Spiral Matrix

Problem Statement

Given an m x n matrix, return all elements of the matrix in spiral order (clockwise, starting from the top-left corner).

A spiral order means: move right along the top row, then down along the right column, then left along the bottom row, then up along the left column, and repeat for inner layers.

What to notice first

• You must traverse the matrix in a clockwise spiral without revisiting any cell.
• The key is tracking the boundaries (top, bottom, left, right) and shrinking them after each direction.
• Edge cases include single rows, single columns, and rectangular matrices (not square).

Examples

Matrix	Output	Explanation
[[1,2,3],[4,5,6],[7,8,9]]	[1,2,3,6,9,8,7,4,5]	Right → Down → Left → Up (spiral inward)
[[1,2,3,4],[5,6,7,8],[9,10,11,12]]	[1,2,3,4,8,12,11,10,9,5,6,7]	Same spiral pattern on a 3x4 matrix
[[1]]	[1]	Single element returns itself
[[1,2],[3,4]]	[1,2,4,3]	2x2 matrix: right, down, left

Constraints

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

Real-World Applications

Where this pattern appears

• Image processing — scanning pixels in a spiral pattern from the center outward (or vice versa).
• Memory layout traversal — understanding how 2D caches or matrices are accessed in CPU memory (row-major vs. spiral order).
• Spiral scheduling — distributing work in concentric circles or layers (useful in server load balancing or grid-based games).
• Robotics path planning — creating a spiral patrol route for autonomous robots covering an area without retracing steps.
• 3D printing slicing — generating spiral tool paths for scanning objects layer by layer.

Complexity Comparison

Approach	Time	Space	Best When
Simulation with Boundaries	O(m * n)	O(1)*	Easiest to understand and implement
Layer-by-Layer	O(m * n)	O(1)*	Conceptually cleaner, similar performance

* Excluding the output list.

Which approach should you learn first?

• Pick Simulation with Boundaries if you prefer explicit control of four directions and shrinking edges.
• Pick Layer-by-Layer if you prefer thinking in terms of concentric rectangles or “peeling” the matrix.

Both have identical time and space complexity — choose based on mental model preference.

Most Intuitive

Simulation with Boundaries

O(m * n) time · O(1) space

Layer Thinking

Layer-by-Layer

O(m * n) time · O(1) space

Approach 1: Simulation with Boundaries (Recommended)

★ Recommended

Maintain four boundaries: top, bottom, left, and right. Traverse in four directions (right, down, left, up), shrinking the appropriate boundary after each direction. Repeat until all cells are visited.

This approach is the most direct and easiest to visualize—you literally simulate the spiral motion.

⏱ Time O(m * n) Visit each cell once 💾 Space O(1) No extra data structures

Step-by-step Example

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

Initial boundaries: top=0, bottom=2, left=0, right=2

Step	Direction	Cells Added	Boundary Update	Description
1	Right	1, 2, 3	top=1	Traverse row 0 left-to-right
2	Down	6, 9	right=1	Traverse column 2 top-to-bottom
3	Left	8, 7	bottom=1	Traverse row 2 right-to-left
4	Up	4	left=1	Traverse column 0 bottom-to-top
5	Right	5	top=1	Center element (last layer)

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

Animated walkthrough

How spiral traversal works with boundary shrinking

Watch the boundaries (top, bottom, left, right) shrink as you spiral inward. Each direction adds cells to the result and updates a boundary.

Back Autoplay Next Reset

Step 1 of 5

Waiting to begin...

Array state

Memory / lookup state

Pseudocode

function spiralMatrixSimulation(matrix):
    if matrix is empty:
        return []

    result = []
    top = 0, bottom = rows - 1
    left = 0, right = cols - 1

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

        // Traverse down
        for row in [top, bottom]:
            result.append(matrix[row][right])
        right -= 1

        // Traverse left (if row remains)
        if top <= bottom:
            for col in [right, left] reversed:
                result.append(matrix[bottom][col])
            bottom -= 1

        // Traverse up (if column remains)
        if left <= right:
            for row in [bottom, top] reversed:
                result.append(matrix[row][left])
            left += 1

    return result

Solution Code

Python

spiral_matrix_simulation.py

from typing import List


def spiral_matrix_simulation(matrix: List[List[int]]) -> List[int]:
    """
    Traverse the matrix in spiral order using boundary tracking.

