<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) {
        
        vector<long long> v;
        v.push_back(0);
        long long n = nums.size();
        int mx = nums[0];
        for(int i=0;i<n;i++) {
            mx = max(mx, nums[i]);
        }

        for(int i=1;i<=n;i++) {
            if(nums[i-1] == mx) {
                v.push_back(1);
            } else {
                v.push_back(0);
            }
            
            v[i] += v[i-1];
            
        }


        long long ans = 0;

        for(int i=1;i<=n;i++) {
            int l = i, r = n + 1;
            while(l < r) {
                long long m = (l + r)/2;
                long long totalOne = v[m]-v[i-1];
                if(totalOne < k) {
                    l = m + 1;
                } else {
                    r = m;
                }
            }
            
            if(r < n + 1) {
                ans += n - r + 1; 
            }
        }

        return ans;
        
    }
};
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