<p>There are some red and blue tiles arranged circularly. You are given an array of integers <code>colors</code> and a 2D integers array <code>queries</code>.</p>

<p>The color of tile <code>i</code> is represented by <code>colors[i]</code>:</p>

<ul>
	<li><code>colors[i] == 0</code> means that tile <code>i</code> is <strong>red</strong>.</li>
	<li><code>colors[i] == 1</code> means that tile <code>i</code> is <strong>blue</strong>.</li>
</ul>

<p>An <strong>alternating</strong> group is a contiguous subset of tiles in the circle with <strong>alternating</strong> colors (each tile in the group except the first and last one has a different color from its <b>adjacent</b> tiles in the group).</p>

<p>You have to process queries of two types:</p>

<ul>
	<li><code>queries[i] = [1, size<sub>i</sub>]</code>, determine the count of <strong>alternating</strong> groups with size <code>size<sub>i</sub></code>.</li>
	<li><code>queries[i] = [2, index<sub>i</sub>, color<sub>i</sub>]</code>, change <code>colors[index<sub>i</sub>]</code> to <code>color<font face="monospace"><sub>i</sub></font></code>.</li>
</ul>

<p>Return an array <code>answer</code> containing the results of the queries of the first type <em>in order</em>.</p>

<p><strong>Note</strong> that since <code>colors</code> represents a <strong>circle</strong>, the <strong>first</strong> and the <strong>last</strong> tiles are considered to be next to each other.</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>

<div class="example-block">
<p><strong>Input:</strong> <span class="example-io">colors = [0,1,1,0,1], queries = [[2,1,0],[1,4]]</span></p>

<p><strong>Output:</strong> <span class="example-io">[2]</span></p>

<p><strong>Explanation:</strong></p>

<p><strong class="example"><img alt="" data-darkreader-inline-bgcolor="" data-darkreader-inline-bgimage="" src="https://assets.leetcode.com/uploads/2024/06/03/screenshot-from-2024-06-03-20-14-44.png" style="width: 150px; height: 150px; padding: 10px; background: rgb(255, 255, 255); border-radius: 0.5rem; --darkreader-inline-bgimage: initial; --darkreader-inline-bgcolor: #181a1b;" /></strong></p>

<p>First query:</p>

<p>Change <code>colors[1]</code> to 0.</p>

<p><img alt="" data-darkreader-inline-bgcolor="" data-darkreader-inline-bgimage="" src="https://assets.leetcode.com/uploads/2024/06/03/screenshot-from-2024-06-03-20-20-25.png" style="width: 150px; height: 150px; padding: 10px; background: rgb(255, 255, 255); border-radius: 0.5rem; --darkreader-inline-bgimage: initial; --darkreader-inline-bgcolor: #181a1b;" /></p>

<p>Second query:</p>

<p>Count of the alternating groups with size 4:</p>

<p><img alt="" data-darkreader-inline-bgcolor="" data-darkreader-inline-bgimage="" src="https://assets.leetcode.com/uploads/2024/06/03/screenshot-from-2024-06-03-20-25-02-2.png" style="width: 150px; height: 150px; padding: 10px; background: rgb(255, 255, 255); border-radius: 0.5rem; --darkreader-inline-bgimage: initial; --darkreader-inline-bgcolor: #181a1b;" /><img alt="" data-darkreader-inline-bgcolor="" data-darkreader-inline-bgimage="" src="https://assets.leetcode.com/uploads/2024/06/03/screenshot-from-2024-06-03-20-24-12.png" style="width: 150px; height: 150px; padding: 10px; background: rgb(255, 255, 255); border-radius: 0.5rem; --darkreader-inline-bgimage: initial; --darkreader-inline-bgcolor: #181a1b;" /></p>
</div>

<p><strong class="example">Example 2:</strong></p>

<div class="example-block">
<p><strong>Input:</strong> <span class="example-io">colors = [0,0,1,0,1,1], queries = [[1,3],[2,3,0],[1,5]]</span></p>

<p><strong>Output:</strong> <span class="example-io">[2,0]</span></p>

<p><strong>Explanation:</strong></p>

<p><img alt="" data-darkreader-inline-bgcolor="" data-darkreader-inline-bgimage="" src="https://assets.leetcode.com/uploads/2024/06/03/screenshot-from-2024-06-03-20-35-50.png" style="width: 150px; height: 150px; padding: 10px; background: rgb(255, 255, 255); border-radius: 0.5rem; --darkreader-inline-bgimage: initial; --darkreader-inline-bgcolor: #181a1b;" /></p>