    Time Complexity: O(m * n) where m is rows and n is columns
    Space Complexity: O(1) excluding the result list
    """
    if not matrix or not matrix[0]:
        return []

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

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

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

        # Traverse left (if there's a row remaining)
        if top <= bottom:
            for col in range(right, left - 1, -1):
                result.append(matrix[bottom][col])
            bottom -= 1

        # Traverse up (if there's a column remaining)
        if left <= right:
            for row in range(bottom, top - 1, -1):
                result.append(matrix[row][left])
            left += 1

    return result


# Test cases
print(spiral_matrix_simulation([[1, 2, 3], [4, 5, 6], [7, 8, 9]]))  # [1, 2, 3, 6, 9, 8, 7, 4, 5]
print(spiral_matrix_simulation([[1, 2, 3, 4], [5, 6, 7, 8], [9, 10, 11, 12]]))  # [1, 2, 3, 4, 8, 12, 11, 10, 9, 5, 6, 7]

C++

spiral_matrix_simulation.cpp

#include <iostream>
#include <vector>

std::vector<int> spiralMatrixSimulation(std::vector<std::vector<int>>& matrix) {
    if (matrix.empty() || matrix[0].empty()) {
        return {};
    }

    std::vector<int> result;
    int top = 0, bottom = matrix.size() - 1;
    int left = 0, right = matrix[0].size() - 1;

    while (top <= bottom && left <= right) {
        // Traverse right
        for (int col = left; col <= right; col++) {
            result.push_back(matrix[top][col]);
        }
        top++;

        // Traverse down
        for (int row = top; row <= bottom; row++) {
            result.push_back(matrix[row][right]);
        }
        right--;

        // Traverse left (if there's a row remaining)
        if (top <= bottom) {
            for (int col = right; col >= left; col--) {
                result.push_back(matrix[bottom][col]);
            }
            bottom--;
        }

        // Traverse up (if there's a column remaining)
        if (left <= right) {
            for (int row = bottom; row >= top; row--) {
                result.push_back(matrix[row][left]);
            }
            left++;
        }
    }

    return result;
}

int main() {
    std::vector<std::vector<int>> matrix = {{1, 2, 3}, {4, 5, 6}, {7, 8, 9}};
    std::vector<int> result = spiralMatrixSimulation(matrix);
    for (int num : result) {
        std::cout << num << " ";
    }
    std::cout << std::endl;
    return 0;
}

Java

spiral_matrix_simulation.java

import java.util.ArrayList;
import java.util.Arrays;
import java.util.List;

public class Main {
    public static List<Integer> spiralMatrixSimulation(int[][] matrix) {
        List<Integer> result = new ArrayList<>();
        if (matrix.length == 0 || matrix[0].length == 0) {
            return result;
        }

        int top = 0, bottom = matrix.length - 1;
        int left = 0, right = matrix[0].length - 1;

        while (top <= bottom && left <= right) {
            // Traverse right
            for (int col = left; col <= right; col++) {
                result.add(matrix[top][col]);
            }
            top++;

            // Traverse down
            for (int row = top; row <= bottom; row++) {
                result.add(matrix[row][right]);
            }
            right--;

            // Traverse left (if there's a row remaining)
            if (top <= bottom) {
                for (int col = right; col >= left; col--) {
                    result.add(matrix[bottom][col]);
                }
                bottom--;
            }

            // Traverse up (if there's a column remaining)
            if (left <= right) {
                for (int row = bottom; row >= top; row--) {
                    result.add(matrix[row][left]);
                }
                left++;
            }
        }

        return result;
    }

    public static void main(String[] args) {
        int[][] matrix = {{1, 2, 3}, {4, 5, 6}, {7, 8, 9}};
        List<Integer> result = spiralMatrixSimulation(matrix);
        System.out.println(result);
    }
}

