<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>& a, int k) {
        // important idea : min decreases/max increases when size of subarray increases
        // nxt[x] : first idx to the right of "k" s.t. a[idx] < x
        // prv[x] : first idx to the left of "k" s.t. a[idx] < x

        int n = a.size();

        vector<int> nxt(a[k]+1 , n-1) , prv(a[k]+1 , 0);

        int done = a[k];
        for(int i=k+1;i<n;i++){
            for(int j=done;j>a[i];j--){
                nxt[j] = i-1;
            }
            done = min(done , a[i]);
        }

        done = a[k];
        for(int i=k-1;i>=0;i--){
            for(int j=done;j>a[i];j--){
                prv[j] = i+1;
            }
            done = min(done , a[i]);
        }

        int ans = 0;
        for(int i=a[k];i>=0;i--){
            // considering i is the minimum
            int l = prv[i];
            int r = nxt[i];
            // [l.. k ..r]
            ans = max(ans , (r-l+1) * i);
        }
        return ans;
    }
};
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