<p>A <strong>peak</strong> in an array <code>arr</code> is an element that is <strong>greater</strong> than its previous and next element in <code>arr</code>.</p>

<p>You are given an integer array <code>nums</code> and a 2D integer array <code>queries</code>.</p>

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

<ul>
	<li><code>queries[i] = [1, l<sub>i</sub>, r<sub>i</sub>]</code>, determine the count of <strong>peak</strong> elements in the <span data-keyword="subarray">subarray</span> <code>nums[l<sub>i</sub>..r<sub>i</sub>]</code>.<!-- notionvc: 73b20b7c-e1ab-4dac-86d0-13761094a9ae --></li>
	<li><code>queries[i] = [2, index<sub>i</sub>, val<sub>i</sub>]</code>, change <code>nums[index<sub>i</sub>]</code> to <code><font face="monospace">val<sub>i</sub></font></code>.</li>
</ul>

<p>Return an array <code>answer</code> containing the results of the queries of the first type in order.<!-- notionvc: a9ccef22-4061-4b5a-b4cc-a2b2a0e12f30 --></p>

<p><strong>Notes:</strong></p>

<ul>
	<li>The <strong>first</strong> and the <strong>last</strong> element of an array or a subarray<!-- notionvc: fcffef72-deb5-47cb-8719-3a3790102f73 --> <strong>cannot</strong> be a peak.</li>
</ul>

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

<div class="example-block">
<p><strong>Input:</strong> <span class="example-io">nums = [3,1,4,2,5], queries = [[2,3,4],[1,0,4]]</span></p>

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

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

<p>First query: We change <code>nums[3]</code> to 4 and <code>nums</code> becomes <code>[3,1,4,4,5]</code>.</p>

<p>Second query: The number of peaks in the <code>[3,1,4,4,5]</code> is 0.</p>
</div>

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

<div class="example-block">
<p><strong>Input:</strong> <span class="example-io">nums = [4,1,4,2,1,5], queries = [[2,2,4],[1,0,2],[1,0,4]]</span></p>

<p><strong>Output:</strong> <span class="example-io">[0,1]</span></p>

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

<p>First query: <code>nums[2]</code> should become 4, but it is already set to 4.</p>

<p>Second query: The number of peaks in the <code>[4,1,4]</code> is 0.</p>

<p>Third query: The second 4 is a peak in the <code>[4,1,4,2,1]</code>.</p>
</div>

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

<ul>
	<li><code>3 &lt;= nums.length &lt;= 10<sup>5</sup></code></li>
	<li><code>1 &lt;= nums[i] &lt;= 10<sup>5</sup></code></li>
	<li><code>1 &lt;= queries.length &lt;= 10<sup>5</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>0 &lt;= queries[i][1] &lt;= queries[i][2] &lt;= nums.length - 1</code></li>
		<li><code>queries[i][0] == 2</code>: <code>0 &lt;= queries[i][1] &lt;= nums.length - 1</code>, <code>1 &lt;= queries[i][2] &lt;= 10<sup>5</sup></code></li>
	</ul>
	</li>
</ul>

class Seg_tree:
    def __init__(self, n):
        self.n = n
        self.t = [0] * (4 * n)

    def build(self, a, v, tl, tr):
        if tl == tr:
            self.t[v] = a[tl]
        else:
            tm = (tl + tr) // 2
            self.build(a, v * 2, tl, tm)
            self.build(a, v * 2 + 1, tm + 1, tr)
            self.t[v] = self.t[v * 2] + self.t[v * 2 + 1]

    def sum(self, v, tl, tr, l, r):
        if l > r:
            return 0
        if l == tl and r == tr:
            return self.t[v]
        tm = (tl + tr) // 2
        return self.sum(v * 2, tl, tm, l, min(r, tm)) + self.sum(v * 2 + 1, tm + 1, tr, max(l, tm + 1), r)

    def update(self, v, tl, tr, pos, new_val):
        if tl == tr:
            self.t[v] = new_val
        else:
            tm = (tl + tr) // 2
            if (pos <= tm):
                self.update(v * 2, tl, tm, pos, new_val)
            else:
                self.update(v * 2 + 1, tm + 1, tr, pos, new_val)
            self.t[v] = self.t[v * 2] + self.t[v * 2 + 1]


class Solution:
    def countOfPeaks(self, nums: List[int], queries: List[List[int]]) -> List[int]:
        n = len(nums)
        peaks = [0] * len(nums)
        def is_peak(i):
            return i > 0 and i + 1 < n and nums[i - 1] < nums[i] and nums[i] > nums[i + 1]
        for i in range(1, n - 1):
            peaks[i] = int(is_peak(i))

        john = Seg_tree(n)
        john.build(peaks, 1, 0, n - 1)

        res = []
        for a, b, c in queries:
            if a == 1:
                res.append(john.sum(1, 0, n - 1, b + 1, c - 1))
                continue
            nums[b] = c
            john.update(1, 0, n - 1, b - 1, int(is_peak(b - 1)))
            john.update(1, 0, n - 1, b, int(is_peak(b)))
            john.update(1, 0, n - 1, b + 1, int(is_peak(b + 1)))

        return res
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