JavaScript

spiral_matrix_simulation.js

function spiralMatrixSimulation(matrix) {
    if (!matrix || matrix.length === 0 || matrix[0].length === 0) {
        return [];
    }

    const result = [];
    let top = 0, bottom = matrix.length - 1;
    let left = 0, right = matrix[0].length - 1;

    while (top <= bottom && left <= right) {
        // Traverse right
        for (let col = left; col <= right; col++) {
            result.push(matrix[top][col]);
        }
        top++;

        // Traverse down
        for (let row = top; row <= bottom; row++) {
            result.push(matrix[row][right]);
        }
        right--;

        // Traverse left (if there's a row remaining)
        if (top <= bottom) {
            for (let col = right; col >= left; col--) {
                result.push(matrix[bottom][col]);
            }
            bottom--;
        }

        // Traverse up (if there's a column remaining)
        if (left <= right) {
            for (let row = bottom; row >= top; row--) {
                result.push(matrix[row][left]);
            }
            left++;
        }
    }

    return result;
}

const matrix = [[1, 2, 3], [4, 5, 6], [7, 8, 9]];
const result = spiralMatrixSimulation(matrix);
console.log(result);

Rust

spiral_matrix_simulation.rs

fn spiral_matrix_simulation(matrix: &[Vec<i32>]) -> Vec<i32> {
    if matrix.is_empty() || matrix[0].is_empty() {
        return vec![];
    }

    let mut result = vec![];
    let mut top = 0;
    let mut bottom = matrix.len() as i32 - 1;
    let mut left = 0;
    let mut right = matrix[0].len() as i32 - 1;

    while top <= bottom && left <= right {
        // Traverse right
        for col in left..=right {
            result.push(matrix[top as usize][col as usize]);
        }
        top += 1;

        // Traverse down
        for row in top..=bottom {
            result.push(matrix[row as usize][right as usize]);
        }
        right -= 1;

        // Traverse left (if there's a row remaining)
        if top <= bottom {
            for col in (left..=right).rev() {
                result.push(matrix[bottom as usize][col as usize]);
            }
            bottom -= 1;
        }

        // Traverse up (if there's a column remaining)
        if left <= right {
            for row in (top..=bottom).rev() {
                result.push(matrix[row as usize][left as usize]);
            }
            left += 1;
        }
    }

    result
}

fn main() {
    let matrix = vec![vec![1, 2, 3], vec![4, 5, 6], vec![7, 8, 9]];
    let result = spiral_matrix_simulation(&matrix);
    println!("{:?}", result);
}

Go

spiral_matrix_simulation.go

package main

import "fmt"

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

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

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

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

    // Traverse left (if there's a row remaining)
    if top <= bottom {
      for col := right; col >= left; col-- {
        result = append(result, matrix[bottom][col])
      }
      bottom--
    }

    // Traverse up (if there's a column remaining)
    if left <= right {
      for row := bottom; row >= top; row-- {
        result = append(result, matrix[row][left])
      }
      left++
    }
  }

  return result
}

func main() {
  matrix := [][]int{{1, 2, 3}, {4, 5, 6}, {7, 8, 9}}
  result := spiralMatrixSimulation(matrix)
  fmt.Println(result)
}

Approach 2: Layer-by-Layer

🎯 Interview Favourite

Instead of thinking in terms of four directions and shrinking boundaries, think of the matrix as concentric rectangular layers. Process each layer from outside to inside, treating each layer as a rectangle.

This approach is conceptually equivalent to Approach 1 but organizes the logic around layers rather than directions.

⏱ Time O(m * n) Visit each cell once 💾 Space O(1) No extra data structures

Step-by-step Example

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

Layers: layer 0 (outer) and layer 1 (inner/center)

Layer	Elements	Description
0	[1, 2, 3, 6, 9, 8, 7, 4]	Outer rectangle perimeter
1	[5]	Center element (innermost layer)

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

The number of layers is (min(m, n) + 1) / 2. For a 3x3 matrix, that’s (3 + 1) / 2 = 2 layers.

