best resource to understand these kinds of problems.

Time Complexity

O(m*n)

Space Complexity

O(1)