LeetCode 59: Spiral Matrix II Solution in Python Explained

Problem Statement

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LeetCode 59, Spiral Matrix II, is a medium-level problem where you’re given an integer n. Your task is to generate an (n \times n) matrix filled with elements from 1 to (n^2) in spiral order, moving clockwise from the top-left corner inward. Return the resulting matrix as a 2D list of integers.

Constraints

  • 1 <= n <= 20: Matrix size is between 1 and 20.

Example

Input: n = 3
Output: [[1,2,3],[8,9,4],[7,6,5]]
Explanation: Spiral order:
1→2→3
↓     ↓
8  9→4
↑     ↓
7←6←5

Input: n = 1
Output: [[1]]
Explanation: 1x1 matrix with value 1.

Input: n = 4
Output: [[1,2,3,4],[12,13,14,5],[11,16,15,6],[10,9,8,7]]
Explanation: 4x4 spiral matrix.

Understanding the Problem

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How do you solve LeetCode 59: Spiral Matrix II in Python? For n = 3, create a 3x3 matrix filled with 1 to 9 in spiral order: [[1,2,3],[8,9,4],[7,6,5]]. We need to place numbers sequentially while moving right, down, left, and up, spiraling inward. We’ll explore two approaches: a Layer-by-Layer Generation Solution (optimal and primary) and an Alternative with Direction Array (intuitive but more complex). The layer-by-layer method runs in O(n^2) time with O(1) extra space, building the matrix directly.

Solution 1: Layer-by-Layer Generation Approach (Primary)

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Explanation

Generate the matrix layer by layer, using four boundaries (top, bottom, left, right) to define each spiral layer. Fill numbers in the order right, down, left, up, incrementing a counter, and shrink the boundaries inward until all (n^2) elements are placed.

Here’s a flow for (n = 3):

Layer 1: 1→2→3→4→5→6→7
Layer 2: 8→9
Result: [[1,2,3],[8,9,4],[7,6,5]]
  1. Initialize Matrix and Boundaries.
  • Create \(n \times n\) matrix, set top = 0, bottom = n-1, left = 0, right = n-1.
  1. Fill Layers.
  • Traverse right, down, left, up, increment value.
  1. Shrink Boundaries.
  • Move inward after each layer.
  1. Return Matrix.

Step-by-Step Example

Example 1: n = 3

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

  • Goal: Generate 3x3 spiral matrix.
  • Result for Reference: [[1,2,3],[8,9,4],[7,6,5]].
  • Start: n = 3, matrix = [[0,0,0],[0,0,0],[0,0,0]], val = 1, top = 0, bottom = 2, left = 0, right = 2.
  • Step 1: First layer.
    • Right: (0,0)=1, (0,1)=2, (0,2)=3, val = 4, top = 1.
    • Down: (1,2)=4, (2,2)=5, val = 6, right = 1.
    • Left: (2,1)=6, (2,0)=7, val = 8, bottom = 1.
    • Up: (1,0)=8, val = 9, left = 1.
    • Matrix = [[1,2,3],[8,0,4],[7,6,5]].
  • Step 2: Second layer.
    • top = 1, bottom = 1, left = 1, right = 1.
    • Right: (1,1)=9, val = 10, top = 2.
    • Matrix = [[1,2,3],[8,9,4],[7,6,5]].
  • Step 3: top > bottom, stop.
  • Finish: Return [[1,2,3],[8,9,4],[7,6,5]].

Example 2: n = 1

We need [[1]].

  • Start: n = 1, matrix = [[0]], val = 1.
  • Step 1: top = 0, bottom = 0, left = 0, right = 0.
    • Right: (0,0)=1, top = 1.
  • Finish: Return [[1]].

