<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> li(n),ri(n);

        vector<int> st(n);
        int stp = 0;
        li[0] = -1;
        st[stp] = 0;
       int i = 1;
        while(stp != -1 && i < n){
            while(stp != -1 && nums[st[stp]] >= nums[i] ){
                stp--;
            }
            if(stp != -1) li[i] = st[stp];
            else li[i] = -1;
            stp++;
            st[stp] = i;
            i++;
        }

        stp = 0;
        ri[n-1] = n;
        st[stp] = n-1;
        i = n-2;
        while(stp != -1 && i>-1){
            while(stp != -1 && nums[st[stp]] >= nums[i] ){
                stp--;
            }
            if(stp != -1) ri[i] = st[stp];
            else ri[i] = n;
            stp++;
            st[stp] = i;
            i--;
        }
        for(auto e:li){
            cout<<e<<" ";
        }
        cout<<endl;
        for(auto e:ri){
            cout<<e<<" ";
        }

        int gs = 0;
        for(int i =0; i < n; i++){
            if(li[i]+1 <= k && ri[i]-1 >= k){
                gs = max(gs, nums[i] * ( (ri[i] - 1) - (li[i] +1) + 1));
            }
        }
        
    return gs;

    }
};
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