<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) {
          stack<int>st;
          int n = nums.size();
          vector<int>left(n,-1), right(n,n);

          for(int i = 0 ; i < n ; i++){
              while(!st.empty() && nums[st.top()] >= nums[i]) {
                st.pop();
              }

              if(!st.empty()) left[i] = st.top();

              st.push(i);
          }

             while (!st.empty()) st.pop();
             for(int i = n-1 ; i >=0 ; i--){
              while(!st.empty() && nums[st.top()] >= nums[i]) {
                st.pop();
              }

              if(!st.empty()) right[i] = st.top();

              st.push(i);
          }


          int ans = 0;

          for(int i = 0 ; i < n ; i++){
            if(left[i] < k && right[i] > k){
                ans = max(ans , (right[i] - left[i]  - 1)* nums[i] );
            }
          }

          return ans;
    }
};
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