<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:
        min_count = 0
        max_count = 0
        left_min, right_min, left_max, right_max = 0,0,0,0
        min_indices = []
        max_indices = []
        left, right = 0,0
        n = len(nums)
        result = 0
        problem = 0

        while right < n:
            if nums[right] == minK:
                min_count += 1
                mindex = right
            if nums[right] == maxK:
                max_count += 1
                maxdex = right
            elif nums[right] > maxK or nums[right] < minK:
                problem = right+1
                left, right = right, right
                min_count = 0
                max_count = 0
                mindex = maxdex = 0
            if max_count > 0 and min_count > 0:
                result += min(maxdex, mindex) - problem + 1
            # print(right_max, right_min, result)

            right+= 1
        return result
            
            
        
30900