LeetCode #2271 — MEDIUM

Maximum White Tiles Covered by a Carpet

Move from brute-force thinking to an efficient approach using array strategy.

Solve on LeetCode

The Problem

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 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 <= l_i <= r_i <= 10⁹
1 <= carpetLen <= 10⁹
The tiles are non-overlapping.

Roadmap

Brute Force Baseline
Core Insight
Algorithm Walkthrough
Edge Cases
Full Annotated Code
Interactive Study Demo
Complexity Analysis

Step 01

Brute Force Baseline

Problem summary: 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.

Baseline thinking

Start with the most direct exhaustive search. That gives a correctness anchor before optimizing.

Pattern signal: Array · Binary Search · Greedy · Sliding Window

Example 1

[[1,5],[10,11],[12,18],[20,25],[30,32]]
10

Example 2

[[10,11],[1,1]]
2

Related Problems

Maximum Number of Vowels in a Substring of Given Length (maximum-number-of-vowels-in-a-substring-of-given-length)

Step 02

Core Insight

What unlocks the optimal approach

Think about the potential placements of the carpet in an optimal solution.
Can we use Prefix Sum and Binary Search to determine how many tiles are covered for a given placement?

Interview move: turn each hint into an invariant you can check after every iteration/recursion step.

Step 03

Algorithm Walkthrough

Iteration Checklist

Define state (indices, window, stack, map, DP cell, or recursion frame).
Apply one transition step and update the invariant.
Record answer candidate when condition is met.
Continue until all input is consumed.

Use the first example testcase as your mental trace to verify each transition.

Step 04

Edge Cases

Minimum Input

Single element / shortest valid input

Validate boundary behavior before entering the main loop or recursion.

Duplicates & Repeats

Repeated values / repeated states

Decide whether duplicates should be merged, skipped, or counted explicitly.

Extreme Constraints

Upper-end input sizes

Re-check complexity target against constraints to avoid time-limit issues.

Invalid / Corner Shape

Empty collections, zeros, or disconnected structures

Handle special-case structure before the core algorithm path.

Step 05

Full Annotated Code

Source-backed implementations are provided below for direct study and interview prep.

// Accepted solution for LeetCode #2271: Maximum White Tiles Covered by a Carpet
class Solution {
    public int maximumWhiteTiles(int[][] tiles, int carpetLen) {
        Arrays.sort(tiles, (a, b) -> a[0] - b[0]);
        int n = tiles.length;
        int s = 0, ans = 0;
        for (int i = 0, j = 0; i < n; ++i) {
            while (j < n && tiles[j][1] - tiles[i][0] + 1 <= carpetLen) {
                s += tiles[j][1] - tiles[j][0] + 1;
                ++j;
            }
            if (j < n && tiles[i][0] + carpetLen > tiles[j][0]) {
                ans = Math.max(ans, s + tiles[i][0] + carpetLen - tiles[j][0]);
            } else {
                ans = Math.max(ans, s);
            }
            s -= (tiles[i][1] - tiles[i][0] + 1);
        }
        return ans;
    }
}

// Accepted solution for LeetCode #2271: Maximum White Tiles Covered by a Carpet
func maximumWhiteTiles(tiles [][]int, carpetLen int) int {
	sort.Slice(tiles, func(i, j int) bool { return tiles[i][0] < tiles[j][0] })
	n := len(tiles)
	s, ans := 0, 0
	for i, j := 0, 0; i < n; i++ {
		for j < n && tiles[j][1]-tiles[i][0]+1 <= carpetLen {
			s += tiles[j][1] - tiles[j][0] + 1
			j++
		}
		if j < n && tiles[i][0]+carpetLen > tiles[j][0] {
			ans = max(ans, s+tiles[i][0]+carpetLen-tiles[j][0])
		} else {
			ans = max(ans, s)
		}
		s -= (tiles[i][1] - tiles[i][0] + 1)
	}
	return ans
}

# Accepted solution for LeetCode #2271: Maximum White Tiles Covered by a Carpet
class Solution:
    def maximumWhiteTiles(self, tiles: List[List[int]], carpetLen: int) -> int:
        tiles.sort()
        n = len(tiles)
        s = ans = j = 0
        for i, (li, ri) in enumerate(tiles):
            while j < n and tiles[j][1] - li + 1 <= carpetLen:
                s += tiles[j][1] - tiles[j][0] + 1
                j += 1
            if j < n and li + carpetLen > tiles[j][0]:
                ans = max(ans, s + li + carpetLen - tiles[j][0])
            else:
                ans = max(ans, s)
            s -= ri - li + 1
        return ans

