<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> nxtmx(n,n),nxtmn(n,n);
        for(int i=n-1;i>=0;i--)
        {
            if(nums[i]==maxK)
                nxtmx[i]=i;
            else if(i+1<n)
                nxtmx[i]=nxtmx[i+1];
            
            if(nums[i]==minK)
                nxtmn[i]=i;
            else if(i+1<n)
                nxtmn[i]=nxtmn[i+1];
        }
        int l=0,r=0;
        long long ans=0;
        while(l<n)
        {
            while(r<n&&nums[r]<=maxK&&nums[r]>=minK)
            {
                r++;
            }
            if(l==r)
            {
                while(l<n&&(nums[l]<minK||nums[l]>maxK))
                {
                    l++;
                }
                r=l;
                continue;
            }
            long long cnt=0;
            for(int i=l;i<r;i++)
            {
                cnt++;
                if(nums[i]==minK)
                {
                    long long rcnt=max(0,r-nxtmx[i]);
                    ans+=(cnt*rcnt);
                    cnt=0;
                }
                else if(nums[i]==maxK)
                {
                    long long rcnt=max(0,r-nxtmn[i]);
                    ans+=(cnt*rcnt);
                    cnt=0;
                }
            }
            l=max(r,l+1);
            r=max(r,l);
        }
    return ans;
    }
};
89800