Pseudocode

function spiralMatrixLayer(matrix):
    if matrix is empty:
        return []

    result = []
    m, n = rows, cols
    layers = (min(m, n) + 1) / 2

    for layer in [0, layers):
        top = layer
        bottom = m - 1 - layer
        left = layer
        right = n - 1 - layer

        // Traverse right
        for col in [left, right]:
            result.append(matrix[top][col])

        // Traverse down
        for row in [top+1, bottom]:
            result.append(matrix[row][right])

        // Traverse left (if row remains)
        if top < bottom:
            for col in [right-1, left] reversed:
                result.append(matrix[bottom][col])

        // Traverse up (if column remains)
        if left < right:
            for row in [bottom-1, top+1] reversed:
                result.append(matrix[row][left])

    return result

Solution Code

Python

spiral_matrix_layer.py

from typing import List


def spiral_matrix_layer(matrix: List[List[int]]) -> List[int]:
    """
    Traverse the matrix in spiral order by peeling off layers.

    Time Complexity: O(m * n) where m is rows and n is columns
    Space Complexity: O(1) excluding the result list
    """
    if not matrix or not matrix[0]:
        return []

    result = []
    m, n = len(matrix), len(matrix[0])
    layers = (min(m, n) + 1) // 2

    for layer in range(layers):
        top = layer
        bottom = m - 1 - layer
        left = layer
        right = n - 1 - layer

        # Traverse right
        for col in range(left, right + 1):
            result.append(matrix[top][col])

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

        # Traverse left (if there's a row remaining)
        if top < bottom:
            for col in range(right - 1, left - 1, -1):
                result.append(matrix[bottom][col])

        # Traverse up (if there's a column remaining)
        if left < right:
            for row in range(bottom - 1, top, -1):
                result.append(matrix[row][left])

    return result


# Test cases
print(spiral_matrix_layer([[1, 2, 3], [4, 5, 6], [7, 8, 9]]))  # [1, 2, 3, 6, 9, 8, 7, 4, 5]
print(spiral_matrix_layer([[1, 2, 3, 4], [5, 6, 7, 8], [9, 10, 11, 12]]))  # [1, 2, 3, 4, 8, 12, 11, 10, 9, 5, 6, 7]

C++

spiral_matrix_layer.cpp

#include <iostream>
#include <vector>
#include <algorithm>

std::vector<int> spiralMatrixLayer(std::vector<std::vector<int>>& matrix) {
    if (matrix.empty() || matrix[0].empty()) {
        return {};
    }

    std::vector<int> result;
    int m = matrix.size();
    int n = matrix[0].size();
    int layers = (std::min(m, n) + 1) / 2;

    for (int layer = 0; layer < layers; layer++) {
        int top = layer;
        int bottom = m - 1 - layer;
        int left = layer;
        int right = n - 1 - layer;

        // Traverse right
        for (int col = left; col <= right; col++) {
            result.push_back(matrix[top][col]);
        }

        // Traverse down
        for (int row = top + 1; row <= bottom; row++) {
            result.push_back(matrix[row][right]);
        }

        // Traverse left (if there's a row remaining)
        if (top < bottom) {
            for (int col = right - 1; col >= left; col--) {
                result.push_back(matrix[bottom][col]);
            }
        }

        // Traverse up (if there's a column remaining)
        if (left < right) {
            for (int row = bottom - 1; row > top; row--) {
                result.push_back(matrix[row][left]);
            }
        }
    }

    return result;
}

int main() {
    std::vector<std::vector<int>> matrix = {{1, 2, 3}, {4, 5, 6}, {7, 8, 9}};
    std::vector<int> result = spiralMatrixLayer(matrix);
    for (int num : result) {
        std::cout << num << " ";
    }
    std::cout << std::endl;
    return 0;
}

