What is the time complexity of Transpose Matrix?

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

What is the space complexity of Transpose Matrix?

The space complexity is O(m*n).

What data structures are used to solve Transpose Matrix?

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

What are the key insights for solving Transpose Matrix?

Insight 1: The dimensions of the transposed matrix are reversed compared to the original matrix (rows become columns, and columns become rows). Insight 2: To compute the transpose, we need to iterate through the original matrix and copy each element to its corresponding position in the transposed matrix.

Which companies ask LeetCode 867. Transpose Matrix in interviews?

This problem is frequently asked at ConsultAdd, Verkada.

JavaScript is required for the full interactive experience. However, all content is accessible without JavaScript.

Transpose Matrix - LeetCode 867 Solution

Q: How to solve Transpose Matrix?

The solution involves creating a new matrix with the dimensions swapped (number of rows = number of columns of original matrix, and number of columns = number of rows of original matrix). Then, we iterate through the original matrix, and for each element matrix[i][j], we place it in the transposed matrix at position result[j][i]. Finally, we return the newly created transposed matrix.

Transpose Matrix - Complete Solution Guide

Transpose Matrix is LeetCode problem 867, a Easy level challenge. This complete guide provides step-by-step explanations, multiple solution approaches, and optimized code in python3.

Problem Statement

Given a 2D integer array matrix , return the transpose of matrix . The transpose of a matrix is the matrix flipped over its main diagonal, switching the matrix's row and column indices. Example 1: Input: matrix = [[1,2,3],[4,5,6],[7,8,9]] Output: [[1,4,7],[2,5,8],[3,6,9]] Example 2: Input: matrix = [[1,2,3],[4,5,6]] Output: [[1,4],[2,5],[3,6]] Constraints: m == matrix.length n == matrix[i].length 1 <= m, n <= 1000 1 <= m * n <= 10 5 -10 9 <= matrix[i][j] <= 10 9

Detailed Explanation

The problem requires us to find the transpose of a given matrix. The transpose of a matrix is obtained by swapping its rows and columns. Essentially, the element at position (i, j) in the original matrix will be at position (j, i) in the transposed matrix. We are given a 2D integer array (matrix) as input, and we must return a new 2D integer array representing the transpose of the input matrix. The constraints specify the dimensions and element values of the matrix.

Solution Approach

The solution involves creating a new matrix with the dimensions swapped (number of rows = number of columns of original matrix, and number of columns = number of rows of original matrix). Then, we iterate through the original matrix, and for each element matrix[i][j], we place it in the transposed matrix at position result[j][i]. Finally, we return the newly created transposed matrix.

Step-by-Step Algorithm

Step 1: Determine the number of rows (rows) and columns (cols) of the input matrix.
Step 2: Create a new matrix called 'result' with dimensions cols x rows (swapped dimensions of the original matrix). Initialize it with zeros.
Step 3: Iterate through the original matrix using nested loops. The outer loop iterates through rows (from 0 to rows - 1), and the inner loop iterates through columns (from 0 to cols - 1).
Step 4: Inside the inner loop, assign the value of matrix[i][j] to result[j][i].
Step 5: After iterating through all elements of the original matrix, return the 'result' matrix, which now holds the transpose of the original matrix.

Key Insights

Insight 1: The dimensions of the transposed matrix are reversed compared to the original matrix (rows become columns, and columns become rows).
Insight 2: To compute the transpose, we need to iterate through the original matrix and copy each element to its corresponding position in the transposed matrix.
Insight 3: A new matrix needs to be created to store the transposed result since we cannot simply transpose in place for non-square matrices, as dimensions change.

Complexity Analysis

Time Complexity: O(m*n)

Space Complexity: O(m*n)

Topics

This problem involves: Array, Matrix, Simulation.

Companies

Asked at: ConsultAdd, Verkada.

Transpose Matrix | LeetCode 867

Easy

Array Matrix Simulation

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

Solution Code

Available in:

View Full Solution on GitHubGitHub Solution View on LeetCodeLeetCode

Problem Info

Description

Problem 867: Transpose Matrix - Detailed explanation and solution approaches.

Topics

Array Matrix Simulation

Companies

ConsultAddVerkada

Stats

Difficulty:Easy

Languages:2

Companies:2

Problem #:867

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

5 steps

1
Step 1: Determine the number of rows (rows) and columns (cols) of the input matrix.
2
Step 2: Create a new matrix called 'result' with dimensions cols x rows (swapped dimensions of the original matrix). Initialize it with zeros.
3
Step 3: Iterate through the original matrix using nested loops. The outer loop iterates through rows (from 0 to rows - 1), and the inner loop iterates through columns (from 0 to cols - 1).
4
Step 4: Inside the inner loop, assign the value of matrix[i][j] to result[j][i].
5
Step 5: After iterating through all elements of the original matrix, return the 'result' matrix, which now holds the transpose of the original matrix.

💡

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?

The time complexity is O(m*n) because we iterate through each element of the input matrix once, where 'm' is the number of rows and 'n' is the number of columns. The space complexity is also O(m*n) because we create a new matrix to store the transposed matrix, which has dimensions n x m, effectively requiring space proportional to m*n.

📖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

In-place transpose (for square matrices only)

For square matrices (m == n), it's possible to transpose in-place by swapping elements across the main diagonal. This avoids the O(m*n) space complexity of creating a new matrix.

Time: O(n^2)Space: O(1)

Trade-offs: This is space-efficient but only works for square matrices. If the matrix is not square, a new matrix needs to be created. The provided solution is more general as it works for all matrix types.

✨

Key Insights

1Insight 1: The dimensions of the transposed matrix are reversed compared to the original matrix (rows become columns, and columns become rows).
2Insight 2: To compute the transpose, we need to iterate through the original matrix and copy each element to its corresponding position in the transposed matrix.
3Insight 3: A new matrix needs to be created to store the transposed result since we cannot simply transpose in place for non-square matrices, as dimensions change.

📚

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: The code gracefully handles an empty matrix (if rows or cols is 0) by returning an empty list.
Edge case 2: Non-square matrix: The algorithm works correctly for both square and non-square matrices, as the transposed matrix is created with the correct dimensions in all cases.

⚠️

Common Mistakes to Avoid

✗Common mistake 1: Trying to transpose in-place for non-square matrices. This leads to incorrect results or out-of-bounds errors.

✗Common mistake 2: Incorrectly calculating the dimensions of the transposed matrix, leading to index errors or incorrect output. Ensure the dimensions are swapped correctly (cols x rows).

✅

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: Transposing matrices is often used in image processing for various operations like rotation or mirroring.

🏦Real-world scenario 2: Data analysis: In data analysis, transposing matrices can be useful for reorienting data for analysis or visualization, especially when rows and columns represent different entities (e.g., users and products in a recommendation system).

💡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

Even for easy problems, discuss time and space complexity. Interviewers want to see you think analytically.

⚠️ 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: October 26, 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

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

@InterviewProJul 13

Try solving this without using extra space first. It's a great exercise for understanding the underlying pattern.

@CodeNinjaJul 6

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

@TechLeadJun 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.

What should I learn from easy problems?

Easy problems introduce core patterns that appear in harder problems. Master basic operations (iteration, conditionals, simple data structures), recognize common patterns (counting, searching, basic transformations), and practice explaining your thought process clearly.

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 "Transpose Matrix"?

For this easy-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.

🤖 Need help with this problem?