<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:
        def divideArray(nums):
            divided = []
            st = []
            for n in nums:
                if minK <= n <= maxK:
                    st.append(n)
                else:
                    divided.append(st)
                    st = []
            if st: 
                divided.append(st)
            return divided 
        
        def helper(arr):
            # all the numbers in the arr are between minK and maxK inclusive 
            minK_idx, maxK_idx = -1, -1
            count = 0 
            for idx, n in enumerate(arr):
                if n == minK: 
                    minK_idx = idx 
                if n == maxK:
                    maxK_idx = idx 
                if maxK_idx != -1 and minK_idx != -1: 
                    # the subarray that ends at this current idx is a valid subarray 
                    # since we have minK and maxK appeared. 
                    count += min(maxK_idx, minK_idx) + 1
            return count 
        
        arr = divideArray(nums)
        count = 0 
        for a in arr: 
            count += helper(a)
        return count

                
        
513