<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 {
public:
    int maximumScore(vector<int>& nums, int k) {

        int N = nums.size();
        vector<int> left_smallest_num(N);
        vector<int> right_smallest_num(N);

        stack<int> l_stc{};
        stack<int> r_stc{};

        for (int i = 0; i < N; i++) {
            while (!l_stc.empty() && nums[l_stc.top()] >= nums[i]) {
                l_stc.pop();
            }
            if (l_stc.empty())
                left_smallest_num[i] = -1;
            else
                left_smallest_num[i] = l_stc.top();

            l_stc.push(i);
        }

        for (int i = N - 1; i >= 0; i--) {
            while (!r_stc.empty() && nums[r_stc.top()] >= nums[i]) {
                r_stc.pop();
            }
            if (r_stc.empty())
                right_smallest_num[i] = N;
            else
                right_smallest_num[i] = r_stc.top();

            r_stc.push(i);
        }

        int max_area = 0;

        for (int i = 0; i < N; i++) {
            if (left_smallest_num[i] < k and right_smallest_num[i] > k) {
                int width = right_smallest_num[i] - left_smallest_num[i] - 1;
                int area = nums[i] * width;
                max_area = max(max_area, area);
            }
        }

        return max_area;
    }
};
103500