<p>You are given a 2D integer array <code>tiles</code> where <code>tiles[i] = [l<sub>i</sub>, r<sub>i</sub>]</code> represents that every tile <code>j</code> in the range <code>l<sub>i</sub> &lt;= j &lt;= r<sub>i</sub></code> is colored white.</p>

<p>You are also given an integer <code>carpetLen</code>, the length of a single carpet that can be placed <strong>anywhere</strong>.</p>

<p>Return <em>the <strong>maximum</strong> number of white tiles that can be covered by the carpet</em>.</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>
<img alt="" src="https://assets.leetcode.com/uploads/2022/03/25/example1drawio3.png" style="width: 644px; height: 158px;" />
<pre>
<strong>Input:</strong> tiles = [[1,5],[10,11],[12,18],[20,25],[30,32]], carpetLen = 10
<strong>Output:</strong> 9
<strong>Explanation:</strong> 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.
</pre>

<p><strong class="example">Example 2:</strong></p>
<img alt="" src="https://assets.leetcode.com/uploads/2022/03/24/example2drawio.png" style="width: 231px; height: 168px;" />
<pre>
<strong>Input:</strong> tiles = [[10,11],[1,1]], carpetLen = 2
<strong>Output:</strong> 2
<strong>Explanation:</strong> Place the carpet starting on tile 10. 
It covers 2 white tiles, so we return 2.
</pre>

<p>&nbsp;</p>
<p><strong>Constraints:</strong></p>

<ul>
	<li><code>1 &lt;= tiles.length &lt;= 5 * 10<sup>4</sup></code></li>
	<li><code>tiles[i].length == 2</code></li>
	<li><code>1 &lt;= l<sub>i</sub> &lt;= r<sub>i</sub> &lt;= 10<sup>9</sup></code></li>
	<li><code>1 &lt;= carpetLen &lt;= 10<sup>9</sup></code></li>
	<li>The <code>tiles</code> are <strong>non-overlapping</strong>.</li>
</ul>

class Solution:
    def maximumWhiteTiles(self, tiles: List[List[int]], carpetLen: int) -> int:
        queue = deque()
        tiles.sort()
        ans = curr = 0
        for s, e in tiles:
            leftBound = e - carpetLen + 1
            while queue and queue[0][1] < leftBound:
                start, end = queue.popleft()
                curr -= end-start+1

            if queue:
                newStart = max(queue[0][0], leftBound)
                curr -= newStart - queue[0][0]
                queue[0][0] = newStart

            queue.append([max(s, leftBound), e])
            curr += queue[-1][1] - queue[-1][0] + 1
            ans = max(ans, curr)

        return ans
41400