<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) {
        int n = a.size();
        map<int,int> mp2;map<int,int> mp1;
        int mi = a[k],ans=a[k];mp2[mi]=k;
        if(n==1){return ans;}
        vector<int> vmi(n,a[k]);
        for(int i=k;i<n;i++){
            mi = min(mi,a[i]);
            mp2[mi]=i;vmi[i]=mi;
        }
        mi = a[k];mp1[mi]=k;
        for(int i=k;i>=0;i--){
            mi= min(mi,a[i]);
            vmi[i] = mi;mp1[mi]=i;
        }
        for(int i=0;i<=k;i++){
            if(i<n-1&&vmi[i]==vmi[i+1])continue;
            int tp = vmi[i];
            auto it = mp2.lower_bound(tp);
            int id = ((it==mp2.end())?k:(*it).second);
            ans = max(ans,tp*(id-mp1[tp]+1));
        }
        for(int i=k;i<n;i++){
            if(i<n-1&&vmi[i]==vmi[i+1])continue;
            int tp = vmi[i];
            auto it = mp1.lower_bound(tp);
            int id = ((it==mp2.end())?k:(*it).second);
            ans = max(ans,tp*(mp2[tp]-id+1));
        }
        return ans;
    }
};
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