<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 Solution:
    def countOfPeaks(self, nums: List[int], queries: List[List[int]]) -> List[int]:
        peaks = []
        for i in range(1,len(nums)-1):
            if nums[i-1] < nums[i] > nums[i+1]:
                peaks.append(1)
            else:
                peaks.append(0)
        peaks = [0]+peaks+[0]
        def build(tree,a,v,al,ar):
            if al==ar:
                tree[v]=a[al]
            else:
                am = (al+ar)//2
                build(tree,a,2*v,al,am)
                build(tree,a,2*v+1,am+1,ar)
                tree[v]= tree[2*v] + tree[2*v+1]
        N = len(nums)
        tree = [0]*(4*N)
        build(tree,peaks,1,0,N-1)   
        def query(tree, v, tl, tr, l, r):
            if r < tl or tr < l:
                return 0
            elif l<=tl and tr<=r:
                return tree[v]
            else:
                m = (tl+tr)//2
                return query(tree,2*v,tl,m,l,r)+query(tree,2*v+1,m+1,tr,l,r)
        #idx: position
        #val: updated(changed) valued s.t. a[idx] = val
        #a: orginal array
        def update(tree, v, tl, tr, idx, val):
            if tl==idx==tr:
                tree[v]=val
                return val
            m = (tl+tr)//2    
            if tl<=idx<=m:
                tree[v]=update(tree,2*v,tl,m,idx,val)+tree[2*v+1]
            else:
                tree[v]=tree[2*v]+update(tree,2*v+1,m+1,tr,idx,val)
            return tree[v]
        res = []
        for t, i, val in queries:
            if t == 1:
                res.append(query(tree,1,0,N-1,i+1,val-1) if i+1<val else 0)
            else:
                if 0<i<N-1:
                    update(tree, 1, 0, N-1, i, nums[i-1]<val>nums[i+1])
                if 0<i-1<N-1:
                    update(tree, 1, 0, N-1, i-1, nums[i-2]<nums[i-1]>val)
                if 0<i+1<N-1:
                    update(tree, 1, 0, N-1, i+1, val<nums[i+1]>nums[i+2])
                nums[i] = val
        return res
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