<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();

        int minV, score = 0, s;
        vector<pair<int, int>> left;
        
        minV = nums[k];
        for (int i = k - 1; i >= 0; i--) {
            if (nums[i] < minV) {
                left.push_back({minV, i + 1});
                minV = nums[i];
            }
        }
        if (left.empty() || minV < left.back().first) left.push_back({minV, 0});
        // for (auto [v, i] : left) cout << i << "(" << v << ")" << endl;
        // cout << endl;
    
        minV = nums[k];
        for (int j = k + 1; j < n; j++) {
            if (nums[j] < minV) {
                for (auto [v1, i] : left) {
                    int s = min(minV, v1) * (j - i);
                    score = max(score, s);
                }
                minV = nums[j];
            }
        }
        int j = n;
        for (auto [v1, i] : left) {
            int s = min(minV, v1) * (j - i);
            score = max(score, s);
        }
        return score;
    }
};
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