<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>

from sortedcontainers import SortedList
class SegTree:
    def __init__(self, array, s, e):
        self.s = s
        self.e = e
        self.array = array
        self.l = self.r = None
        if self.s == self.e:
            self.v = array[s]
            return
        mid = (s+e) // 2
        self.l = SegTree(array, s, mid)
        self.r = SegTree(array, mid+1, e)
        self.v = self.l.v + self.r.v
        
    def update(self, i, v):
        if i == self.s == self.e:
            self.v = v
            self.array[i] = v
            return
        mid = (self.s+self.e) // 2
        if self.s <= i <= mid:
            self.l.update(i, v)
        elif mid + 1 <= i <= self.e:
            self.r.update(i, v)
        self.v = self.l.v + self.r.v

    def query(self, s, e):
        if self.s == s and self.e == e:
            return self.v
        ans = 0
        mid = (self.s+self.e) // 2
        if self.s <= s <= e <= mid:
            ans += self.l.query(s, e)
        elif mid + 1 <= s <= e <= self.e:
            ans += self.r.query(s, e)
        else:
            ans += self.l.query(s, mid)
            ans += self.r.query(mid+1, e)
        return ans


class Solution:
    def numberOfAlternatingGroups(self, colors: List[int], queries: List[List[int]]) -> List[int]:
        count = SortedList()
        count.add(len(colors))
        tile = SegTree([0] * (len(colors) + 1), 0, len(colors))
        tile.update(len(colors), len(colors))
        sl = SortedList()

        def update(length, c):
            length = length % len(colors)
            if length == 0:
                length = len(colors)
            if c > 0:
                count.add(length)
            else:
                count.remove(length)
            tile.update(length, tile.array[length] + c*length)
            
        def merge_or_break(index):
            index = index % len(colors)
            if index in sl:
                # merge
                if len(sl) == 1:
                    sl.remove(index)
                    return
                ii = sl.bisect_left(index)
                prev = sl[ii-1]
                after = sl[(ii+1) % len(sl)]
                sl.remove(index)
                update(index - prev, -1)
                update(after - index, -1)
                update(after - prev, 1)
            else:
                # break
                if len(sl) == 0:
                    sl.add(index)
                    return
                ii = sl.bisect_left(index)
                prev = sl[ii-1]
                after = sl[(ii) % len(sl)]
                sl.add(index)
                update(index - prev, 1)
                update(after - index, 1)
                update(after - prev, -1)
                
        for i in range(len(colors)):
            if colors[i] == colors[i-1]:
                merge_or_break(i-1)

        res = []
        for q in queries:
            if q[0] == 1:
                if len(sl) == 0:
                    res.append(len(colors))
                    continue
                l = q[1]
                res.append(tile.query(l, len(colors)) - (l - 1) * (len(count) - count.bisect_left(l) ))
            elif colors[q[1]] != q[2]:
                colors[q[1]] = q[2]
                merge_or_break(q[1])
                merge_or_break((q[1] - 1) % len(colors))
        return res
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