LeetCode #885 — MEDIUM

Spiral Matrix III

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

The Problem

Problem Statement

You start at the cell (rStart, cStart) of an rows x cols grid facing east. The northwest corner is at the first row and column in the grid, and the southeast corner is at the last row and column.

You will walk in a clockwise spiral shape to visit every position in this grid. Whenever you move outside the grid's boundary, we continue our walk outside the grid (but may return to the grid boundary later.). Eventually, we reach all rows * cols spaces of the grid.

Return an array of coordinates representing the positions of the grid in the order you visited them.

Example 1:

Input: rows = 1, cols = 4, rStart = 0, cStart = 0
Output: [[0,0],[0,1],[0,2],[0,3]]

Example 2:

Input: rows = 5, cols = 6, rStart = 1, cStart = 4
Output: [[1,4],[1,5],[2,5],[2,4],[2,3],[1,3],[0,3],[0,4],[0,5],[3,5],[3,4],[3,3],[3,2],[2,2],[1,2],[0,2],[4,5],[4,4],[4,3],[4,2],[4,1],[3,1],[2,1],[1,1],[0,1],[4,0],[3,0],[2,0],[1,0],[0,0]]

Constraints:

1 <= rows, cols <= 100
0 <= rStart < rows
0 <= cStart < cols

Step 01

Brute Force Baseline

Problem summary: You start at the cell (rStart, cStart) of an rows x cols grid facing east. The northwest corner is at the first row and column in the grid, and the southeast corner is at the last row and column. You will walk in a clockwise spiral shape to visit every position in this grid. Whenever you move outside the grid's boundary, we continue our walk outside the grid (but may return to the grid boundary later.). Eventually, we reach all rows * cols spaces of the grid. Return an array of coordinates representing the positions of the grid in the order you visited them.

Baseline thinking

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

Pattern signal: Array

Example 1

Example 2

Core Insight

What unlocks the optimal approach

No official hints in dataset. Start from constraints and look for a monotonic or reusable state.

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

Source-backed implementations are provided below for direct study and interview prep.

// Accepted solution for LeetCode #885: Spiral Matrix III
class Solution {
    public int[][] spiralMatrixIII(int rows, int cols, int rStart, int cStart) {
        int cnt = rows * cols;
        int[][] ans = new int[cnt][2];
        ans[0] = new int[] {rStart, cStart};
        if (cnt == 1) {
            return ans;
        }
        for (int k = 1, idx = 1;; k += 2) {
            int[][] dirs = new int[][] {{0, 1, k}, {1, 0, k}, {0, -1, k + 1}, {-1, 0, k + 1}};
            for (int[] dir : dirs) {
                int r = dir[0], c = dir[1], dk = dir[2];
                while (dk-- > 0) {
                    rStart += r;
                    cStart += c;
                    if (rStart >= 0 && rStart < rows && cStart >= 0 && cStart < cols) {
                        ans[idx++] = new int[] {rStart, cStart};
                        if (idx == cnt) {
                            return ans;
                        }
                    }
                }
            }
        }
    }
}

// Accepted solution for LeetCode #885: Spiral Matrix III
func spiralMatrixIII(rows int, cols int, rStart int, cStart int) [][]int {
	cnt := rows * cols
	ans := [][]int{[]int{rStart, cStart}}
	if cnt == 1 {
		return ans
	}
	for k := 1; ; k += 2 {
		dirs := [][]int{{0, 1, k}, {1, 0, k}, {0, -1, k + 1}, {-1, 0, k + 1}}
		for _, dir := range dirs {
			r, c, dk := dir[0], dir[1], dir[2]
			for dk > 0 {
				rStart += r
				cStart += c
				if rStart >= 0 && rStart < rows && cStart >= 0 && cStart < cols {
					ans = append(ans, []int{rStart, cStart})
					if len(ans) == cnt {
						return ans
					}
				}
				dk--
			}
		}
	}
}

# Accepted solution for LeetCode #885: Spiral Matrix III
class Solution:
    def spiralMatrixIII(
        self, rows: int, cols: int, rStart: int, cStart: int
    ) -> List[List[int]]:
        ans = [[rStart, cStart]]
        if rows * cols == 1:
            return ans
        k = 1
        while True:
            for dr, dc, dk in [[0, 1, k], [1, 0, k], [0, -1, k + 1], [-1, 0, k + 1]]:
                for _ in range(dk):
                    rStart += dr
                    cStart += dc
                    if 0 <= rStart < rows and 0 <= cStart < cols:
                        ans.append([rStart, cStart])
                        if len(ans) == rows * cols:
                            return ans
            k += 2

// Accepted solution for LeetCode #885: Spiral Matrix III
struct Solution;

impl Solution {
    fn spiral_matrix_iii(r: i32, c: i32, r0: i32, c0: i32) -> Vec<Vec<i32>> {
        let mut res = vec![];
        let total = (r * c) as usize;
        let directions = vec![(0, 1), (1, 0), (0, -1), (-1, 0)];
        let mut d = 0;
        let mut step = 1;
        let mut i = r0;
        let mut j = c0;
        let mut k = 0;
        while res.len() < total {
            if i >= 0 && i < r && j >= 0 && j < c {
                res.push(vec![i, j]);
            }
            i += directions[d].0;
            j += directions[d].1;
            k += 1;
            if k == step {
                d += 1;
                d %= 4;
                k = 0;
                if d % 2 == 0 {
                    step += 1;
                }
            }
        }
        res
    }
}

#[test]
fn test() {
    let r = 1;
    let c = 4;
    let r0 = 0;
    let c0 = 0;
    let res = vec_vec_i32![[0, 0], [0, 1], [0, 2], [0, 3]];
    assert_eq!(Solution::spiral_matrix_iii(r, c, r0, c0), res);
    let r = 5;
    let c = 6;
    let r0 = 1;
    let c0 = 4;
    let res = vec_vec_i32![
        [1, 4],
        [1, 5],
        [2, 5],
        [2, 4],
        [2, 3],
        [1, 3],
        [0, 3],
        [0, 4],
        [0, 5],
        [3, 5],
        [3, 4],
        [3, 3],
        [3, 2],
        [2, 2],
        [1, 2],
        [0, 2],
        [4, 5],
        [4, 4],
        [4, 3],
        [4, 2],
        [4, 1],
        [3, 1],
        [2, 1],
        [1, 1],
        [0, 1],
        [4, 0],
        [3, 0],
        [2, 0],
        [1, 0],
        [0, 0]
    ];
    assert_eq!(Solution::spiral_matrix_iii(r, c, r0, c0), res);
}

// Accepted solution for LeetCode #885: Spiral Matrix III
function spiralMatrixIII(rows: number, cols: number, rStart: number, cStart: number): number[][] {
    // prettier-ignore
    const dir = [[1,0],[0,1],[-1,0],[0,-1]]
    let [x, y, i, size] = [cStart, rStart, 0, 0];
    const ans: number[][] = [[y, x]];
    const total = rows * cols;

    while (ans.length < total) {
        if (i % 2 === 0) size++;

        for (let j = 0; ans.length < total && j < size; j++) {
            x += dir[i][0];
            y += dir[i][1];

            if (0 <= x && x < cols && 0 <= y && y < rows) {
                ans.push([y, x]);
            }
        }

        i = (i + 1) % 4;
    }

    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(n)

Space

O(1)

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.