<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 {
public:
    long long countSubarrays(vector<int>& nums, int minK, int maxK) {
        int n = nums.size();
        int out = -1, lower = -1, upper = -1; // pointers to last seen out-of-range element, minK, and maxK
        long long result = 0;
        for (int i = 0; i < n; i++) {
            if (nums[i] == minK) {
                lower = i; // update last seen index of minK
            }
            if (nums[i] == maxK) {
                upper = i; // update last seen index of maxK
            }
            if (nums[i] < minK || nums[i] > maxK) {
                out = i; // update last seen index of out-of-range element
            }
            int left = min(lower, upper); // find the index of the later of minK and maxK
            if (out < left) { 
                result += left - out; // if both minK and maxK have been seen more recently than any out-of-range element, add the count of valid subarrays
            }
        }
        return result;
    }
};
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