<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>& a, int minK, int maxK) {
        int n = a.size();
        
        deque<int> minDeque;
        deque<int> maxDeque;
        
        int i=0,j=0;
        long ans=0,curr=0;
        
        while(j<n){
            while(!minDeque.empty() && a[minDeque.back()] >= a[j])
                minDeque.pop_back();
            
            minDeque.push_back(j);
            
            while(!maxDeque.empty() && a[maxDeque.back()] <= a[j])
                maxDeque.pop_back();
            
            maxDeque.push_back(j);
            
            if(a[minDeque.front()] == minK && a[maxDeque.front()] == maxK){
                while(i != minDeque.front() && i != maxDeque.front())
                    curr++, i++;
                
                ans += curr+1;
            }
            
            else if(a[minDeque.front()] < minK || a[maxDeque.front()] > maxK){
                i = j+1, curr=0;
                minDeque.clear();
                maxDeque.clear();
            }
            
            j++;
        }
        
        return ans;
    }
};
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