<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) {
        deque<int> maxq, minq;
        int i = 0, n = nums.size();
        long long count = 0;
        int validStart = -1;
        int lastMinK = -1, lastMaxK = -1;

        for (int j = 0; j < n; ++j) {
            while (!maxq.empty() && nums[j] >= nums[maxq.back()]) 
                maxq.pop_back();
            maxq.push_back(j);
            
            while (!minq.empty() && nums[j] <= nums[minq.back()]) 
                minq.pop_back();
            minq.push_back(j);
            
            if (nums[j] < minK || nums[j] > maxK) {
                validStart = j;
            }
            
            if (nums[j] == minK) lastMinK = j;
            if (nums[j] == maxK) lastMaxK = j;

            while (!maxq.empty() && nums[maxq.front()] > maxK) 
                maxq.pop_front();
            while (!minq.empty() && nums[minq.front()] < minK) 
                minq.pop_front();

            if (lastMinK != -1 && lastMaxK != -1) {
                int newStart = min(lastMinK, lastMaxK);
                if (newStart > validStart) {
                    count += newStart - validStart;
                }
            }
        }

        return count;
    }
};

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