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🟨 Spiral Matrix (#54)

🔗 Problem Link​

📋 Problem Statement​

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

💡 Examples​

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]

🔑 Key Insights & Approach​

Core Observation: Process layer by layer using four boundaries (top, right, bottom, left).

Why Layer by Layer?

  • O(m * n) time, O(1) extra space
  • Track and shrink boundaries
  • Process: right → down → left → up

Pattern: "Spiral Traversal" with boundary tracking.

🐍 Solution: Python​

Approach: Boundary Tracking​

Time Complexity: O(m * n) | Space Complexity: O(1)

from typing import List

class Solution:
    def spiralOrder(self, matrix: List[List[int]]) -> List[int]:
        result = []
        if not matrix:
            return result

        top, bottom = 0, len(matrix) - 1
        left, right = 0, len(matrix[0]) - 1

        while top <= bottom and left <= right:
            # Right
            for col in range(left, right + 1):
                result.append(matrix[top][col])
            top += 1

            # Down
            for row in range(top, bottom + 1):
                result.append(matrix[row][right])
            right -= 1

            # Left (if still rows)
            if top <= bottom:
                for col in range(right, left - 1, -1):
                    result.append(matrix[bottom][col])
                bottom -= 1

            # Up (if still cols)
            if left <= right:
                for row in range(bottom, top - 1, -1):
                    result.append(matrix[row][left])
                left += 1

        return result

🔵 Solution: Golang​

Approach: Boundary Tracking​

Time Complexity: O(m * n) | Space Complexity: O(1)

func spiralOrder(matrix [][]int) []int {
    if len(matrix) == 0 {
        return []int{}
    }

    result := []int{}
    top, bottom := 0, len(matrix)-1
    left, right := 0, len(matrix[0])-1

    for top <= bottom && left <= right {
        for col := left; col <= right; col++ {
            result = append(result, matrix[top][col])
        }
        top++

        for row := top; row <= bottom; row++ {
            result = append(result, matrix[row][right])
        }
        right--

        if top <= bottom {
            for col := right; col >= left; col-- {
                result = append(result, matrix[bottom][col])
            }
            bottom--
        }

        if left <= right {
            for row := bottom; row >= top; row-- {
                result = append(result, matrix[row][left])
            }
            left++
        }
    }

    return result
}

📚 References​