<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 Solution:
    def numberOfAlternatingGroups(self, colors: List[int], queries: List[List[int]]) -> List[int]:
        sl = SortedList()
        n = len(colors)
        l = []
        for i in range(n):
            l.append(int(colors[i] != colors[(i+1)%n]))
        p = []
        cts = {}
        
        def addseg(q):
            sl.add(q)
            t = q[1] - q[0] + 1
            if t not in cts:
                cts[t] = 0
            cts[t] += 1
            
        def rmseg(q):
            sl.remove(q)
            t = q[1] - q[0] + 1
            cts[t] -= 1
            if cts[t] == 0:
                del cts[t]
        
        
        for i in range(n):
            if l[i] and not p:
                p.append(i)
            if not l[i] and p:
                p.append(i-1)
                addseg(tuple(p))
                p = []
        if p:
            p.append(n-1)
            addseg(tuple(p))
        
        def fin(v):
            if not 0 <= v < n:
                return None
            a = sl.bisect_left((v, v))
            for x in [a, a-1]:
                if 0 <= x < len(sl):
                    g, h = sl[x]
                    if g <= v <= h:
                        return (g, h)
            return None
        
        
        
        def flip(idx):
            l[idx] = 1 - l[idx]
            if l[idx] == 1:
                a = fin(idx - 1)
                c = fin(idx + 1)
                if not a and not c:
                    addseg((idx, idx))
                elif a and c:
                    rmseg(a)
                    rmseg(c)
                    addseg((a[0], c[1]))
                elif a:
                    rmseg(a)
                    addseg((a[0], idx))
                else:
                    rmseg(c)
                    addseg((idx, c[1]))
            else:
                a, b = fin(idx)
                if a == b:
                    rmseg((a, b))
                elif a < idx < b:
                    rmseg((a, b))
                    addseg((a, idx-1))
                    addseg((idx+1, b))
                elif a == idx:
                    rmseg((a, b))
                    addseg((a+1, b))
                else:
                    rmseg((a, b))
                    addseg((a, b-1))
        
        
        ret = []
        for q in queries:
            if len(q) == 3:
                _, x, y = q
                if colors[x] != y:
                    colors[x] = y
                    flip(x)
                    flip((x-1)%n)
            else:
                _, p = q
                if len(sl) == 1 and sl[0] == (0, n-1):
                    ret.append(n)
                else:
                    cur = 0
                    for t in cts:
                        if t >= p - 1:
                            v = cts[t]
                            cur += v * (t - p + 2)
                    if sl and sl[0][0] == 0 and sl[-1][1] == n-1:
                        a = sl[0][1] - sl[0][0] + 1
                        b = sl[-1][1] - sl[-1][0] + 1
                        if a + b >= p - 1:
                            cur += a + b - p + 2
                        if a >= p - 1:
                            cur -= a - p + 2
                        if b >= p - 1:
                            cur -= b - p + 2
                    ret.append(cur)
        return ret
                            

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