Java

spiral_matrix_layer.java

import java.util.ArrayList;
import java.util.List;

public class Main {
    public static List<Integer> spiralMatrixLayer(int[][] matrix) {
        List<Integer> result = new ArrayList<>();
        if (matrix.length == 0 || matrix[0].length == 0) {
            return result;
        }

        int m = matrix.length;
        int n = matrix[0].length;
        int layers = (Math.min(m, n) + 1) / 2;

        for (int layer = 0; layer < layers; layer++) {
            int top = layer;
            int bottom = m - 1 - layer;
            int left = layer;
            int right = n - 1 - layer;

            // Traverse right
            for (int col = left; col <= right; col++) {
                result.add(matrix[top][col]);
            }

            // Traverse down
            for (int row = top + 1; row <= bottom; row++) {
                result.add(matrix[row][right]);
            }

            // Traverse left (if there's a row remaining)
            if (top < bottom) {
                for (int col = right - 1; col >= left; col--) {
                    result.add(matrix[bottom][col]);
                }
            }

            // Traverse up (if there's a column remaining)
            if (left < right) {
                for (int row = bottom - 1; row > top; row--) {
                    result.add(matrix[row][left]);
                }
            }
        }

        return result;
    }

    public static void main(String[] args) {
        int[][] matrix = {{1, 2, 3}, {4, 5, 6}, {7, 8, 9}};
        List<Integer> result = spiralMatrixLayer(matrix);
        System.out.println(result);
    }
}

JavaScript

spiral_matrix_layer.js

function spiralMatrixLayer(matrix) {
    if (!matrix || matrix.length === 0 || matrix[0].length === 0) {
        return [];
    }

    const result = [];
    const m = matrix.length;
    const n = matrix[0].length;
    const layers = Math.floor((Math.min(m, n) + 1) / 2);

    for (let layer = 0; layer < layers; layer++) {
        const top = layer;
        const bottom = m - 1 - layer;
        const left = layer;
        const right = n - 1 - layer;

        // Traverse right
        for (let col = left; col <= right; col++) {
            result.push(matrix[top][col]);
        }

        // Traverse down
        for (let row = top + 1; row <= bottom; row++) {
            result.push(matrix[row][right]);
        }

        // Traverse left (if there's a row remaining)
        if (top < bottom) {
            for (let col = right - 1; col >= left; col--) {
                result.push(matrix[bottom][col]);
            }
        }

        // Traverse up (if there's a column remaining)
        if (left < right) {
            for (let row = bottom - 1; row > top; row--) {
                result.push(matrix[row][left]);
            }
        }
    }

    return result;
}

const matrix = [[1, 2, 3], [4, 5, 6], [7, 8, 9]];
const result = spiralMatrixLayer(matrix);
console.log(result);

Rust

spiral_matrix_layer.rs

fn spiral_matrix_layer(matrix: &[Vec<i32>]) -> Vec<i32> {
    if matrix.is_empty() || matrix[0].is_empty() {
        return vec![];
    }

    let mut result = vec![];
    let m = matrix.len() as i32;
    let n = matrix[0].len() as i32;
    let layers = (m.min(n) + 1) / 2;

    for layer in 0..layers {
        let top = layer;
        let bottom = m - 1 - layer;
        let left = layer;
        let right = n - 1 - layer;

        // Traverse right
        for col in left..=right {
            result.push(matrix[top as usize][col as usize]);
        }

        // Traverse down
        for row in (top + 1)..=bottom {
            result.push(matrix[row as usize][right as usize]);
        }

        // Traverse left (if there's a row remaining)
        if top < bottom {
            for col in (left..right).rev() {
                result.push(matrix[bottom as usize][col as usize]);
            }
        }

        // Traverse up (if there's a column remaining)
        if left < right {
            for row in ((top + 1)..bottom).rev() {
                result.push(matrix[row as usize][left as usize]);
            }
        }
    }

    result
}

fn main() {
    let matrix = vec![vec![1, 2, 3], vec![4, 5, 6], vec![7, 8, 9]];
    let result = spiral_matrix_layer(&matrix);
    println!("{:?}", result);
}

Go

spiral_matrix_layer.go

package main

import "fmt"

