<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:
        # form the prefix sums and flatten the 2d list
        tiles.sort()
        n = len(tiles)
        n2 = 2* n
        prefix, aux = [0] * n2, [0] * n2
        aux[0], aux[1] = tiles[0][0], tiles[0][1]
        prefix[0], prefix[1] = 1, tiles[0][1] - tiles[0][0] + 1
        idx = 2
        for i in range(1, n):
            aux[idx] = tiles[i][0]
            prefix[idx] = prefix[idx-1] + 1
            idx += 1
            aux[idx] = tiles[i][1]
            prefix[idx] = tiles[i][1]-tiles[i][0] + prefix[idx-1]
            idx += 1
        # algo begins here
        ans = 0
        for i in range(n2):
            # place the carpet at aux[i]
            right = bisect_left(aux, aux[i]+carpetLen-1)
            if right == n2:
                right = n2-1
                return max(ans, prefix[n2-1] - prefix[i]+1)
            if aux[i]+carpetLen-1 == aux[right]:
                ans = max(ans, prefix[right] - prefix[i]+1)
            else:
                # check if the additional interval is to the left side or the right side
                # if right%2 == 0, then it is after an interval's begingging
                # else: it is after an interval's end
                if right%2:
                    ans = max(ans, prefix[right-1]-prefix[i] + aux[i]+carpetLen-aux[right-1])
                else:
                    ans = max(ans, prefix[right-1] - prefix[i] + 1)
        return ans
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