What is the time complexity of Maximum White Tiles Covered by a Carpet?

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

What data structures are used to solve Maximum White Tiles Covered by a Carpet?

This problem uses Array, Binary Search, Greedy, Sliding Window, Sorting, Prefix Sum. Understanding these concepts is essential for solving this Medium problem efficiently.

What are the key insights for solving Maximum White Tiles Covered by a Carpet?

Insight 1: Sorting the tiles allows us to efficiently use a sliding window approach. Insight 2: The optimal placement of the carpet ends on the right edge of one of the tiles. This significantly reduces the search space.

Which companies ask LeetCode 2271. Maximum White Tiles Covered by a Carpet in interviews?

This problem is frequently asked at LTI.

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Maximum White Tiles Covered by a Carpet - LeetCode 2271 Solution

Q: How to solve Maximum White Tiles Covered by a Carpet?

The solution uses a sliding window approach after sorting the tiles based on their starting positions. For each tile ending at `rj`, we consider placing the carpet so that its right end is at `rj`. This ensures that at least one of the tile endings is fully utilized by the carpet. We then slide a window from the beginning to find all the tiles that can be covered by the carpet. While doing this, we keep track of the current coverage. When a tile is only partially covered by the carpet, we calculate that partial coverage separately.

Maximum White Tiles Covered by a Carpet - Complete Solution Guide

Maximum White Tiles Covered by a Carpet is LeetCode problem 2271, 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 a 2D integer array tiles where tiles[i] = [l i , r i ] represents that every tile j in the range l i <= j <= r i is colored white. You are also given an integer carpetLen , the length of a single carpet that can be placed anywhere . Return the maximum number of white tiles that can be covered by the carpet . Example 1: Input: tiles = [[1,5],[10,11],[12,18],[20,25],[30,32]], carpetLen = 10 Output: 9 Explanation: Place the carpet starting on tile 10. It covers 9 white tiles, so we re

Detailed Explanation

The problem asks us to find the maximum number of white tiles that can be covered by a carpet of a given length. We are given a list of tile ranges, where each range [l, r] indicates that all tiles from l to r (inclusive) are white. We need to place the carpet such that it covers the maximum number of white tiles. Tiles are non-overlapping, which simplifies the problem. The carpet can be placed anywhere on the number line.

Solution Approach

The solution uses a sliding window approach after sorting the tiles based on their starting positions. For each tile ending at `rj`, we consider placing the carpet so that its right end is at `rj`. This ensures that at least one of the tile endings is fully utilized by the carpet. We then slide a window from the beginning to find all the tiles that can be covered by the carpet. While doing this, we keep track of the current coverage. When a tile is only partially covered by the carpet, we calculate that partial coverage separately.

Step-by-Step Algorithm

Step 1: Sort the `tiles` array based on the starting position of each tile (i.e., the first element of each tile array).
Step 2: Initialize `i` to 0 (left pointer), `current_coverage` to 0, and `max_coverage` to 0.
Step 3: Iterate through the sorted `tiles` array using a right pointer `j`.
Step 4: For each `j`, calculate `carpet_start = rj - carpetLen + 1`, where `rj` is the ending position of the current tile.
Step 5: Update `current_coverage` by adding the length of the current tile (`rj - lj + 1`).
Step 6: Move the left pointer `i` forward until `tiles[i][0] >= carpet_start`. While doing this, subtract the length of the tiles that are no longer fully covered (`ri - li + 1`) from `current_coverage`.
Step 7: Calculate `fully_covered_len` as the current `current_coverage` which represents coverage from `tiles[i]` to `tiles[j]`.
Step 8: Calculate `partial_coverage`. Check if `i > 0`. If so, determine how much of the tile at `tiles[i-1]` is covered by the carpet that starts at `carpet_start`, but ends at the current `rj`. Add that to the `partial_coverage`.
Step 9: Update `max_coverage = max(max_coverage, fully_covered_len + partial_coverage)`. This stores the maximum coverage we have seen so far.
Step 10: After iterating through all tiles, return `max_coverage`.

Key Insights

Insight 1: Sorting the tiles allows us to efficiently use a sliding window approach.
Insight 2: The optimal placement of the carpet ends on the right edge of one of the tiles. This significantly reduces the search space.
Insight 3: We need to handle partially covered tiles, meaning the carpet might not fully cover every tile within the sliding window.

Complexity Analysis

Time Complexity: O(n log n)

Space Complexity: O(1)

Topics

This problem involves: Array, Binary Search, Greedy, Sliding Window, Sorting, Prefix Sum.

Companies

Asked at: LTI.

Maximum White Tiles Covered by a Carpet | LeetCode 2271

Medium

Array Binary Search Greedy Sliding Window Sorting Prefix Sum

Time: O(n log n)Space: O(1)

Solution Code

Available in:

import java.util.Arrays;

class Solution {
    public int maximumWhiteTiles(int[][] tiles, int carpetLen) {
        Arrays.sort(tiles, (a, b) -> a[0] - b[0]);
        int n = tiles.length;
        
        int i = 0;
        int current_coverage = 0;
        int max_coverage = 0;
        
        for (int j = 0; j < n; j++) {
            int lj = tiles[j][0];
            int rj = tiles[j][1];
            current_coverage += (rj - lj + 1);
            
            // Consider a carpet that ends exactly at rj
            int carpet_start = rj - carpetLen + 1;
            
