<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();
        vector<int> stop;
        stop.push_back(-1);
        stack<int> st_max, st_min;
        vector<int> next_max(n,n), next_min(n,n);
        for(int i = 0; i < n; i++)
        {
            if(nums[i] > maxK || nums[i] < minK)
            {
                stop.push_back(i);
            }
            else
            {
                if(nums[i] == maxK)
                {
                    next_max[i] = i;
                    while(!st_max.empty())
                    {
                        next_max[st_max.top()] = i;
                        st_max.pop();
                    }
                    st_min.push(i);
                }
                if(nums[i] == minK)
                {
                    next_min[i] = i;
                    while(!st_min.empty())
                    {
                        next_min[st_min.top()] = i;
                        st_min.pop();
                    }
                    if(maxK != minK) st_max.push(i);
                }
                if(nums[i] != maxK && nums[i] != minK)
                {
                    st_max.push(i);
                    st_min.push(i);
                }
            }
        }
        stop.push_back(n);
        long long ans = 0;
        for(int i = 0; i < n; i++)
        {
            if(nums[i] > maxK || nums[i] < minK) continue;
            int next_max_idx = next_max[i];
            int next_min_idx = next_min[i];
            int limit_idx = lower_bound(stop.begin(), stop.end(), i) - stop.begin();
            int limit = stop[limit_idx];
           
            if(max(next_max_idx, next_min_idx) >= limit) continue;
            
            ans += limit - max(next_max_idx, next_min_idx);
        }

        return ans;
    }
};
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