What is the time complexity of Spiral Matrix II?

The optimal solution has a time complexity of O(n^2).

What is the space complexity of Spiral Matrix II?

The space complexity is O(n^2).

What data structures are used to solve Spiral Matrix II?

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 II?

Insight 1: Simulating the spiral path by defining directions (right, down, left, up) and changing the direction when encountering a boundary or an already filled cell is crucial. Insight 2: Using a direction array (dr, dc) to represent the change in row and column indices for each direction simplifies the movement within the matrix.

Which companies ask LeetCode 59. Spiral Matrix II in interviews?

This problem is frequently asked at Adobe, Amazon, Bloomberg, Google, Meta and 5 more companies.

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Spiral Matrix II - LeetCode 59 Solution

Q: How to solve Spiral Matrix II?

The solution simulates the spiral traversal of the matrix. It initializes the matrix with zeros and maintains a current position (row, column) and a current direction. It iterates from 1 to n*n, filling each cell with the current number. After filling a cell, it calculates the next position based on the current direction. If the next position is out of bounds or already filled, it changes the direction to the next direction in the spiral (right -> down -> left -> up) before updating the current position.

Spiral Matrix II - Complete Solution Guide

Spiral Matrix II is LeetCode problem 59, 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

Given a positive integer n , generate an n x n matrix filled with elements from 1 to n 2 in spiral order. Example 1: Input: n = 3 Output: [[1,2,3],[8,9,4],[7,6,5]] Example 2: Input: n = 1 Output: [[1]] Constraints: 1 <= n <= 20

Detailed Explanation

The problem asks us to generate a square matrix of size `n x n` and fill it with numbers from 1 to `n^2` in a spiral order, starting from the top-left corner and moving clockwise. This means filling the first row from left to right, then the last column from top to bottom, then the last row from right to left, and finally the first column from bottom to top. This pattern repeats inwards until the entire matrix is filled.

Solution Approach

The solution simulates the spiral traversal of the matrix. It initializes the matrix with zeros and maintains a current position (row, column) and a current direction. It iterates from 1 to n*n, filling each cell with the current number. After filling a cell, it calculates the next position based on the current direction. If the next position is out of bounds or already filled, it changes the direction to the next direction in the spiral (right -> down -> left -> up) before updating the current position.

Step-by-Step Algorithm

Step 1: Initialize an n x n matrix with zeros.
Step 2: Initialize the row (r) and column (c) to 0 (top-left corner).
Step 3: Initialize the direction to 0 (right). The direction array (dr, dc) contains the row and column increments for each direction.
Step 4: Iterate from num = 1 to n*n.
Step 5: Assign the current number 'num' to the current cell matrix[r][c].
Step 6: Calculate the next row (next_r) and next column (next_c) based on the current direction.
Step 7: Check if the next cell (next_r, next_c) is within bounds (0 <= next_r < n and 0 <= next_c < n) and if it is empty (matrix[next_r][next_c] == 0).
Step 8: If the next cell is out of bounds or already filled, change the direction to the next direction in the spiral (direction = (direction + 1) % 4).
Step 9: Update the current row and column based on the current direction (r += dr[direction], c += dc[direction]).
Step 10: Repeat steps 5-9 until all cells are filled.

Key Insights

Insight 1: Simulating the spiral path by defining directions (right, down, left, up) and changing the direction when encountering a boundary or an already filled cell is crucial.
Insight 2: Using a direction array (dr, dc) to represent the change in row and column indices for each direction simplifies the movement within the matrix.
Insight 3: Keeping track of the current position (row, column) and updating it based on the current direction is key to the algorithm's success.

Complexity Analysis

Time Complexity: O(n^2)

Space Complexity: O(n^2)

Topics

This problem involves: Array, Matrix, Simulation.

Companies

Asked at: Adobe, Amazon, Bloomberg, Google, Meta, Microsoft, Roblox, TikTok, Uber, Zoho.

