<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:
        # https://www.youtube.com/watch?v=z6LwIkEn9qc
        # https://algo.monster/liteproblems/2444
        # check for every subarray ending at index i
        # can that be valid subarray
        # to check subarray is valid
        # check value at index i is less than minK or greater than maxK
        # if it is then that's the culprit


        n = len(nums)
        minPos = -1
        maxPos = -1
        culpritPos = -1
        ans = 0

        # iterate ove every index
        # subarray ending at index i
        for i in range(n):
            
            # check i out of the range of mink and maxk
            if nums[i] < minK or nums[i] > maxK:
                culpritPos = i

            if nums[i] == minK:
                minPos = i
            
            if nums[i] == maxK:
                maxPos = i
            
            smallerIndx = min(minPos, maxPos)

            temp = smallerIndx - culpritPos

            if temp >= 0:
                ans += temp


        return ans


        
575