<p>You are given an array of integers <code>nums</code> <strong>(0-indexed)</strong> and an integer <code>k</code>.</p>

<p>The <strong>score</strong> of a subarray <code>(i, j)</code> is defined as <code>min(nums[i], nums[i+1], ..., nums[j]) * (j - i + 1)</code>. A <strong>good</strong> subarray is a subarray where <code>i &lt;= k &lt;= j</code>.</p>

<p>Return <em>the maximum possible <strong>score</strong> of a <strong>good</strong> subarray.</em></p>

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

<pre>
<strong>Input:</strong> nums = [1,4,3,7,4,5], k = 3
<strong>Output:</strong> 15
<strong>Explanation:</strong> The optimal subarray is (1, 5) with a score of min(4,3,7,4,5) * (5-1+1) = 3 * 5 = 15. 
</pre>

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

<pre>
<strong>Input:</strong> nums = [5,5,4,5,4,1,1,1], k = 0
<strong>Output:</strong> 20
<strong>Explanation:</strong> The optimal subarray is (0, 4) with a score of min(5,5,4,5,4) * (4-0+1) = 4 * 5 = 20.
</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>1 &lt;= nums[i] &lt;= 2 * 10<sup>4</sup></code></li>
	<li><code>0 &lt;= k &lt; nums.length</code></li>
</ul>

class Solution {
private:
    void stack_op(int i, stack<int> &st, vector<int> &heights, vector<int> &nse){
        if (st.empty()){
            st.push(i);
        }
        else{
            while (!st.empty() && heights[st.top()] > heights[i]){
                nse[st.top()] = i;
                st.pop();
            }
            st.push(i);
        }
    }

    void find_nse(vector<int> &nse, bool forward, vector<int> &heights, int n){
        stack<int> st;
        if (forward){
            for (int i = 0; i < n; i++){
                stack_op(i, st, heights, nse);
            }
        }
        else{
            for (int i = n - 1; i >= 0; i--){
                stack_op(i, st, heights, nse);
            }
        }
    }

    int largestRectangleArea(vector<int>& heights, int k) {
        int n = heights.size();
        vector<int> nse_forward(n, -1), nse_backward(n, -1);
        find_nse(nse_forward, true, heights, n);
        find_nse(nse_backward, false, heights, n);

        int max_area = -1;
        for (int i = 0; i < n; i++){
            int cur_area = 0;
            int right_index = nse_forward[i], left_index = nse_backward[i];
            if (left_index == -1 && right_index == -1){
                left_index = -1;
                right_index = n;
                cur_area = heights[i] * n;
            }
            else if (left_index == -1){
                left_index = -1;
                cur_area = heights[i] * (right_index);
            }
            else if (right_index == -1){
                right_index = n;
                cur_area = heights[i] * (i - left_index + n - 1 - i);
            }
            else
                cur_area = heights[i] * (right_index - left_index - 1);
            
            if (left_index + 1 > k || right_index - 1 < k)
                continue;
            
            max_area = max(max_area, cur_area);
        }

        return max_area;
    }

public:
    int maximumScore(vector<int>& nums, int k) {
        return largestRectangleArea(nums, k);
    }
};
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