What is the time complexity of Reshape the Matrix?

The optimal solution has a time complexity of O(r*c).

What is the space complexity of Reshape the Matrix?

The space complexity is O(r*c).

What data structures are used to solve Reshape the 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 Reshape the Matrix?

Insight 1: The reshaping is possible only if the product of the original matrix dimensions (m * n) equals the product of the desired reshaped matrix dimensions (r * c). Insight 2: The elements of the original matrix need to be traversed in row-major order and mapped to the new matrix also in row-major order.

Which companies ask LeetCode 566. Reshape the Matrix in interviews?

This problem is frequently asked at MathWorks.

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Reshape the Matrix - LeetCode 566 Solution

Reshape the Matrix - Complete Solution Guide

Reshape the Matrix is LeetCode problem 566, a Easy level challenge. This complete guide provides step-by-step explanations, multiple solution approaches, and optimized code in python3, java, cpp, c.

Problem Statement

In MATLAB, there is a handy function called reshape which can reshape an m x n matrix into a new one with a different size r x c keeping its original data. You are given an m x n matrix mat and two integers r and c representing the number of rows and the number of columns of the wanted reshaped matrix. The reshaped matrix should be filled with all the elements of the original matrix in the same row-traversing order as they were. If the reshape operation with given parameters is possible and lega

Detailed Explanation

The problem asks us to reshape a given matrix into a new matrix with specified dimensions (number of rows `r` and number of columns `c`). The reshaping operation should maintain the original data in a row-traversing order. If the reshaping is possible (i.e., the number of elements in the original matrix equals the number of elements in the reshaped matrix), return the new reshaped matrix. Otherwise, return the original matrix. The input is a 2D integer array `mat` and two integers `r` and `c`. The output is either the reshaped matrix or the original matrix based on whether the reshaping is valid.

Solution Approach

The solution first checks if the reshaping is possible by comparing the product of the dimensions of the original matrix and the desired matrix. If it's not possible, the original matrix is returned. Otherwise, a new matrix with the desired dimensions is created. The elements of the original matrix are then traversed in row-major order, and each element is placed into the new matrix in row-major order as well. The conversion of the row-major index in the original matrix to the row and column indices for placement in the reshaped matrix is performed with integer division and modulo operator.

Step-by-Step Algorithm

Step 1: Determine the dimensions (m, n) of the input matrix `mat`.
Step 2: Check if the reshaping is valid: `if m * n != r * c`, return the original matrix.
Step 3: Create a new matrix `res` with dimensions (r, c).
Step 4: Initialize a counter `k` to 0. This represents the current element index.
Step 5: Iterate through the rows of the reshaped matrix from `i = 0` to `r - 1`.
Step 6: Iterate through the columns of the reshaped matrix from `j = 0` to `c - 1`.
Step 7: Place the element from the original matrix at index `k` into `res[i][j]`. Calculate the original row and column indices using `mat[k // n][k % n]`.
Step 8: Increment `k` to move to the next element.
Step 9: After all elements are processed, return the reshaped matrix `res`.

Key Insights

Insight 1: The reshaping is possible only if the product of the original matrix dimensions (m * n) equals the product of the desired reshaped matrix dimensions (r * c).
Insight 2: The elements of the original matrix need to be traversed in row-major order and mapped to the new matrix also in row-major order.
Insight 3: A single index can be used to traverse the matrix by calculating row and column indices using division and modulo operations.

Complexity Analysis

Time Complexity: O(r*c)

Space Complexity: O(r*c)

Topics

This problem involves: Array, Matrix, Simulation.

Companies

Asked at: MathWorks.

Reshape the Matrix | LeetCode 566

Easy

Array Matrix Simulation

Time: O(r*c)Space: O(r*c)

Solution Code

Available in:

class Solution {
    public int[][] matrixReshape(int[][] mat, int r, int c) {
        int m = mat.length;
        int n = mat[0].length;
        if (m * n != r * c) {
            return mat;
        }
        int[][] reshapedMat = new int[r][c];
        int row = 0;
        int col = 0;
        for (int i = 0; i < m; i++) {
            for (int j = 0; j < n; j++) {
                reshapedMat[row][col] = mat[i][j];
                col++;
                if (col == c) {
                    col = 0;
                    row++;
                }
            }
        }
        return reshapedMat;
    }
}

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Problem Info

Description

Problem 566: Reshape the Matrix - Detailed explanation and solution approaches.

