<p>You are given an array of <strong>positive</strong> integers <code>nums</code>.</p>

<p>Return the number of <span data-keyword="subarray-nonempty">subarrays</span> of <code>nums</code>, where the <strong>first</strong> and the <strong>last</strong> elements of the subarray are <em>equal</em> to the <strong>largest</strong> element in the subarray.</p>

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

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

<p><strong>Output:</strong> <span class="example-io">6</span></p>

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

<p>There are 6 subarrays which have the first and the last elements equal to the largest element of the subarray:</p>

<ul>
	<li>subarray <code>[<strong><u>1</u></strong>,4,3,3,2]</code>, with its largest element 1. The first element is 1 and the last element is also 1.</li>
	<li>subarray <code>[1,<u><strong>4</strong></u>,3,3,2]</code>, with its largest element 4. The first element is 4 and the last element is also 4.</li>
	<li>subarray <code>[1,4,<u><strong>3</strong></u>,3,2]</code>, with its largest element 3. The first element is 3 and the last element is also 3.</li>
	<li>subarray <code>[1,4,3,<u><strong>3</strong></u>,2]</code>, with its largest element 3. The first element is 3 and the last element is also 3.</li>
	<li>subarray <code>[1,4,3,3,<u><strong>2</strong></u>]</code>, with its largest element 2. The first element is 2 and the last element is also 2.</li>
	<li>subarray <code>[1,4,<u><strong>3,3</strong></u>,2]</code>, with its largest element 3. The first element is 3 and the last element is also 3.</li>
</ul>

<p>Hence, we return 6.</p>
</div>

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

<div class="example-block">
<p><strong>Input:</strong> <span class="example-io">nums = [3,3,3]</span></p>

<p><strong>Output:</strong> <span class="example-io">6</span></p>

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

<p>There are 6 subarrays which have the first and the last elements equal to the largest element of the subarray:</p>

<ul>
	<li>subarray <code>[<u><strong>3</strong></u>,3,3]</code>, with its largest element 3. The first element is 3 and the last element is also 3.</li>
	<li>subarray <code>[3,<strong><u>3</u></strong>,3]</code>, with its largest element 3. The first element is 3 and the last element is also 3.</li>
	<li>subarray <code>[3,3,<u><strong>3</strong></u>]</code>, with its largest element 3. The first element is 3 and the last element is also 3.</li>
	<li>subarray <code>[<strong><u>3,3</u></strong>,3]</code>, with its largest element 3. The first element is 3 and the last element is also 3.</li>
	<li>subarray <code>[3,<u><strong>3,3</strong></u>]</code>, with its largest element 3. The first element is 3 and the last element is also 3.</li>
	<li>subarray <code>[<u><strong>3,3,3</strong></u>]</code>, with its largest element 3. The first element is 3 and the last element is also 3.</li>
</ul>

<p>Hence, we return 6.</p>
</div>

<p><strong class="example">Example 3:</strong></p>

<div class="example-block">
<p><strong>Input:</strong> <span class="example-io">nums = [1]</span></p>

<p><strong>Output:</strong> <span class="example-io">1</span></p>

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

<p>There is a single subarray of <code>nums</code> which is <code>[<strong><u>1</u></strong>]</code>, with its largest element 1. The first element is 1 and the last element is also 1.</p>

<p>Hence, we return 1.</p>
</div>

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

<ul>
	<li><code>1 &lt;= nums.length &lt;= 10<sup>5</sup></code></li>
	<li><code>1 &lt;= nums[i] &lt;= 10<sup>9</sup></code></li>
</ul>

class Solution:
    def numberOfSubarrays(self, nums: List[int]) -> int:
        stack = [] # [number, frequency]
        ans = 0
        for n in nums:
            while stack and n > stack[-1][0]:
                _, f = stack.pop()
                ans += (f * (f + 1)) // 2
            if stack and stack[-1][0] == n:
                stack[-1][1] += 1
            else:
                stack.append([n, 1])
        return ans + sum((f * (f + 1)) // 2 for _, f in stack)
876