<p>You are given an integer array <code>nums</code> and two integers <code>minK</code> and <code>maxK</code>.</p>

<p>A <strong>fixed-bound subarray</strong> of <code>nums</code> is a subarray that satisfies the following conditions:</p>

<ul>
	<li>The <strong>minimum</strong> value in the subarray is equal to <code>minK</code>.</li>
	<li>The <strong>maximum</strong> value in the subarray is equal to <code>maxK</code>.</li>
</ul>

<p>Return <em>the <strong>number</strong> of fixed-bound subarrays</em>.</p>

<p>A <strong>subarray</strong> is a <strong>contiguous</strong> part of an array.</p>

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

<pre>
<strong>Input:</strong> nums = [1,3,5,2,7,5], minK = 1, maxK = 5
<strong>Output:</strong> 2
<strong>Explanation:</strong> The fixed-bound subarrays are [1,3,5] and [1,3,5,2].
</pre>

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

<pre>
<strong>Input:</strong> nums = [1,1,1,1], minK = 1, maxK = 1
<strong>Output:</strong> 10
<strong>Explanation:</strong> Every subarray of nums is a fixed-bound subarray. There are 10 possible subarrays.
</pre>

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

<ul>
	<li><code>2 &lt;= nums.length &lt;= 10<sup>5</sup></code></li>
	<li><code>1 &lt;= nums[i], minK, maxK &lt;= 10<sup>6</sup></code></li>
</ul>

class Solution:
    def countSubarrays(self, nums: List[int], minK: int, maxK: int) -> int:
        k = 0
        i = 0
        l = []
        h = []
        cnt = [0]*len(nums)
        cnt2 = 0
        ret = 0

        while k < len(nums) and i < len(nums):
            if nums[k] < minK or nums[k] > maxK:
                k += 1
                i = k
                continue
            
            if nums[i] == minK:
                l.append(i)
            
            if nums[i] == maxK:
                h.append(i)

            if nums[i] < minK or nums[i] > maxK:
                k = i+1
                ret += cnt2
                cnt2 = 0
                l = []
                h = []
            elif len(l) > 0 and len(h) > 0:
                # if i == l[0] or i == h[0]:
                #     cnt2 += len(l)*len(h)
                #     max_i = max(l[0], h[0])
                min_i = min(l[-1], h[-1])
                cnt[i] = 1 + min_i - k
                # else:
                #     cnt[i] = cnt[i-1]

            i += 1


        return sum(cnt)
763