<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 count(self,nums,l,r,minK,maxK):
        if r-l<=1:
            return 0
        mink_ind=[]
        maxk_ind=[]
        for i,n in enumerate(nums[l+1:r]):
            if n==minK:
                mink_ind.append(i)
            if n==maxK:
                maxk_ind.append(i)
        p_min=0
        p_max=0
        ans=0
        seg_length=r-(l+1)
        for i,n in enumerate(nums[l+1:r]):
            while p_min<len(mink_ind) and mink_ind[p_min]<i:
                p_min+=1
            if p_min==len(mink_ind):
                return ans
            while p_max<len(maxk_ind) and maxk_ind[p_max]<i:
                p_max+=1
            if p_max==len(maxk_ind):
                return ans
            left_most_right=max(mink_ind[p_min],maxk_ind[p_max])
            ans+=seg_length-1-left_most_right+1
        return ans
    def countSubarrays(self, nums: List[int], minK: int, maxK: int) -> int:
        l=-1
        ans=0
        for i, n in enumerate(nums):
            if n<minK or n>maxK:
                ans+=self.count(nums,l,i,minK,maxK)
                l=i
        ans+=self.count(nums,l,len(nums),minK,maxK)
        return ans
781