<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 SegmentTree:
    def __init__(self, nums):
        self.n = len(nums)
        self.tree = [0] * (4 * self.n)
        self.nums = nums
        self.build(0, 0, self.n - 1)
    
    def build(self, node, start, end):
        if start == end:
            self.tree[node] = 1 if self.is_peak(start) else 0
            return
        
        mid = (start + end) // 2
        self.build(2*node+1, start, mid)
        self.build(2*node+2, mid+1, end)
        self.tree[node] = self.tree[2*node+1] + self.tree[2*node+2]
    
    def is_peak(self, idx):
        if 0 < idx < self.n - 1:
            return self.nums[idx-1] < self.nums[idx] > self.nums[idx+1]
        return False
    
    def update(self, node, start, end, idx, val):
        if start == end:
            self.nums[idx] = val
            self.tree[node] = 1 if self.is_peak(idx) else 0
            return
        
        mid = (start + end) // 2
        if idx <= mid:
            self.update(2*node+1, start, mid, idx, val)
        else:
            self.update(2*node+2, mid+1, end, idx, val)
        self.tree[node] = self.tree[2*node+1] + self.tree[2*node+2]
    
    def query(self, node, start, end, l, r):
        if r < start or end < l:
            return 0
        if l <= start and end <= r:
            return self.tree[node]
        
        mid = (start + end) // 2
        left = self.query(2*node+1, start, mid, l, r)
        right = self.query(2*node+2, mid+1, end, l, r)
        return left + right
class Solution:
 def countOfPeaks(self,nums: List[int], queries: List[List[int]]) -> List[int]:
    st = SegmentTree(nums)
    ans = []
    
    for q in queries:
        if q[0] == 1:  # Count peaks in range
            l, r = q[1], q[2]
            ans.append(st.query(0, 0, len(nums)-1, l+1, r-1))
        else:  # Update value
            idx, val = q[1], q[2]
            st.update(0, 0, len(nums)-1, idx, val)
            if idx > 0:
                st.update(0, 0, len(nums)-1, idx-1, nums[idx-1])
            if idx < len(nums) - 1:
                st.update(0, 0, len(nums)-1, idx+1, nums[idx+1])
    
    return ans

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