What is the time complexity of Rotating the Box?

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

What is the space complexity of Rotating the Box?

The space complexity is O(m*n).

What data structures are used to solve Rotating the Box?

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

What are the key insights for solving Rotating the Box?

Insight 1: The rotation can be achieved by transposing the matrix and then reversing each row. Insight 2: Applying gravity can be simulated by iterating through each column and tracking the lowest available position for a stone to fall.

Which companies ask LeetCode 1861. Rotating the Box in interviews?

This problem is frequently asked at Block, Capital One, Commvault, Roblox, Visa.

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Rotating the Box - LeetCode 1861 Solution

Rotating the Box - Complete Solution Guide

Rotating the Box is LeetCode problem 1861, 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 are given an m x n matrix of characters boxGrid representing a side-view of a box. Each cell of the box is one of the following: A stone '#' A stationary obstacle '*' Empty '.' The box is rotated 90 degrees clockwise , causing some of the stones to fall due to gravity. Each stone falls down until it lands on an obstacle, another stone, or the bottom of the box. Gravity does not affect the obstacles' positions, and the inertia from the box's rotation does not affect the stones' horizontal pos

Detailed Explanation

The problem asks us to simulate rotating a 2D grid (representing a box) by 90 degrees clockwise and then applying gravity, causing stones ('#') to fall to the bottom of the box until they encounter an obstacle ('*'), another stone, or the bottom edge. We need to return the resulting grid after this rotation and gravity simulation. The input is a 2D array of characters, and the output is a new 2D array of characters representing the rotated and adjusted box.

Solution Approach

The solution involves two main steps: first, rotating the input grid 90 degrees clockwise, and second, applying gravity to the rotated grid. The rotation is done by creating a new grid with dimensions transposed and reversing each row. Gravity is applied by iterating through the columns of the rotated grid. For each column, a `write_row` variable tracks the lowest empty position where a stone can fall. As we iterate upwards, if we encounter a stone, we move it to the `write_row` position and update the `write_row`. If we encounter an obstacle, it blocks the fall, and we update `write_row` to be right above the obstacle.

Step-by-Step Algorithm

Step 1: Determine the dimensions of the input `boxGrid` (m rows, n columns).
Step 2: Create a new `rotated_box` grid with dimensions n x m. This is where the rotated result will be stored.
Step 3: Populate `rotated_box` by transposing `boxGrid` and reversing each row. Specifically, `rotated_box[j][m-1-i] = boxGrid[i][j]` does the rotation efficiently.
Step 4: Iterate through each column `c` of `rotated_box` (from 0 to m-1).
Step 5: For each column `c`, initialize `write_row` to the bottom row (`n-1`). `write_row` tracks where the next stone should be placed due to gravity.
Step 6: Iterate upwards through each row `r` in column `c` (from n-1 to 0).
Step 7: If `rotated_box[r][c]` is an obstacle ('*'), update `write_row` to `r-1`. The obstacle blocks the fall of any stones below.
Step 8: If `rotated_box[r][c]` is a stone ('#'), move it to the `write_row` position. Set `rotated_box[r][c]` to '.' and `rotated_box[write_row][c]` to '#', and then decrement `write_row`.
Step 9: Return the `rotated_box`.

Key Insights

Insight 1: The rotation can be achieved by transposing the matrix and then reversing each row.
Insight 2: Applying gravity can be simulated by iterating through each column and tracking the lowest available position for a stone to fall.
Insight 3: Obstacles ('*') remain in place and act as stopping points for the falling stones.

Complexity Analysis

Time Complexity: O(m*n)

Space Complexity: O(m*n)

Topics

This problem involves: Array, Two Pointers, Matrix.

Companies

Asked at: Block, Capital One, Commvault, Roblox, Visa.

Rotating the Box | LeetCode 1861

Medium

Array Two Pointers Matrix

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

Solution Code

Available in:

class Solution {
    public char[][] rotateTheBox(char[][] boxGrid) {
        int m = boxGrid.length;
        int n = boxGrid[0].length;

        // Step 1: Rotate the box 90 degrees clockwise.
        // This is equivalent to transposing and then reversing each new row.
        // The new grid will have dimensions n x m.
        char[][] rotated_box = new char[n][m];
        for (int i = 0; i < m; i++) {
            for (int j = 0; j < n; j++) {
                rotated_box[j][m - 1 - i] = boxGrid[i][j];
            }
        }


        // Get dimensions of the rotated box.
        int rows = n;
        int cols = m;

        // Step 2: Apply gravity. Stones fall downwards in each column.
        for (int c = 0; c < cols; c++) {
            // 'write_row' points to the lowest available spot for a stone.
            // It starts at the bottom of the column.
            int write_row = rows - 1;

            // Iterate upwards from the bottom of the current column.
            for (int r = rows - 1; r >= 0; r--) {
                if (rotated_box[r][c] == '*') {
                    // An obstacle blocks the fall. The new "floor" is right above it.
                    write_row = r - 1;
                } else if (rotated_box[r][c] == '#') {
                    // A stone is found.
                    // Move the stone to the available 'write_row' position.
                    // First, make the current position empty.
                    rotated_box[r][c] = '.';
                    // Then place the stone at the landing spot.
                    rotated_box[write_row][c] = '#';
                    // The next stone will land on top of the current one.
                    write_row--;
                }
            }
        }

        return rotated_box;
    }
}

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Problem Info

Description

Problem 1861: Rotating the Box - Detailed explanation and solution approaches.

