<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<pair<int,int>>v;
        int prev = -1;
        for(int i = 0;i < nums.size(); i++){
            if(nums[i] < minK || nums[i] > maxK){
                v.push_back({prev + 1,i - 1});
                prev = i;
            }
        }

        if(prev + 1< nums.size()) v.push_back({prev + 1,n -1});
        auto f = [&](int x,int y)->long long{
            cout << x <<" " << y << endl;
            int cntMx = 0;
            int cntMn = 0;
            int i = x,j = x;
            long long ans = 0;
            while(j <= y){
                if(cntMx > 0 && cntMn > 0){
                    ans += y - j + 2;
                    if(maxK == nums[i]) cntMx--;
                    if(minK == nums[i]) cntMn--;
                    i++;
                }else{
                    if(maxK == nums[j]) cntMx++;
                    if(minK == nums[j]) cntMn++;
                    j++;
                }
            }
            while(i <= y){
                if(cntMx > 0 && cntMn > 0){
                    ans += y - j + 2;
                    if(maxK == nums[i]) cntMx--;
                    if(minK == nums[i]) cntMn--;
                    i++;
                }
                else break;
            }
            return ans;
        };
        long long k = 0;
        for(int i = 0;i < v.size(); i++)   k += f(v[i].first,v[i].second);
        return k;
    }
};
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