<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) {

        vector<int> mina;
        vector<int> maxa; 
        long long int ans = 0; 
        for(int i=0; i<nums.size(); i++) {
            if(nums[i] == minK) {
                mina.push_back(i);
            }

            if(nums[i] == maxK) {   
                maxa.push_back(i);
            }
        }

        for(int i=0; i<nums.size(); i++) {
            if(nums[i] < minK || nums[i] > maxK) {
                continue; 
            }

            int l = i; 
            while( i < nums.size() && nums[i] >= minK && nums[i] <= maxK) {
                i++; 
            }

            i--; 
            int last = l-1; 
            for(int j= l; j<=i; j++) {

                if(nums[j] == minK) {
                    int next = lower_bound(maxa.begin(), maxa.end(), j) - maxa.begin();
                    if(next == maxa.size()) continue; 
                    long long cnt1 = j - last; 
                    long int cnt2 = i - maxa[next] + 1;
                    if(cnt2>0) {
                    ans+= cnt1* cnt2;  
                    last = j; 
                    }

                }

                if(nums[j] == maxK) {

                    int next = lower_bound(mina.begin(), mina.end(), j) - mina.begin();
                    if(next == mina.size()) continue; 
                    long long cnt1 = j - last; 
                    long int cnt2 = i - mina[next] + 1;
                    if(cnt2>0) {
                    ans+= cnt1* cnt2;  
                    last = j; 
                    }

                }
            }
        }
        return ans; 
    }
};

// 4 -- 4
/*
1, 5,2,5,5, 3, 1, 1, 4

*/


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