What is the time complexity of Spiral Matrix?

The optimal solution has a time complexity of O(m*n).

What is the space complexity of Spiral Matrix?

The space complexity is O(m*n).

What data structures are used to solve Spiral Matrix?

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?

Recognizing the spiral traversal can be broken down into four directional traversals: right, down, left, and up. Maintaining boundaries (top, bottom, left, right) to keep track of the current layer of the spiral.

Which companies ask LeetCode 54. Spiral Matrix in interviews?

This problem is frequently asked at AMD, Accenture, Adobe, Amazon, Anduril and 40 more companies.

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Spiral Matrix - LeetCode 54 Solution

Q: How to solve Spiral Matrix?

The solution uses a boundary-based approach. It initializes `top`, `bottom`, `left`, and `right` pointers to represent the boundaries of the current spiral layer. It then iterates through the matrix in a spiral order by moving right, then down, then left, and then up. After each directional traversal, the corresponding boundary is updated to move inwards. The iteration continues until either `left > right` or `top > bottom`, indicating that all elements have been traversed.

Spiral Matrix - Complete Solution Guide

Spiral Matrix is LeetCode problem 54, 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 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

Detailed Explanation

The problem asks us to traverse a given m x n matrix in a spiral order, starting from the outermost elements and moving inwards layer by layer, and return a list of elements in that order. For example, given a 3x3 matrix [[1,2,3],[4,5,6],[7,8,9]], the spiral order is [1,2,3,6,9,8,7,4,5].

Solution Approach

The solution uses a boundary-based approach. It initializes `top`, `bottom`, `left`, and `right` pointers to represent the boundaries of the current spiral layer. It then iterates through the matrix in a spiral order by moving right, then down, then left, and then up. After each directional traversal, the corresponding boundary is updated to move inwards. The iteration continues until either `left > right` or `top > bottom`, indicating that all elements have been traversed.

Step-by-Step Algorithm

Step 1: Initialize `top` to 0, `bottom` to `m-1`, `left` to 0, and `right` to `n-1`, where `m` is the number of rows and `n` is the number of columns in the matrix.
Step 2: While `left <= right` and `top <= bottom`, perform the following steps:
Step 3: Traverse from left to right along the `top` row and add the elements to the result list. Increment `top`.
Step 4: Traverse from top to bottom along the `right` column and add the elements to the result list. Decrement `right`.
Step 5: If `top <= bottom`, traverse from right to left along the `bottom` row and add the elements to the result list. Decrement `bottom`.
Step 6: If `left <= right`, traverse from bottom to top along the `left` column and add the elements to the result list. Increment `left`.
Step 7: Repeat steps 3-6 until the condition in step 2 is no longer met.
Step 8: Return the result list.

Key Insights

Recognizing the spiral traversal can be broken down into four directional traversals: right, down, left, and up.
Maintaining boundaries (top, bottom, left, right) to keep track of the current layer of the spiral.
Updating the boundaries after each traversal to move inwards to the next layer.
Handling edge cases where the matrix might not be a square (m != n), and stopping the iteration when no more elements are left to traverse.

Complexity Analysis

Time Complexity: O(m*n)

Space Complexity: O(m*n)

Topics

This problem involves: Array, Matrix, Simulation.

Companies

Asked at: AMD, Accenture, Adobe, Amazon, Anduril, Apple, Bloomberg, Capital One, Cisco, CrowdStrike, DE Shaw, Darwinbox, Databricks, Deutsche Bank, Epic Systems, Goldman Sachs, IBM, Infosys, Intuit, J.P. Morgan, Lenskart, Meta, Microsoft, Netflix, Nordstrom, Nutanix, Nvidia, Oracle, PayPal, RBC, Roblox, SIG, Tekion, TikTok, Trexquant, Turing, Uber, Visa, Walmart Labs, WatchGuard, Wells Fargo, Yahoo, Zoho, eBay, tcs.

Spiral Matrix | LeetCode 54

Medium

Array Matrix Simulation

Time: O(m*n)Space: O(m*n)

Solution Code

Available in:

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

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

        int m = matrix.length;
        int n = matrix[0].length;
        List<Integer> result = new ArrayList<>();
        int top = 0, bottom = m - 1;
        int left = 0, right = n - 1;

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

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

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

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

        return result;
    }
}

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Problem Info

Description

Problem 54: Spiral Matrix - Detailed explanation and solution approaches.