How the Code Works (Layer-by-Layer)

Here’s the Python code with line-by-line explanation:

def generateMatrix(n: int) -> list[list[int]]:
    # Line 1: Initialize matrix and variables
    matrix = [[0] * n for _ in range(n)]
    top, bottom = 0, n - 1
    left, right = 0, n - 1
    val = 1

    # Line 2: Fill layers
    while val <= n * n:
        # Line 3: Fill right
        for j in range(left, right + 1):
            matrix[top][j] = val
            val += 1
        top += 1

        # Line 4: Fill down
        for i in range(top, bottom + 1):
            matrix[i][right] = val
            val += 1
        right -= 1

        # Line 5: Fill left (if still valid)
        if top <= bottom:
            for j in range(right, left - 1, -1):
                matrix[bottom][j] = val
                val += 1
            bottom -= 1

        # Line 6: Fill up (if still valid)
        if left <= right:
            for i in range(bottom, top - 1, -1):
                matrix[i][left] = val
                val += 1
            left += 1

    # Line 7: Return completed matrix
    return matrix

Alternative: Direction Array Approach

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Explanation

Use a direction array to define movement (right, down, left, up) and fill the matrix by changing direction when hitting a boundary or filled cell. Track the current position and value, continuing until (n^2) elements are placed.

  1. Initialize Matrix and Direction.
  2. Fill with Directions.
  • Move, place value, change direction if needed.

3. Return Matrix.

Step-by-Step Example (Alternative)

For (n = 3):

  • Start: matrix = [[0,0,0],[0,0,0],[0,0,0]], dir = [(0,1),(1,0),(0,-1),(-1,0)], i = 0, j = 0, d = 0, val = 1.
  • Step 1: Right: [1,2,3], i = 0, j = 2, d = 1, val = 4.
  • Step 2: Down: [4,5], i = 2, j = 2, d = 2, val = 6.
  • Step 3: Left: [6,7], i = 2, j = 0, d = 3, val = 8.
  • Step 4: Up: [8,9], i = 1, j = 0, done.
  • Finish: Matrix = [[1,2,3],[8,9,4],[7,6,5]], return it.

How the Code Works (Direction Array)

def generateMatrixDir(n: int) -> list[list[int]]:
    # Line 1: Initialize matrix and variables
    matrix = [[0] * n for _ in range(n)]
    directions = [(0, 1), (1, 0), (0, -1), (-1, 0)]  # Right, down, left, up
    i, j = 0, 0
    d = 0  # Direction index
    val = 1

    # Line 2: Fill matrix
    while val <= n * n:
        # Line 3: Place current value
        matrix[i][j] = val
        val += 1

        # Line 4: Next position
        next_i, next_j = i + directions[d][0], j + directions[d][1]

        # Line 5: Change direction if needed
        if (next_i < 0 or next_i >= n or 
            next_j < 0 or next_j >= n or 
            matrix[next_i][next_j] != 0):
            d = (d + 1) % 4
            next_i, next_j = i + directions[d][0], j + directions[d][1]

        # Line 6: Move to next position
        i, j = next_i, next_j

    # Line 7: Return completed matrix
    return matrix

Complexity

  • Layer-by-Layer:
    • Time: O(n^2) – Fill each of \(n^2\) elements once.
    • Space: O(1) – Only boundaries (excluding output matrix).
  • Direction Array:
    • Time: O(n^2) – Fill each element once.
    • Space: O(1) – Only variables (excluding output).

Efficiency Notes

Both are O(n^2) time and O(1) space (excluding the output matrix), suitable for LeetCode 59. Layer-by-Layer is more structured and easier to follow, while Direction Array is flexible but requires careful boundary checking.

Key Insights

Tips to master LeetCode 59:

  • Layer Logic: Fill boundaries systematically.
  • Boundary Shrink: Move inward after each layer.
  • Spiral Pattern: Right, down, left, up sequence.

Additional Example

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For (n = 2):

  • Goal: [[1,2],[4,3]].
  • Layer: Right [1,2], down [3], left [], up [4].
  • Result: [[1,2],[4,3]].

Edge Cases

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  • \(n = 1\): [[1]].
  • \(n = 2\): [[1,2],[4,3]].
  • \(n = 20\): Fill 400 elements spirally.

Final Thoughts

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The Layer-by-Layer Generation solution is a top choice for LeetCode 59 in Python—clean, efficient, and perfect for spiral creation. For a related challenge, try LeetCode 54: Spiral Matrix to traverse an existing matrix! Ready to spiral? Solve LeetCode 59 on LeetCode now and test it with (n = 3) or (n = 4) to master matrix generation. Dive in and spin it up!