Topics

Array Two Pointers Matrix

Companies

BlockCapital OneCommvaultRobloxVisa

Stats

Difficulty:Medium

Languages:4

Companies:5

Problem #:1861

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 of the input `boxGrid` (m rows, n columns).
2
Step 2: Create a new `rotated_box` grid with dimensions n x m. This is where the rotated result will be stored.
3
Step 3: Populate `rotated_box` by transposing `boxGrid` and reversing each row. Specifically, `rotated_box[j][m-1-i] = boxGrid[i][j]` does the rotation efficiently.
4
Step 4: Iterate through each column `c` of `rotated_box` (from 0 to m-1).
5
Step 5: For each column `c`, initialize `write_row` to the bottom row (`n-1`). `write_row` tracks where the next stone should be placed due to gravity.
6
Step 6: Iterate upwards through each row `r` in column `c` (from n-1 to 0).
7
Step 7: If `rotated_box[r][c]` is an obstacle ('*'), update `write_row` to `r-1`. The obstacle blocks the fall of any stones below.
8
Step 8: If `rotated_box[r][c]` is a stone ('#'), move it to the `write_row` position. Set `rotated_box[r][c]` to '.' and `rotated_box[write_row][c]` to '#', and then decrement `write_row`.
9
Step 9: Return the `rotated_box`.

💡

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 algorithm iterates through each cell of the input grid twice: once for rotation and once for applying gravity. Both these operations take O(m*n) time. A new grid of size n x m is created to store the rotated box, leading to O(m*n) 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

In-place rotation

Attempt to rotate the matrix in-place. This is not practical here because the input and output matrix dimensions are different (m x n becomes n x m). An in-place transpose would be possible if m==n, but additional logic would be needed for gravity, further complicating things.

Time: Difficult to achieve O(m*n) for both rotation and gravity in-placeSpace: O(1) if in-place rotation was achievable

Trade-offs: It's generally not recommended for this specific problem due to the varying dimensions and gravity application requirement, making it overly complex and prone to errors.

✨

Key Insights

1Insight 1: The rotation can be achieved by transposing the matrix and then reversing each row.
2Insight 2: Applying gravity can be simulated by iterating through each column and tracking the lowest available position for a stone to fall.
3Insight 3: Obstacles ('*') remain in place and act as stopping points for the falling stones.

📚

Key Terms Explained

For Students

Array

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

Two Pointers

Using two indices to traverse data from different positions simultaneously.

Edge Cases

Edge case 1: Empty boxGrid. The solution will create an empty rotated box, which is the correct behavior.
Edge case 2: A boxGrid containing only obstacles. The algorithm will execute without errors, correctly returning the rotated box filled with obstacles.
Edge case 3: A boxGrid containing only empty cells. The algorithm will execute without errors, correctly returning the rotated box filled with empty cells.
Edge case 4: A boxGrid where a stone is already at the bottom. The algorithm places it in the same location, which is the correct behavior.

⚠️

Common Mistakes to Avoid

✗Common mistake 1: Incorrectly implementing the rotation logic (e.g., not reversing the rows after transposing).

✗Common mistake 2: Not handling the obstacles correctly when applying gravity, causing stones to fall through them.

✗Common mistake 3: Forgetting to set the original stone position to '.' after moving it due to gravity.

✗Common mistake 4: Incorrectly calculating the new `write_row` position or not updating it properly.

✅

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: Game development: Simulating physics and object movements in a 2D environment, such as blocks falling in a puzzle game.

🏦Real-world scenario 2: Logistics and warehouse management: Optimizing the arrangement of items in a container or on shelves based on gravity and other constraints.

💡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

If stuck, try working through a small example by hand. The pattern often reveals itself.

⚠️ 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: August 1, 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

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

@CodeNinjaJun 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 do I avoid infinite loops with two pointers?

Always ensure your loop condition guarantees termination (typically left < right or slow != fast). Move at least one pointer in each iteration, and handle edge cases where pointers might not move (like skipping duplicates). Draw out pointer movement on paper first.

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 "Rotating the Box"?

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 two pointers techniques can optimize the solution. Always handle edge cases and validate your approach with examples before coding.

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