<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 p=INT_MIN;
       for(int t:nums)
       {
        p=max(p,t);
       }
        vector<long long int>vec1;
     for(int i=0;i<nums.size();i++)
     {
        if(nums[i]==p)
        vec1.push_back(i);
     }
   long long  int ans=0;   
for(int i=0;i<vec1.size();i++)
{
    long long int x;
    if(i+k-1<vec1.size())
    {
        x=nums.size()-vec1[i+k-1];
      long long int x1;
        if(i==0)
x1=0;
else
x1=vec1[i-1]+1;
long long int s=(vec1[i]-x1+1)*x;
ans=ans+s;
    }
}
return ans;
    }
};
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