What is the time complexity of Spiral Matrix III?

The optimal solution has a time complexity of O(rows * cols).

What is the space complexity of Spiral Matrix III?

The space complexity is O(rows * cols).

What data structures are used to solve Spiral Matrix III?

This problem uses Array, Matrix, Simulation. Understanding these concepts is essential for solving this Medium problem efficiently.

What are the key insights for solving Spiral Matrix III?

Insight 1: The spiral path can be simulated by iteratively changing direction and increasing the number of steps taken in each direction after every two direction changes (East/West or South/North). Insight 2: We don't need to physically create the entire grid; we only need to keep track of the current position and check if it's within the boundaries of the grid.

Which companies ask LeetCode 885. Spiral Matrix III in interviews?

This problem is frequently asked at Dataminr.

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Spiral Matrix III - LeetCode 885 Solution

Q: How to solve Spiral Matrix III?

The solution simulates the spiral path using a `while` loop that continues until all cells in the grid have been visited (i.e., the size of the result list is equal to rows * cols). Inside the loop, the algorithm iterates through four directions (East, South, West, North), incrementing the number of steps to take in each direction every two direction changes. In each step, the current row and column are updated, and if the new position is within the grid boundaries, its coordinates are added to the result list. The direction is updated using modulo arithmetic to cycle through the four directions.

Spiral Matrix III - Complete Solution Guide

Spiral Matrix III is LeetCode problem 885, a Medium level challenge. This complete guide provides step-by-step explanations, multiple solution approaches, and optimized code in python3, java, cpp, c.

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 repre

Detailed Explanation

The problem requires generating a spiral path within a grid of size `rows x cols`, starting from a given cell `(rStart, cStart)`. The spiral moves clockwise, and the path should include every cell in the grid, even if it initially steps outside the grid boundaries. The output should be a list of coordinates representing the visited cells in the order they were visited, until all cells in the grid have been visited at least virtually. The grid boundaries are `0 <= row < rows` and `0 <= col < cols`.

Solution Approach

The solution simulates the spiral path using a `while` loop that continues until all cells in the grid have been visited (i.e., the size of the result list is equal to rows * cols). Inside the loop, the algorithm iterates through four directions (East, South, West, North), incrementing the number of steps to take in each direction every two direction changes. In each step, the current row and column are updated, and if the new position is within the grid boundaries, its coordinates are added to the result list. The direction is updated using modulo arithmetic to cycle through the four directions.

Step-by-Step Algorithm

Step 1: Initialize the result list, current position (r, c), and direction vectors (dr, dc) for East, South, West, and North.
Step 2: Add the starting position (rStart, cStart) to the result list.
Step 3: Initialize the number of steps to take in each direction to 0, the current direction to 0 (East), and the total number of cells to rows * cols.
Step 4: Enter a `while` loop that continues until the result list contains all cells inside the grid. (res.size() < rows * cols)
Step 5: Inside the loop, check if the current direction is East (0) or West (2). If so, increment the number of steps.
Step 6: Iterate 'steps' number of times within inner for loop to move current position.
Step 7: Update the current row and column by adding the corresponding direction vectors (dr[direction] and dc[direction]).
Step 8: Check if the updated position is within the grid boundaries (0 <= r < rows and 0 <= c < cols). If it is, add the position to the result list.
Step 9: After completing 'steps' iterations, update the current direction to the next direction by taking the modulo of (direction + 1) with 4.
Step 10: Repeat steps 4-9 until all cells inside grid are recorded (res.size() == rows * cols)
Step 11: Return the result list.

Key Insights

Insight 1: The spiral path can be simulated by iteratively changing direction and increasing the number of steps taken in each direction after every two direction changes (East/West or South/North).
Insight 2: We don't need to physically create the entire grid; we only need to keep track of the current position and check if it's within the boundaries of the grid.
Insight 3: Since we need to visit every cell (virtually) even when out of bound, the loop continues till the number of visited cells (within the bounds) equals rows * cols.

Complexity Analysis

Time Complexity: O(rows * cols)

Space Complexity: O(rows * cols)

Topics

This problem involves: Array, Matrix, Simulation.

Companies

Asked at: Dataminr.