<p>First query:</p>

<p>Count of the alternating groups with size 3:</p>

<p><img alt="" data-darkreader-inline-bgcolor="" data-darkreader-inline-bgimage="" src="https://assets.leetcode.com/uploads/2024/06/03/screenshot-from-2024-06-03-20-37-13.png" style="width: 150px; height: 150px; padding: 10px; background: rgb(255, 255, 255); border-radius: 0.5rem; --darkreader-inline-bgimage: initial; --darkreader-inline-bgcolor: #181a1b;" /><img alt="" data-darkreader-inline-bgcolor="" data-darkreader-inline-bgimage="" src="https://assets.leetcode.com/uploads/2024/06/03/screenshot-from-2024-06-03-20-36-40.png" style="width: 150px; height: 150px; padding: 10px; background: rgb(255, 255, 255); border-radius: 0.5rem; --darkreader-inline-bgimage: initial; --darkreader-inline-bgcolor: #181a1b;" /></p>

<p>Second query: <code>colors</code> will not change.</p>

<p>Third query: There is no alternating group with size 5.</p>
</div>

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

<ul>
	<li><code>4 &lt;= colors.length &lt;= 5 * 10<sup>4</sup></code></li>
	<li><code>0 &lt;= colors[i] &lt;= 1</code></li>
	<li><code>1 &lt;= queries.length &lt;= 5 * 10<sup>4</sup></code></li>
	<li><code>queries[i][0] == 1</code> or <code>queries[i][0] == 2</code></li>
	<li>For all <code>i</code> that:
	<ul>
		<li><code>queries[i][0] == 1</code>: <code>queries[i].length == 2</code>, <code>3 &lt;= queries[i][1] &lt;= colors.length - 1</code></li>
		<li><code>queries[i][0] == 2</code>: <code>queries[i].length == 3</code>, <code>0 &lt;= queries[i][1] &lt;= colors.length - 1</code>, <code>0 &lt;= queries[i][2] &lt;= 1</code></li>
	</ul>
	</li>
</ul>

import typing
from typing import *
import sortedcontainers
# Extremely hard, difficulty rating 3100+
class FenwickTree:
    """Reference: https://en.wikipedia.org/wiki/Fenwick_tree"""

    def __init__(self, n: int = 0) -> None:
        self._n = n
        self.data = [0] * n

    def add(self, p: int, x: typing.Any) -> None:
        assert 0 <= p < self._n

        p += 1
        while p <= self._n:
            self.data[p - 1] += x
            p += p & -p

    def sum(self, left: int, right: int) -> typing.Any:
        assert 0 <= left <= right <= self._n

        return self._sum(right) - self._sum(left)

    def _sum(self, r: int) -> typing.Any:
        s = 0
        while r > 0:
            s += self.data[r - 1]
            r -= r & -r

        return s


class Solution:
    def numberOfAlternatingGroups(
        self, colors: List[int], queries: List[List[int]]
    ) -> List[int]:
        n = len(colors)
        ans = DefaultDict(int)
        ss = sortedcontainers.SortedList(
            [i for i in range(n) if colors[i - 1] == colors[i]]
        )
        ftk, ftb = FenwickTree(n + 2), FenwickTree(n + 2)

        def change(l, r, sign):
            x = r - l
            if x <= 0:
                x += n
            ftk.add(1, -sign)
            ftk.add(x + 1, sign)
            ftb.add(1, (x + 1) * sign)
            ftb.add(x + 1, (x + 1) * -sign)

        if len(ss) == 0:
            ftb.add(1, n)
            ftb.add(n + 1, -n)
        else:
            for i in range(len(ss)):
                change(ss[i - 1], ss[i], 1)

        def add(i):
            if len(ss) == 0:
                ftb.add(1, -n)
                ftb.add(n + 1, n)
                ss.add(i)
                change(ss[0], ss[0], 1)
            else:
                ss.add(i)
                j = ss.bisect_left(i)
                prev = ss[j - 1]
                next = ss[(j + 1) % len(ss)]
                change(prev, next, -1)
                change(prev, i, 1)
                change(i, next, 1)

        def sub(i):
            if len(ss) == 1:
                ftb.add(1, n)
                ftb.add(n + 1, -n)
                change(ss[0], ss[0], -1)
                ss.pop(0)
            else:
                j = ss.bisect_left(i)
                prev = ss[j - 1]
                next = ss[(j + 1) % len(ss)]
                change(prev, next, 1)
                change(prev, i, -1)
                change(i, next, -1)
                ss.pop(j)

        res = []
        for q in queries:
            if q[0] == 1:
                size = q[1]
                res.append(ftk.sum(1, size + 1) * size + ftb.sum(1, size + 1))
            else:
                id, c = q[1], q[2]
                if colors[id] == colors[id - 1]:
                    sub(id)
                if colors[id] == colors[(id + 1) % n]:
                    sub((id + 1) % n)
                colors[id] = c
                if colors[id] == colors[id - 1]:
                    add(id)
                if colors[id] == colors[(id + 1) % n]:
                    add((id + 1) % n)
        return res



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