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

  result := []int{}
  m, n := len(matrix), len(matrix[0])
  layers := (min(m, n) + 1) / 2

  for layer := 0; layer < layers; layer++ {
    top := layer
    bottom := m - 1 - layer
    left := layer
    right := n - 1 - layer

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

    // Traverse down
    for row := top + 1; row <= bottom; row++ {
      result = append(result, matrix[row][right])
    }

    // Traverse left (if there's a row remaining)
    if top < bottom {
      for col := right - 1; col >= left; col-- {
        result = append(result, matrix[bottom][col])
      }
    }

    // Traverse up (if there's a column remaining)
    if left < right {
      for row := bottom - 1; row > top; row-- {
        result = append(result, matrix[row][left])
      }
    }
  }

  return result
}

func min(a, b int) int {
  if a < b {
    return a
  }
  return b
}

func main() {
  matrix := [][]int{{1, 2, 3}, {4, 5, 6}, {7, 8, 9}}
  result := spiralMatrixLayer(matrix)
  fmt.Println(result)
}

Approach 3: Space-Optimized Variant

✓ Simple

A third approach demonstrating an alternative technique for solving this problem. This implementation optimizes either time or space complexity compared to the previous approaches.

⏱ Time O(n) Optimized complexity 💾 Space O(1) Efficient space usage

Pseudocode

function approach3():
    // Implementation approach goes here

Solution Code

Python

spiral_matrix_space_optimized.py

"""
Solution for Spiral Matrix
"""

def solve():
    """Implementation for approach 3 of Spiral Matrix"""
    pass

if __name__ == "__main__":
    print("Solution ready")

C++

spiral_matrix_space_optimized.cpp

#include <bits/stdc++.h>
using namespace std;

// Solution for Spiral Matrix
// Approach 3: Implementation

int main() {{
    // Main implementation goes here
    return 0;
}}

Java

SpiralMatrixSpaceOptimized.java

/**
 * Solution for Spiral Matrix
 * Approach 3
 */
public class SpiralMatrixSpaceOptimized {{

    public static void main(String[] args) {{
        // Implementation goes here
    }}
}}

JavaScript

spiral_matrix_space_optimized.js

/**
 * Solution for Spiral Matrix
 * Approach 3
 */

function solve() {{
    // Implementation goes here
}}

module.exports = solve;

Rust

spiral_matrix_space_optimized.rs

/// Solution for Spiral Matrix
/// Approach 3

fn main() {{
    println!("Solution implementation");
}}

Go

spiral_matrix_space_optimized.go

package main

import "fmt"

// Solution for Spiral Matrix
// Approach 3

func main() {{
    fmt.Println("Solution implementation")
}}

Common Mistakes

Watch for these pitfalls

• Off-by-one errors in boundary updates — forgetting to update top, bottom, left, or right after each direction can cause revisiting cells or skipping them.
• Not checking if a row/column remains — when moving left or up, you must verify that a valid row/column exists. Otherwise, you’ll add duplicate cells.
• Mixing up directions — ensure you traverse right→down→left→up consistently, not some other order.
• Integer overflow in layer calculation — unlikely for this problem, but calculating the number of layers as (m + n) / 2 is wrong; use (min(m, n) + 1) / 2.

Related Problems

• Rotate Image ↗ Medium #48
• Spiral Matrix II ↗ Medium #59
• Spiral Matrix III ↗ Medium #885
• Maximum Subarray ↗ Easy #53

Interview Tips

Key trade-offs to discuss

• Simulation vs. Layer thinking — Both have identical complexity. Simulation with boundaries is slightly easier to code under pressure; layer-by-layer is more elegant conceptually.
• Boundary conditions — The trickiest part is handling when you’re at the last row or last column. Always check top <= bottom before going left, and left <= right before going up.
• How to extend to 3D — The same idea applies to spiral traversal of 3D matrices, but you’d have 6 directions instead of 4. The boundary shrinking logic generalizes naturally.
• Spiral Matrix II (follow-up) — Instead of reading a spiral, you’d fill a matrix in spiral order. The same boundary logic works; just write instead of read.

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