<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:
        max_q = deque()
        min_q = deque()
        ret = 0
        l, r = 0, 0
        bad_idx = -1

        while r < len(nums):
            while max_q and nums[r] >= nums[max_q[-1]]:
                max_q.pop()

            max_q.append(r)

            while min_q and nums[r] <= nums[min_q[-1]]:
                min_q.pop()

            min_q.append(r)

            if nums[max_q[0]] == maxK and nums[min_q[0]] == minK:
                ret += max(0, min(max_q[0], min_q[0]) - bad_idx)

            if max_q and nums[max_q[0]] > maxK:
                while max_q and nums[max_q[0]] > maxK:
                    idx = max_q.popleft()

                while min_q and min_q[0] <= idx:
                    min_q.popleft()

                bad_idx = max(bad_idx, idx)

            if min_q and nums[min_q[0]] < minK:
                while min_q and nums[min_q[0]] < minK:
                    idx = min_q.popleft()

                while max_q and max_q[0] <= idx:
                    max_q.popleft()

                bad_idx = max(bad_idx, idx)
            
            # if nums[max_q[0]] > maxK or nums[min_q[0]] < minK:
            #     if max_q[0] == l:
            #         max_q.popleft()

            #     if min_q[0] == l:
            #         min_q.popleft()

            #     bad_idx = l
            #     l += 1

            r += 1

        return ret

            
835