<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:
#define ll long long
/*
    [mink, maxk]
    A = no of subarrays <= maxk
    B = no of subarrays <= minK
*/

    ll get(vector<int>& A, int m, int n, int l)
    {
        int ismin = -1, ismax=-1;
        ll prev=0;
        ll ans=0;
        int h=l;
        while(h<A.size())
        {
            //cout<<ans<<" "<<h<<" "<<prev<< " "<<ismin<<" "<<ismax<<endl;
            if(A[h]==m)
            {
                ismin=h;
            }
            if(A[h]==n)
            {
                ismax=h;
            }
            if(A[h]==m && ismax!=-1)
            {
                prev=ismax-l+1;
                //cout<<prev<<endl;
                ismin=h;
            }
            if(A[h]==n && ismin!=-1)
            {
                prev=ismin-l+1;
                //cout<<prev<<endl;
                ismax=h;
            }
            if(A[h]<m || A[h]>n)
            {
                return ans+get(A,m,n,h+1);
            }
            ans+=prev;
            h++;
        }
        return ans;
    }

    long long countSubarrays(vector<int>& A, int m, int n) {
     return get(A,m,n,0);   
    }
};
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