Topics

Array Matrix Simulation

Companies

MathWorks

Stats

Difficulty:Easy

Languages:4

Companies:1

Problem #:566

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

9 steps

1
Step 1: Determine the dimensions (m, n) of the input matrix `mat`.
2
Step 2: Check if the reshaping is valid: `if m * n != r * c`, return the original matrix.
3
Step 3: Create a new matrix `res` with dimensions (r, c).
4
Step 4: Initialize a counter `k` to 0. This represents the current element index.
5
Step 5: Iterate through the rows of the reshaped matrix from `i = 0` to `r - 1`.
6
Step 6: Iterate through the columns of the reshaped matrix from `j = 0` to `c - 1`.
7
Step 7: Place the element from the original matrix at index `k` into `res[i][j]`. Calculate the original row and column indices using `mat[k // n][k % n]`.
8
Step 8: Increment `k` to move to the next element.
9
Step 9: After all elements are processed, return the reshaped matrix `res`.

💡

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(r*c)

💾

Space Complexity

O(r*c)

📝

Why?

The algorithm iterates through all the elements of the new matrix (r rows and c columns) exactly once, performing constant-time operations for each element. Therefore, the time complexity is O(r * c). A new matrix with r rows and c columns is created to store the reshaped result, which leads to O(r * c) space complexity.

📖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 1D array for intermediate storage

First, flatten the original matrix into a 1D array. Then, iterate through the 1D array and populate the new matrix.

Time: O(m * n), equivalent to O(r * c) when reshaping is valid.Space: O(m * n), equivalent to O(r * c) when reshaping is valid.

Trade-offs: This approach might be more readable for some, as it separates the flattening and reshaping processes. However, it doesn't offer significant performance advantages since it still needs to iterate through all the elements.

✨

Key Insights

1Insight 1: The reshaping is possible only if the product of the original matrix dimensions (m * n) equals the product of the desired reshaped matrix dimensions (r * c).
2Insight 2: The elements of the original matrix need to be traversed in row-major order and mapped to the new matrix also in row-major order.
3Insight 3: A single index can be used to traverse the matrix by calculating row and column indices using division and modulo operations.

📚

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: The product of original matrix's dimensions is not equal to the product of the new matrix's dimensions. In this case, the original matrix is returned, as reshaping is not possible.
Edge case 2: Empty input matrix. While the constraints specify m, n >= 1, handling an empty matrix (m = 0 or n = 0) gracefully would be a good practice. The given code will likely throw an exception due to accessing mat[0].length when m is 0. An explicit check for an empty matrix could be added.

⚠️

Common Mistakes to Avoid

✗Common mistake 1: Not checking if the reshaping operation is valid (i.e., m * n == r * c). This will lead to incorrect results or exceptions.

✗Common mistake 2: Incorrectly calculating the indices of the original matrix's elements when mapping them to the new matrix. Off-by-one errors are common here when using modulo and division operators.

✅

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 where images (represented as matrices) need to be resized or reshaped for different display resolutions or input requirements of machine learning models.

🏦Real-world scenario 2: Data manipulation in databases where data is stored in a tabular format and needs to be reshaped for analysis or reporting purposes. For example, pivoting tables.

💡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

Don't rush. Explain your thought process clearly. Easy problems are often about communication.

⚠️ 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: June 27, 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

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

@AlgoMasterJul 13

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

@InterviewProJul 6

❓

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.

How should I approach easy-level algorithm problems?

Easy problems build foundational skills. Focus on: 1) Understanding the problem completely before coding, 2) Writing clean, readable code, 3) Handling all edge cases (empty input, single element, etc.), 4) Analyzing time and space complexity. Don't skip complexity analysis even for simple solutions.

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 "Reshape the 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?