<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 ans=0;
        stack<int> st;
        stack<int> st_r;
        vector<int> NSL;
        vector<int> NSR(n,0);
        for(int i=0;i<n;i++){
            while(!st.empty() && nums[i]<=nums[st.top()]){
                st.pop();
            }
            if(st.empty()) NSL.push_back(-1);
            else NSL.push_back(st.top());
            st.push(i);
        }
        // st.clear();
        for(int i=n-1;i>=0;i--){ // decreasing monotonic stack
            while(!st_r.empty() && nums[i]<=nums[st_r.top()]){
                st_r.pop();
            }
            if(st_r.empty()) NSR[i]=n;
            else NSR[i]=st_r.top();
            st_r.push(i);
        }
        for(int i=0;i<n;i++){
            if(NSL[i]+1<=k && NSR[i]-1>=k){ // i<=k && j>=k // here i = NSL[i]+1. and j=NSR[i]-1
                ans=max(ans,nums[i]*(NSR[i]-NSL[i]-1));
            }
        }
    return ans;
    }
};
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