Topics

Array Matrix Simulation

Companies

AMDAccentureAdobeAmazonAndurilAppleBloombergCapital One

Stats

Difficulty:Medium

Languages:4

Companies:45

Problem #:54

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

8 steps

1
Step 1: Initialize `top` to 0, `bottom` to `m-1`, `left` to 0, and `right` to `n-1`, where `m` is the number of rows and `n` is the number of columns in the matrix.
2
Step 2: While `left <= right` and `top <= bottom`, perform the following steps:
3
Step 3: Traverse from left to right along the `top` row and add the elements to the result list. Increment `top`.
4
Step 4: Traverse from top to bottom along the `right` column and add the elements to the result list. Decrement `right`.
5
Step 5: If `top <= bottom`, traverse from right to left along the `bottom` row and add the elements to the result list. Decrement `bottom`.
6
Step 6: If `left <= right`, traverse from bottom to top along the `left` column and add the elements to the result list. Increment `left`.
7
Step 7: Repeat steps 3-6 until the condition in step 2 is no longer met.
8
Step 8: 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(m*n)

💾

Space Complexity

O(m*n)

📝

Why?

Each element in the matrix is added to the result exactly once. Therefore, the number of operations is proportional to the number of elements in the matrix (m*n), leading to O(m*n) time complexity. We are only using a few integer variables to store the boundaries, so the space complexity is O(1).

📖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

Recursion

A recursive approach can be used where we process the outer layer and then recursively call the function for the inner matrix. This can be done by creating a smaller submatrix and passing it into the recursive call.

Time: O(m*n)Space: O(m*n) due to the recursion stack in the worst case (e.g., when m == n and is large), and the need to copy the submatrix in each recursive call.

Trade-offs: The recursive approach is generally less efficient due to the overhead of function calls and submatrix creation. The iterative approach is generally preferred due to its better space complexity.

✨

Key Insights

1Recognizing the spiral traversal can be broken down into four directional traversals: right, down, left, and up.
2Maintaining boundaries (top, bottom, left, right) to keep track of the current layer of the spiral.
3Updating the boundaries after each traversal to move inwards to the next layer.
4Handling edge cases where the matrix might not be a square (m != n), and stopping the iteration when no more elements are left to traverse.

📚

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: Empty matrix (matrix is null or has no rows or columns). The code handles this case by returning an empty list immediately.
Edge case 2: Matrix with only one row or one column. The algorithm correctly handles this case due to the `while` loop condition and boundary updates. The algorithm handles rectangular matrices (m != n) without issue.

⚠️

Common Mistakes to Avoid

✗Common mistake 1: Incorrectly updating the boundary indices after each traversal, leading to infinite loops or incorrect output. It's crucial to update the boundaries (top, bottom, left, right) in the correct order and direction.

✗Common mistake 2: Forgetting to check `top <= bottom` and `left <= right` before traversing the bottom row and the left column. If these checks are omitted, the same elements can be added to the result multiple times, leading to an incorrect output.

✗Common mistake 3: Not handling the empty matrix edge case.

✅

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. Spiral order traversal can be used for specific image processing tasks where data needs to be accessed in a spiral pattern, such as applying certain filters or compression algorithms.

🏦Real-world scenario 2: Data compression. Some compression algorithms may benefit from processing data in a spiral order for better locality or pattern detection.

💡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.

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 25, 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 14

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

@DevExpert92Jul 7

❓

Frequently Asked Questions

What is the most efficient way to solve array problems?

Most array problems can be solved efficiently using techniques like two-pointer approach, sliding window, or hash maps. The key is identifying whether the array is sorted (enabling binary search) or if tracking seen elements (using hash maps) can help achieve O(n) time complexity.

What techniques are commonly tested in medium-difficulty problems?

Medium problems often require combining multiple techniques: two-pointer with hash maps, BFS/DFS with state tracking, dynamic programming with optimization, or divide-and-conquer approaches. They test your ability to recognize patterns and choose the right combination of techniques.

How should I practice algorithm problems effectively?

Focus on understanding patterns rather than memorizing solutions. After solving a problem, review optimal solutions and understand the intuition. Group similar problems to recognize patterns. Practice explaining your approach out loud. Review problems after a few days to test retention.

What is the optimal approach for "Spiral Matrix"?

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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