Introduction

In this blog post, we will explore how to solve the LeetCode problem "2271" using AI. LeetCode is a popular platform for preparing for coding interviews, and with the help of AI tools like Chatmagic, we can generate solutions quickly and efficiently - helping you pass the interviews and get the job offer without having to study for months.

Problem Statement

You are given a 2D integer array tiles where tiles[i] = [li, ri] represents that every tile j in the range li <= j <= ri 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 return 9. Note that there may be other places where the carpet covers 9 white tiles. It can be shown that the carpet cannot cover more than 9 white tiles. Example 2: Input: tiles = [[10,11],[1,1]], carpetLen = 2 Output: 2 Explanation: Place the carpet starting on tile 10. It covers 2 white tiles, so we return 2. Constraints: 1 <= tiles.length <= 5 * 104 tiles[i].length == 2 1 <= li <= ri <= 109 1 <= carpetLen <= 109 The tiles are non-overlapping.

Explanation

Here's a breakdown of the solution:

• Sort Tiles: Sort the tiles based on their starting positions to efficiently iterate through them.
• Sliding Window: Use a sliding window approach. The window represents the carpet's position. Slide the window across the tiles, calculating the covered white tiles in each position.

• Calculate Coverage: For each window position, efficiently calculate the number of covered white tiles by considering partial and full tile overlaps with the carpet.

• Runtime Complexity: O(N log N), where N is the number of tiles (due to sorting). Storage Complexity: O(1)

Code

    def maximumWhiteTiles(tiles, carpetLen):
    tiles.sort()
    n = len(tiles)
    max_covered = 0
    current_covered = 0
    left = 0
    for right in range(n):
        current_covered += tiles[right][1] - tiles[right][0] + 1
        while tiles[right][1] - tiles[left][0] + 1 > carpetLen:
            covered_by_left = tiles[left][1] - tiles[left][0] + 1
            overlap = max(0, tiles[right][1] - tiles[left][0] + 1 - carpetLen)
            current_covered -= (covered_by_left - overlap)
            left += 1
        max_covered = max(max_covered, current_covered)
    return max_covered