<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:
        ans = 0
        min_pos = -1
        max_pos = -1
        left_bound = -1

        for i, num in enumerate(nums):
            # Numebr is outside of the range minK to maxK
            if num < minK or num > maxK:
                left_bound = i

            if num == minK:
                min_pos = i
            
            if num == maxK:
                max_pos = i

            # If both satisfy being more than left bound, then we form new subarrays
            # based on left bound and smaller of two positions to gaurantee 
            # we have min and max
            # This is because the number of elements between left_bound and the min
            # of the positions is the number of new subarrays we can form each time we 
            # find a new element that is within the bounds and we have already found min
            # and max
            if left_bound < min_pos and left_bound < max_pos:
                ans += min(min_pos, max_pos) - left_bound
        return ans





            

405