Spiral Matrix II | LeetCode 59

Medium

Array Matrix Simulation

Time: O(n^2)Space: O(n^2)

Solution Code

Available in:

class Solution {
    public int[][] generateMatrix(int n) {
        int[][] matrix = new int[n][n];
        int r = 0, c = 0, direction = 0;
        int[] dr = {0, 1, 0, -1};
        int[] dc = {1, 0, -1, 0};

        for (int num = 1; num <= n * n; num++) {
            matrix[r][c] = num;
            int next_r = r + dr[direction];
            int next_c = c + dc[direction];

            if (!(0 <= next_r && next_r < n && 0 <= next_c && next_c < n && matrix[next_r][next_c] == 0)) {
                direction = (direction + 1) % 4;
            }
            
            r += dr[direction];
            c += dc[direction];
        }

        return matrix;
    }
}

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Problem Info

Description

Problem 59: Spiral Matrix II - Detailed explanation and solution approaches.

Topics

Array Matrix Simulation

Companies

AdobeAmazonBloombergGoogleMetaMicrosoftRobloxTikTok

Stats

Difficulty:Medium

Languages:4

Companies:10

Problem #:59

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

10 steps

1
Step 1: Initialize an n x n matrix with zeros.
2
Step 2: Initialize the row (r) and column (c) to 0 (top-left corner).
3
Step 3: Initialize the direction to 0 (right). The direction array (dr, dc) contains the row and column increments for each direction.
4
Step 4: Iterate from num = 1 to n*n.
5
Step 5: Assign the current number 'num' to the current cell matrix[r][c].
6
Step 6: Calculate the next row (next_r) and next column (next_c) based on the current direction.
7
Step 7: Check if the next cell (next_r, next_c) is within bounds (0 <= next_r < n and 0 <= next_c < n) and if it is empty (matrix[next_r][next_c] == 0).
8
Step 8: If the next cell is out of bounds or already filled, change the direction to the next direction in the spiral (direction = (direction + 1) % 4).
9
Step 9: Update the current row and column based on the current direction (r += dr[direction], c += dc[direction]).
10
Step 10: Repeat steps 5-9 until all cells are filled.

💡

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(n^2)

🐢 Quadratic - Gets slow with large inputs. Consider optimization!

💾

Space Complexity

O(n^2)

📚 Quadratic space - Uses lots of memory!

📝

Why?

The algorithm visits each cell in the n x n matrix exactly once to fill it with a value. Therefore, the time complexity is proportional to the number of cells, which is n^2. The space complexity is also O(n^2) because we are creating an n x n matrix to store the spiral.

📖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

Layer-by-Layer Iteration

Instead of tracking the direction, you can iterate through the matrix layer by layer. In each layer, you fill the top row, right column, bottom row, and left column. The layer's start row and start column increment for each subsequent inner layer.

Time: O(n^2)Space: O(n^2)

Trade-offs: This approach avoids the direction array but might be slightly harder to implement correctly due to the multiple loops required for each layer. The time and space complexities remain the same.

✨

Key Insights

1Insight 1: Simulating the spiral path by defining directions (right, down, left, up) and changing the direction when encountering a boundary or an already filled cell is crucial.
2Insight 2: Using a direction array (dr, dc) to represent the change in row and column indices for each direction simplifies the movement within the matrix.
3Insight 3: Keeping track of the current position (row, column) and updating it based on the current direction is key to the algorithm's success.

📚

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: n = 1. The algorithm correctly handles this case by creating a 1x1 matrix and filling it with the value 1.
Edge case 2: Input validation is not strictly necessary since the constraint 1 <= n <= 20. The core logic handles filling the matrix correctly regardless within that range. If n was 0, the code would still execute without error, returning an empty matrix. But it would violate the problem constraints.

⚠️

Common Mistakes to Avoid

✗Common mistake 1: Incorrectly updating the direction, leading to an infinite loop or out-of-bounds access. The modulo operator (%) is essential to cycle through the directions correctly.

✗Common mistake 2: Failing to check for out-of-bounds or already-filled cells before updating the current position, resulting in overwriting existing values or accessing memory outside the matrix bounds.

✗Common mistake 3: Off-by-one errors when calculating the indices for filling each side of the spiral. Careful attention to the start and end indices is required.

✅

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: Image processing - certain image compression or encryption algorithms might use spiral traversal of pixel data.

🏦Real-world scenario 2: Memory allocation in certain embedded systems might follow a spiral pattern for improved locality of reference in specific scenarios.

💡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

Consider multiple approaches before coding. Mentioning alternatives shows breadth of knowledge.

⚠️ Watch Out

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

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: July 30, 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

In a real interview, always clarify constraints first. Ask about input size and value ranges.

@TechLeadJul 13

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

@DevExpert92Jul 6

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

@AlgoMasterJun 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 II"?

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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