Spiral Matrix III | LeetCode 885

Medium

Array Matrix Simulation

Time: O(rows * cols)Space: O(rows * cols)

Solution Code

Available in:

class Solution {
    public int[][] spiralMatrixIII(int rows, int cols, int rStart, int cStart) {
        List<int[]> resList = new ArrayList<>();
        int[] dr = {0, 1, 0, -1};  // E, S, W, N
        int[] dc = {1, 0, -1, 0};

        int r = rStart, c = cStart;
        resList.add(new int[]{r, c});

        int steps = 0;
        int direction = 0;

        int total_cells = rows * cols;

        while (resList.size() < total_cells) {
            if (direction == 0 || direction == 2) {
                steps++;
            }

            for (int i = 0; i < steps; i++) {
                r += dr[direction];
                c += dc[direction];

                if (0 <= r && r < rows && 0 <= c && c < cols) {
                    resList.add(new int[]{r, c});
                }
            }

            direction = (direction + 1) % 4;
        }

        int[][] res = new int[resList.size()][2];
        for (int i = 0; i < resList.size(); i++) {
            res[i][0] = resList.get(i)[0];
            res[i][1] = resList.get(i)[1];
        }

        return res;
    }
}

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Problem Info

Description

Problem 885: Spiral Matrix III - Detailed explanation and solution approaches.

Topics

Array Matrix Simulation

Companies

Dataminr

Stats

Difficulty:Medium

Languages:4

Companies:1

Problem #:885

Problem Explanation

💡

Think Before You Code

🤔 Ask Yourself

→What is the input? What output do I need?
→Can I solve a smaller version of this problem first?
→What patterns have I seen in similar problems?

✏️ Try This First

Before looking at the solution, try to solve it on paper with a simple example!

🚀 Try on LeetCode

Solution Approach

📋

Step-by-Step Solution

11 steps

1
Step 1: Initialize the result list, current position (r, c), and direction vectors (dr, dc) for East, South, West, and North.
2
Step 2: Add the starting position (rStart, cStart) to the result list.
3
Step 3: Initialize the number of steps to take in each direction to 0, the current direction to 0 (East), and the total number of cells to rows * cols.
4
Step 4: Enter a `while` loop that continues until the result list contains all cells inside the grid. (res.size() < rows * cols)
5
Step 5: Inside the loop, check if the current direction is East (0) or West (2). If so, increment the number of steps.
6
Step 6: Iterate 'steps' number of times within inner for loop to move current position.
7
Step 7: Update the current row and column by adding the corresponding direction vectors (dr[direction] and dc[direction]).
8
Step 8: Check if the updated position is within the grid boundaries (0 <= r < rows and 0 <= c < cols). If it is, add the position to the result list.
9
Step 9: After completing 'steps' iterations, update the current direction to the next direction by taking the modulo of (direction + 1) with 4.
10
Step 10: Repeat steps 4-9 until all cells inside grid are recorded (res.size() == rows * cols)
11
Step 11: Return the result list.

💡

Pro Tip for Students

Try tracing through each step with a small example on paper before coding. Understanding "why" each step happens is more important than memorizing the code!

⚡

Complexity Analysis

⏱️

Time Complexity

O(rows * cols)

💾

Space Complexity

O(rows * cols)

📝

Why?

The algorithm visits each cell inside the grid once within the boundary. The outer loop runs until the number of valid cells visited equals rows * cols. The inner loop runs a number of steps proportional to the spiral layer being processed, which is also ultimately bounded by O(rows * cols). The space complexity is also O(rows * cols) because the output list stores the coordinates of each visited cell within bounds.

📖New to Big O? Click to learn more

Big O Notation - Quick Guide

O(1)Best - Instant

O(log n)Excellent - Binary Search

O(n)Good - One loop

O(n log n)OK - Sorting

O(n²)Slow - Nested loops

O(2ⁿ)Very Slow - Recursion

Alternative Approaches

Using a Set to Track Visited Cells

Instead of simply checking the size of the result list, maintain a set of visited cells to avoid revisiting the same cell. This approach will ensure that the final list contains unique visited cell coordinates.

Time: O(rows * cols)Space: O(rows * cols)

Trade-offs: The trade-off is that this approach has a slightly higher overhead due to set operations. It might be helpful if you need to ensure that each cell is visited only once in the result, which isn't explicitly mentioned in the problem, but is a reasonable interpretation.

✨

Key Insights

1Insight 1: The spiral path can be simulated by iteratively changing direction and increasing the number of steps taken in each direction after every two direction changes (East/West or South/North).
2Insight 2: We don't need to physically create the entire grid; we only need to keep track of the current position and check if it's within the boundaries of the grid.
3Insight 3: Since we need to visit every cell (virtually) even when out of bound, the loop continues till the number of visited cells (within the bounds) equals rows * cols.

