NUHMAN.com

Introduction

LeetCode 54: Spiral Matrix is a neat array traversal problem where the goal is to return all elements of a 2D matrix in spiral order. This kind of problem is commonly asked in coding interviews to test understanding of boundaries, matrix dimensions, and directional iteration.

Problem Statement

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

Example:

go

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

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

Constraints

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

Approach: Layer-by-Layer Traversal

To move in a spiral, we simulate the movement by using four pointers:

• top – starting row
• bottom – ending row
• left – starting column
• right – ending column

We loop until top > bottom or left > right, and traverse in this order:

1. Left to Right (Top Row)
2. Top to Bottom (Right Column)
3. Right to Left (Bottom Row) – if still in bounds
4. Bottom to Top (Left Column) – if still in bounds

Go Implementation

go

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

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

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

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

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

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

    return result
}

Example Walkthrough

Given:

go

[
  [1, 2, 3],
  [4, 5, 6],
  [7, 8, 9]
]

Step-by-step result:

• → 1 2 3
• ↓ 6 9
• ← 8 7
• ↑ 4
• → 5

Final output: [1 2 3 6 9 8 7 4 5]

Time and Space Complexity

MetricComplexityTime ComplexityO(m × n)Space ComplexityO(1) (excluding output)

Edge Cases

• Single row or single column: Algorithm still works.
• Empty matrix: Return empty array.
• Non-square matrix: Handled due to flexible bounds.

Key Takeaways

• Use four pointers (top, bottom, left, right) to track the layer boundaries.
• Loop while top <= bottom and left <= right.
• Elegant simulation technique without needing visited flags or direction arrays.