<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 BIT:
    def __init__(self, n):
        self.n = n
        self.tree = [0] * (n + 1)
        
    def update(self, i, val):
        while i <= self.n:
            self.tree[i] += val
            i += self.lowbit(i)
    
    def query(self, i):
        res = 0
        while i > 0:
            res += self.tree[i]
            i -= self.lowbit(i)
        return res
    
    def lowbit(self, i):
        return i & (- i)
    
class Solution:
    def countOfPeaks(self, nums, queries):
        n = len(nums)

        dp = [0]*n

        tree = BIT(n)

        for i in range(1, n - 1):
            if nums[i - 1] < nums[i] and nums[i + 1] < nums[i]:
                dp[i] = 1
                tree.update(i + 1, 1)
        res = []
        for a, i, j in queries:
            if a == 2:
                if i > 1:
                    if dp[i - 1] == 1:
                        if nums[i - 1] <= j:
                            dp[i - 1] = 0
                            tree.update(i, -1)
                    else:
                        if nums[i - 1] > nums[i - 2] and nums[i - 1] > j:
                            dp[i - 1] = 1
                            tree.update(i, 1)
                if i < n - 2:
                    if dp[i + 1] == 1:
                        if nums[i + 1] <= j:
                            dp[i + 1] = 0
                            tree.update(i + 2, -1)
                    else:
                        if nums[i + 1] > nums[i + 2] and nums[i + 1] > j:
                            dp[i + 1] = 1
                            tree.update(i + 2, 1)
                if i > 0 and i < n - 1:
                    if dp[i] == 1:
                        if j <= nums[i - 1] or j <= nums[i + 1]:
                            dp[i] = 0
                            tree.update(i + 1, -1)
                    else:
                        if j > nums[i - 1] and j > nums[i + 1]:
                            dp[i] = 1
                            tree.update(i + 1, 1)
                nums[i] = j
            else:
                if j - i < 2:
                    res.append(0)
                elif i + 1 == j:
                    if nums[i + 1] > nums[i] and nums[i + 1] > nums[i + 1]:
                        res.append(1)
                    else:
                        res.append(0)
                else:
                    res.append(tree.query(j) - tree.query(i + 1))
               
        return res
        
        
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