<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();
        queue<int> minQ,maxQ;
        long long ans=0;
        if(minK==maxK){
            int ct=0;
            for(int i=0;i<n;++i){
                if(nums[i]==minK){
                    ct++;
                    ans+=ct;
                }
                else ct=0;
            }
            return ans;
        }
        int j=0;
        for(int i=0;i<=n;i++){
            while(i<n && nums[i]>=minK && nums[i]<=maxK){
                if(nums[i]==minK) minQ.push(i);
                if(nums[i]==maxK) maxQ.push(i);
                i++;
            }

            while(!minQ.empty() && !maxQ.empty()){
                int ct=max(minQ.front(),maxQ.front());
                ans+=(i-ct);
                if(nums[j]==minK) minQ.pop();
                if(nums[j]==maxK) maxQ.pop();
                j++;
            }
            while(!minQ.empty()) minQ.pop();
            while(!maxQ.empty()) maxQ.pop();
            j=i+1;
        }
        return ans;
    }   
};
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