📚

Key Terms Explained

For Students

Array

A collection of items stored at contiguous memory locations. Like a row of boxes.

Edge Cases

Edge case 1: `rows = 1, cols = 1, rStart = 0, cStart = 0`. The solution handles this case correctly by adding the initial position to the result and terminating the loop since all cells are visited.
Edge case 2: `rStart` and `cStart` are near the edge of the matrix. The solution correctly handles the scenario when the path initially goes out of bounds. The condition inside the `if` statement ensures that only coordinates that are within the grid boundaries are added to the result.

⚠️

Common Mistakes to Avoid

✗Common mistake 1: Incorrectly calculating the number of steps in each direction. Failing to increment the number of steps after every two direction changes will result in an incorrect spiral path.

✗Common mistake 2: Incorrectly updating the current row and column. Using the wrong direction vectors will result in an incorrect spiral path.

✗Common mistake 3: Not checking the grid boundaries before adding a coordinate to the result list. This will result in coordinates that are out of bounds being added to the result list, leading to an incorrect result.

✗Common mistake 4: Terminating the loop too early or too late. The loop must continue until the result list contains all cells within the grid (rows * cols). A premature exit will yield an incomplete spiral.

✅

Learning Tip

Every mistake is a learning opportunity! When you make an error, understand why it happened before moving on.

🌍

Where is This Used in Real Life?

💼Real-world scenario 1: Path planning for robots or drones in a grid-based environment. The robot or drone needs to visit all locations within a certain area in a systematic manner.

🏦Real-world scenario 2: Image processing algorithms that need to traverse a 2D image in a spiral pattern, such as for compression or feature extraction. Spiral traversal allows for localized processing near a center point.

💡Understanding real-world applications helps you see the "why" behind algorithms. It's not just about passing interviews—these concepts power the apps and websites you use every day!

Author's Notes

💡 Key Insight

When dealing with arrays, always consider if sorting would simplify the problem. Many array problems have O(n) solutions using hash maps or two-pointer technique.

🎯 Interview Tip

Medium problems are the bread and butter of technical interviews. Practice explaining your approach before coding.

⚠️ Watch Out

Off-by-one errors are the most common mistake. Double-check your loop bounds and array indices.

👨‍💻 From Experience

This problem pattern appears frequently in real codebases. I've used similar logic when building production systems.

Written by Saswata Mondal

Software Engineer • Algorithms Enthusiast • Interview Coach

👨‍💻

About the Author

👨‍💻

Saswata MondalAuthor400+ LC Problems

B.Tech CSE, IEM Kolkata | LeetCode Rating: 1672 (Top 15%)

Certified: IBM Java Developer, NPTEL Programming in Java

Last updated: November 13, 2024LeetCode Profile GitHub

This solution has been crafted based on my experience solving 400+ LeetCode problems. Each explanation is designed to help you understand the underlying concepts, not just memorize the code.

Community Tips & Insights

Don't forget to handle the edge case when the input is empty. Many candidates miss this in interviews.

@DevExpert92Jul 13

The key insight here is recognizing the pattern. Once you see it, the solution becomes straightforward.

@AlgoMasterJul 6

I was asked this exact problem at my Google interview. The interviewer wanted me to optimize space complexity.

@InterviewProJun 29

❓

Frequently Asked Questions

When should I use in-place modification vs creating a new array?

Use in-place modification when space complexity matters (O(1) space requirement) and the original array can be modified. Create a new array when you need to preserve the original data or when the problem involves significant restructuring that would complicate in-place logic.

How much time should I spend on a medium problem in interviews?

In a 45-minute interview, aim for: 5-10 minutes understanding and clarifying, 5-10 minutes discussing approach, 20-25 minutes coding, and 5-10 minutes testing. If stuck after 15 minutes of thinking, verbalize your thought process and ask for hints.

What is the importance of time and space complexity analysis?

Complexity analysis is crucial because: 1) Interviewers always ask about it, 2) It helps you choose between approaches, 3) It demonstrates CS fundamentals. Always state both time and space complexity, and be prepared to explain how you derived them.

What is the optimal approach for "Spiral Matrix III"?

For this medium-level Array problem, the key is to identify the core pattern. Analyze the input constraints to determine acceptable time complexity. Consider whether matrix techniques can optimize the solution. Always handle edge cases and validate your approach with examples before coding.

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