// Accepted solution for LeetCode #2271: Maximum White Tiles Covered by a Carpet
/**
 * [2271] Maximum White Tiles Covered by a Carpet
 *
 * 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:
 * <img alt="" src="https://assets.leetcode.com/uploads/2022/03/25/example1drawio3.png" style="width: 644px; height: 158px;" />
 * 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:
 * <img alt="" src="https://assets.leetcode.com/uploads/2022/03/24/example2drawio.png" style="width: 231px; height: 168px;" />
 * 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 * 10^4
 * 	tiles[i].length == 2
 * 	1 <= li <= ri <= 10^9
 * 	1 <= carpetLen <= 10^9
 * 	The tiles are non-overlapping.
 *
 */
pub struct Solution {}

// problem: https://leetcode.com/problems/maximum-white-tiles-covered-by-a-carpet/
// discuss: https://leetcode.com/problems/maximum-white-tiles-covered-by-a-carpet/discuss/?currentPage=1&orderBy=most_votes&query=

// submission codes start here

impl Solution {
    pub fn maximum_white_tiles(tiles: Vec<Vec<i32>>, carpet_len: i32) -> i32 {
        0
    }
}

// submission codes end

#[cfg(test)]
mod tests {
    use super::*;

    #[test]
    #[ignore]
    fn test_2271_example_1() {
        let tiles = vec![
            vec![1, 5],
            vec![10, 11],
            vec![12, 18],
            vec![20, 25],
            vec![30, 32],
        ];
        let carpet_len = 10;

        let result = 9;

        assert_eq!(Solution::maximum_white_tiles(tiles, carpet_len), result);
    }

    #[test]
    #[ignore]
    fn test_2271_example_2() {
        let tiles = vec![vec![10, 11], vec![1, 1]];
        let carpet_len = 10;

        let result = 2;

        assert_eq!(Solution::maximum_white_tiles(tiles, carpet_len), result);
    }
}

// Accepted solution for LeetCode #2271: Maximum White Tiles Covered by a Carpet
// Auto-generated TypeScript example from java.
function exampleSolution(): void {
}
// Reference (java):
// // Accepted solution for LeetCode #2271: Maximum White Tiles Covered by a Carpet
// class Solution {
//     public int maximumWhiteTiles(int[][] tiles, int carpetLen) {
//         Arrays.sort(tiles, (a, b) -> a[0] - b[0]);
//         int n = tiles.length;
//         int s = 0, ans = 0;
//         for (int i = 0, j = 0; i < n; ++i) {
//             while (j < n && tiles[j][1] - tiles[i][0] + 1 <= carpetLen) {
//                 s += tiles[j][1] - tiles[j][0] + 1;
//                 ++j;
//             }
//             if (j < n && tiles[i][0] + carpetLen > tiles[j][0]) {
//                 ans = Math.max(ans, s + tiles[i][0] + carpetLen - tiles[j][0]);
//             } else {
//                 ans = Math.max(ans, s);
//             }
//             s -= (tiles[i][1] - tiles[i][0] + 1);
//         }
//         return ans;
//     }
// }

Step 06

Interactive Study Demo

Use this to step through a reusable interview workflow for this problem.

Custom checkpoints (one per line)

Press Step or Run All to begin.

Step 07

Complexity Analysis

Time

O(log n)

Space

O(1)

Approach Breakdown

LINEAR SCAN

O(n) time

O(1) space

Check every element from left to right until we find the target or exhaust the array. Each comparison is O(1), and we may visit all n elements, giving O(n). No extra space needed.

BINARY SEARCH

O(log n) time

O(1) space

Each comparison eliminates half the remaining search space. After k comparisons, the space is n/2ᵏ. We stop when the space is 1, so k = log₂ n. No extra memory needed — just two pointers (lo, hi).

Shortcut: Halving the input each step → O(log n). Works on any monotonic condition, not just sorted arrays.

Coach Notes

Common Mistakes

Review these before coding to avoid predictable interview regressions.

Off-by-one on range boundaries

Wrong move: Loop endpoints miss first/last candidate.

Usually fails on: Fails on minimal arrays and exact-boundary answers.

Fix: Re-derive loops from inclusive/exclusive ranges before coding.

Boundary update without `+1` / `-1`

Wrong move: Setting `lo = mid` or `hi = mid` can stall and create an infinite loop.

Usually fails on: Two-element ranges never converge.

Fix: Use `lo = mid + 1` or `hi = mid - 1` where appropriate.

Using greedy without proof

Wrong move: Locally optimal choices may fail globally.

Usually fails on: Counterexamples appear on crafted input orderings.

Fix: Verify with exchange argument or monotonic objective before committing.

Shrinking the window only once

Wrong move: Using `if` instead of `while` leaves the window invalid for multiple iterations.

Usually fails on: Over-limit windows stay invalid and produce wrong lengths/counts.

Fix: Shrink in a `while` loop until the invariant is valid again.