<p>Given an integer array <code>nums</code> and two integers <code>left</code> and <code>right</code>, return <em>the number of contiguous non-empty <strong>subarrays</strong> such that the value of the maximum array element in that subarray is in the range </em><code>[left, right]</code>.</p>

<p>The test cases are generated so that the answer will fit in a <strong>32-bit</strong> integer.</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>

<pre>
<strong>Input:</strong> nums = [2,1,4,3], left = 2, right = 3
<strong>Output:</strong> 3
<strong>Explanation:</strong> There are three subarrays that meet the requirements: [2], [2, 1], [3].
</pre>

<p><strong class="example">Example 2:</strong></p>

<pre>
<strong>Input:</strong> nums = [2,9,2,5,6], left = 2, right = 8
<strong>Output:</strong> 7
</pre>

<p>&nbsp;</p>
<p><strong>Constraints:</strong></p>

<ul>
	<li><code>1 &lt;= nums.length &lt;= 10<sup>5</sup></code></li>
	<li><code>0 &lt;= nums[i] &lt;= 10<sup>9</sup></code></li>
	<li><code>0 &lt;= left &lt;= right &lt;= 10<sup>9</sup></code></li>
</ul>

class Solution {
public:
    // **** Monotonic Stack (Non - Strictly) - O(N), O(N) ****
    vector<int> nextGreaterElementIndex(vector<int>& nums) {
        int n = nums.size();
        vector<int> res(n, n);
        stack<int> decStack;
        for (int i = 0; i < n; i++)
        {
            while(decStack.size() && nums[decStack.top()] <= nums[i]) {
                res[decStack.top()] = i;
                decStack.pop();
            }
            decStack.push(i);
        }
        return res;
    }

    vector<int> prevGreaterElementIndex(vector<int>& nums) {
        int n = nums.size();
        vector<int> res(n, -1);
        stack<int> decStack;
        for (int i = n - 1; i >= 0; i--)
        {
            while(decStack.size() && nums[decStack.top()] < nums[i]) {
                res[decStack.top()] = i;
                decStack.pop();
            }
            decStack.push(i);
        }
        return res;
    }

    int numSubarrayBoundedMax(vector<int>& nums, int left, int right) {
        int res = 0;
        vector<int> next = nextGreaterElementIndex(nums);
        vector<int> prev = prevGreaterElementIndex(nums);
        for(int i = 0; i < nums.size(); i++) {
            if(left <= nums[i] && nums[i] <= right) {
                res += (i - prev[i]) * (next[i] - i);
            }
        }
        return res;
    }
};
62000