<p>You are given an integer array <code>nums</code> and a <strong>positive</strong> integer <code>k</code>.</p>

<p>Return <em>the number of subarrays where the <strong>maximum</strong> element of </em><code>nums</code><em> appears <strong>at least</strong> </em><code>k</code><em> times in that subarray.</em></p>

<p>A <strong>subarray</strong> is a contiguous sequence of elements within an array.</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>

<pre>
<strong>Input:</strong> nums = [1,3,2,3,3], k = 2
<strong>Output:</strong> 6
<strong>Explanation:</strong> The subarrays that contain the element 3 at least 2 times are: [1,3,2,3], [1,3,2,3,3], [3,2,3], [3,2,3,3], [2,3,3] and [3,3].
</pre>

<p><strong class="example">Example 2:</strong></p>

<pre>
<strong>Input:</strong> nums = [1,4,2,1], k = 3
<strong>Output:</strong> 0
<strong>Explanation:</strong> No subarray contains the element 4 at least 3 times.
</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;= 10<sup>6</sup></code></li>
	<li><code>1 &lt;= k &lt;= 10<sup>5</sup></code></li>
</ul>

class Solution {
public:
    long long countSubarrays(vector<int>& nums, int k) {
      int max=nums[0];
      int n=nums.size();
      for(int i=1;i<n;i++)
            if(max<nums[i]) max=nums[i];
      vector<int> max_index;
      max_index.push_back(-1);
      for(int i=0;i<n;i++)
            if(max==nums[i]) max_index.push_back(i);
      int index_size=max_index.size();
      if(index_size-1<k)
             return 0;
      long long subarray=0;
      vector<long long> sub;
      for(int i=1;i<=index_size-k;i++){
        long long a=(max_index[i]-max_index[i-1]);
        long long b=(n-max_index[i+k-1]);
        subarray+=a*b;
    }
 
    return subarray;
    }
    
};
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