<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]:
        arr = [0 for i in range(4*len(nums))]
        def build(v,l,r):
            if l == r:
                arr[v] = 0
            else:
                m = (l+r)//2
                newPeak = 0
                if m-1 >= l and m+1 <= r:
                    newPeak = nums[m] > nums[m-1] and nums[m] > nums[m+1]
                if m+2 <= r:
                    newPeak = newPeak or (nums[m+1] > nums[m] and nums[m+1] > nums[m+2])
                build(2*v,l,m)
                build(2*v+1,m+1,r)
                arr[v] = arr[2*v] + arr[2*v+1] + newPeak
        def search(v,tl,tr,l,r):
            if l > r:
                return 0
            if tl == l and tr == r:
                return arr[v]
            tm = (tl+tr)//2
            newPeak = 0
            if r > tm and l <= tm:
                if tm-1 >= l and tm+1 <= r:
                    newPeak = nums[tm] > nums[tm-1] and nums[tm] > nums[tm+1]
                if tm+2 <= r:
                    newPeak = newPeak or (nums[tm+1] > nums[tm] and nums[tm+1] > nums[tm+2])     
            return search(2*v,tl,tm,l,min(r,tm)) + search(2*v+1,tm+1,tr,max(l,tm+1),r) + newPeak
        def update(v,tl,tr,pos,val):
            if tl == tr:
                nums[pos] = val
            else:
                tm = (tl+tr)//2
                if pos <= tm:
                    update(2*v,tl,tm,pos,val)
                else:
                    update(2*v+1,tm+1,tr,pos,val)
                newPeak = 0                
                if tm-1 >= tl and tm+1 <= tr:
                    newPeak = nums[tm] > nums[tm-1] and nums[tm] > nums[tm+1]
                if tm+2 <= tr:
                    newPeak = newPeak or (nums[tm+1] > nums[tm] and nums[tm+1] > nums[tm+2])
                arr[v] = arr[2*v] + arr[2*v+1] + newPeak
        res = []
        build(1,0,len(nums)-1)
        for t,l,r in queries:
            if t == 1:
                res.append(search(1,0,len(nums)-1,l,r))
            else:
                update(1,0,len(nums)-1,l,r)
        return res
    
    
                
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