            // Shrink the window from the left by advancing pointer i.
            // After this loop, tiles[i] will be the first tile whose start
            // is >= carpet_start.
            while (i < n && tiles[i][0] < carpet_start) {
                int li = tiles[i][0];
                int ri = tiles[i][1];
                current_coverage -= (ri - li + 1);
                i++;
            }
            
            // At this point, current_coverage is the sum of lengths of tiles
            // in the window [i, j]. All these tiles start at or after carpet_start,
            // so they are fully covered by the carpet ending at rj.
            int fully_covered_len = current_coverage;
            
            // The tile at i-1 (if it exists) might be partially covered.
            int partial_coverage = 0;
            if (i > 0) {
                int prev_li = tiles[i-1][0];
                int prev_ri = tiles[i-1][1];
                if (prev_ri >= carpet_start) {
                    partial_coverage = prev_ri - carpet_start + 1;
                }
            }
            
            max_coverage = Math.max(max_coverage, fully_covered_len + partial_coverage);
            
        }
        
        return max_coverage;
    }
}

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Problem Info

Description

Problem 2271: Maximum White Tiles Covered by a Carpet - Detailed explanation and solution approaches.

Topics

Array Binary Search Greedy Sliding Window Sorting Prefix Sum

Companies

LTI

Stats

Difficulty:Medium

Languages:4

Companies:1

Problem #:2271

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: Sort the `tiles` array based on the starting position of each tile (i.e., the first element of each tile array).
2
Step 2: Initialize `i` to 0 (left pointer), `current_coverage` to 0, and `max_coverage` to 0.
3
Step 3: Iterate through the sorted `tiles` array using a right pointer `j`.
4
Step 4: For each `j`, calculate `carpet_start = rj - carpetLen + 1`, where `rj` is the ending position of the current tile.
5
Step 5: Update `current_coverage` by adding the length of the current tile (`rj - lj + 1`).
6
Step 6: Move the left pointer `i` forward until `tiles[i][0] >= carpet_start`. While doing this, subtract the length of the tiles that are no longer fully covered (`ri - li + 1`) from `current_coverage`.
7
Step 7: Calculate `fully_covered_len` as the current `current_coverage` which represents coverage from `tiles[i]` to `tiles[j]`.
8
Step 8: Calculate `partial_coverage`. Check if `i > 0`. If so, determine how much of the tile at `tiles[i-1]` is covered by the carpet that starts at `carpet_start`, but ends at the current `rj`. Add that to the `partial_coverage`.
9
Step 9: Update `max_coverage = max(max_coverage, fully_covered_len + partial_coverage)`. This stores the maximum coverage we have seen so far.
10
Step 10: After iterating through all tiles, return `max_coverage`.

💡

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 log n)

✨ Log-linear - Great for sorting algorithms!

💾

Space Complexity

O(1)

💪 Constant space - Uses same memory regardless of input!

📝

Why?

The time complexity is primarily determined by the sorting operation. The sliding window iterates through the tiles array once, so it is linear. The space complexity is constant because we only use a few integer variables to keep track of the pointers and coverage.

📖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

Prefix Sum with Binary Search

Calculate prefix sums of tile lengths. For each tile ending at rj, use binary search to find the starting tile that is within carpetLen distance. Use prefix sums to calculate coverage.

Time: O(n log n)Space: O(n)

Trade-offs: This approach also has O(n log n) time complexity due to binary search. It requires O(n) space for prefix sums. The sliding window approach is generally preferred due to its O(1) space complexity.

✨

Key Insights

1Insight 1: Sorting the tiles allows us to efficiently use a sliding window approach.
2Insight 2: The optimal placement of the carpet ends on the right edge of one of the tiles. This significantly reduces the search space.
3Insight 3: We need to handle partially covered tiles, meaning the carpet might not fully cover every tile within the sliding window.

📚

Key Terms Explained

For Students

Array

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

Binary Search

Halving the search space each step. Like finding a word in a dictionary!

Greedy Algorithm

Making the best choice at each step without looking ahead.

Sliding Window

A moving "window" over data to process subsets efficiently.

Sorting

Arranging elements in a specific order (ascending/descending).

Edge Cases

Edge case 1: `tiles` is empty. The algorithm will correctly return 0 since the loop won't execute.
Edge case 2: `carpetLen` is larger than the entire range of all tiles. The algorithm will correctly compute total length of all tiles in that case
Edge case 3: Tiles are already sorted. The sorting step will still execute, but its complexity is maintained.

⚠️

Common Mistakes to Avoid

✗Common mistake 1: Not sorting the tiles first. This will lead to incorrect results as the sliding window approach relies on sorted input.

✗Common mistake 2: Not handling partially covered tiles correctly. Forgetting to account for the portion of the first tile in the window that is not fully covered.

✗Common mistake 3: Incorrectly calculating the start position of the carpet. It should be `rj - carpetLen + 1` not `rj - carpetLen`.

✅

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: Resource allocation. Imagine a server farm and you want to allocate contiguous memory blocks of a fixed size (carpetLen) to maximize the amount of utilized memory (white tiles). The tiles represent used memory ranges.

🏦Real-world scenario 2: Scheduling. Imagine you are trying to schedule a meeting of a certain duration (`carpetLen`) and you have a set of available time slots (`tiles`). You want to maximize the amount of time covered by the meeting within the available slots.

💡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: September 20, 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.

Can binary search be applied to non-sorted arrays?

Binary search requires a monotonic property, not necessarily sorted data. It can be applied to any search space where you can determine if the answer is in the left or right half. Examples include searching rotated arrays or finding minimum in optimization problems.

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 "Maximum White Tiles Covered by a Carpet"?

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

🤖 Need help with this problem?