LeetCode #54 — MEDIUM

Spiral Matrix

Move from brute-force thinking to an efficient approach using array strategy.

The Problem

Problem Statement

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

Step 01

Brute Force Baseline

Problem summary: 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]

Baseline thinking

Start with the most direct exhaustive search. That gives a correctness anchor before optimizing.

Pattern signal: Array

Example 1

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

Example 2

[[1,2,3,4],[5,6,7,8],[9,10,11,12]]

Core Insight

What unlocks the optimal approach

Well for some problems, the best way really is to come up with some algorithms for simulation. Basically, you need to simulate what the problem asks us to do.
We go boundary by boundary and move inwards. That is the essential operation. First row, last column, last row, first column, and then we move inwards by 1 and repeat. That's all. That is all the simulation that we need.
Think about when you want to switch the progress on one of the indexes. If you progress on i out of [i, j], you'll shift in the same column. Similarly, by changing values for j, you'd be shifting in the same row. Also, keep track of the end of a boundary so that you can move inwards and then keep repeating. It's always best to simulate edge cases like a single column or a single row to see if anything breaks or not.

Interview move: turn each hint into an invariant you can check after every iteration/recursion step.

Step 03

Algorithm Walkthrough

Iteration Checklist

Define state (indices, window, stack, map, DP cell, or recursion frame).
Apply one transition step and update the invariant.
Record answer candidate when condition is met.
Continue until all input is consumed.

Use the first example testcase as your mental trace to verify each transition.

Step 04

Edge Cases

Minimum Input

Single element / shortest valid input

Validate boundary behavior before entering the main loop or recursion.

Duplicates & Repeats

Repeated values / repeated states

Decide whether duplicates should be merged, skipped, or counted explicitly.

Extreme Constraints

Upper-end input sizes

Re-check complexity target against constraints to avoid time-limit issues.

Invalid / Corner Shape

Empty collections, zeros, or disconnected structures

Handle special-case structure before the core algorithm path.

Step 05

Full Annotated Code

import java.util.*;

class Solution {
    public List<Integer> spiralOrder(int[][] matrix) {
        List<Integer> ans = new ArrayList<>();
        if (matrix.length == 0) return ans;

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

        while (top <= bottom && left <= right) {
            for (int c = left; c <= right; c++) ans.add(matrix[top][c]);
            top++;

            for (int r = top; r <= bottom; r++) ans.add(matrix[r][right]);
            right--;

            if (top <= bottom) {
                for (int c = right; c >= left; c--) ans.add(matrix[bottom][c]);
                bottom--;
            }

            if (left <= right) {
                for (int r = bottom; r >= top; r--) ans.add(matrix[r][left]);
                left++;
            }
        }

        return ans;
    }
}

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

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

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

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

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

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

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

        top, bottom = 0, len(matrix) - 1
        left, right = 0, len(matrix[0]) - 1
        ans: List[int] = []

        while top <= bottom and left <= right:
            for c in range(left, right + 1):
                ans.append(matrix[top][c])
            top += 1

            for r in range(top, bottom + 1):
                ans.append(matrix[r][right])
            right -= 1

            if top <= bottom:
                for c in range(right, left - 1, -1):
                    ans.append(matrix[bottom][c])
                bottom -= 1

            if left <= right:
                for r in range(bottom, top - 1, -1):
                    ans.append(matrix[r][left])
                left += 1

        return ans

impl Solution {
    pub fn spiral_order(matrix: Vec<Vec<i32>>) -> Vec<i32> {
        if matrix.is_empty() {
            return vec![];
        }

        let mut top: i32 = 0;
        let mut bottom: i32 = matrix.len() as i32 - 1;
        let mut left: i32 = 0;
        let mut right: i32 = matrix[0].len() as i32 - 1;
        let mut ans = Vec::new();

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

            for r in top..=bottom {
                ans.push(matrix[r as usize][right as usize]);
            }
            right -= 1;

            if top <= bottom {
                for c in (left..=right).rev() {
                    ans.push(matrix[bottom as usize][c as usize]);
                }
                bottom -= 1;
            }

            if left <= right {
                for r in (top..=bottom).rev() {
                    ans.push(matrix[r as usize][left as usize]);
                }
                left += 1;
            }
        }

        ans
    }
}

function spiralOrder(matrix: number[][]): number[] {
  if (matrix.length === 0) return [];

  let top = 0;
  let bottom = matrix.length - 1;
  let left = 0;
  let right = matrix[0].length - 1;
  const ans: number[] = [];

  while (top <= bottom && left <= right) {
    for (let c = left; c <= right; c++) ans.push(matrix[top][c]);
    top++;

    for (let r = top; r <= bottom; r++) ans.push(matrix[r][right]);
    right--;

    if (top <= bottom) {
      for (let c = right; c >= left; c--) ans.push(matrix[bottom][c]);
      bottom--;
    }

    if (left <= right) {
      for (let r = bottom; r >= top; r--) ans.push(matrix[r][left]);
      left++;
    }
  }

  return ans;
}

Step 06

Interactive Study Demo

Use this to step through a reusable interview workflow for this problem.

Custom checkpoints (one per line)

Press Step or Run All to begin.

Step 07

Complexity Analysis

Time

O(m × n)

Space

O(m × n)

Approach Breakdown

BRUTE FORCE

O(n²) time

O(1) space

Two nested loops check every pair or subarray. The outer loop fixes a starting point, the inner loop extends or searches. For n elements this gives up to n²/2 operations. No extra space, but the quadratic time is prohibitive for large inputs.

OPTIMIZED

O(n) time

O(1) space

Most array problems have an O(n²) brute force (nested loops) and an O(n) optimal (single pass with clever state tracking). The key is identifying what information to maintain as you scan: a running max, a prefix sum, a hash map of seen values, or two pointers.

Shortcut: If you are using nested loops on an array, there is almost always an O(n) solution. Look for the right auxiliary state.

Coach Notes

Common Mistakes

Review these before coding to avoid predictable interview regressions.

Off-by-one on range boundaries

Wrong move: Loop endpoints miss first/last candidate.

Usually fails on: Fails on minimal arrays and exact-boundary answers.

Fix: Re-derive loops from inclusive/exclusive ranges before coding.