Count Complete Subarrays in an Array | Leetcode 2799

This video explains Count Complete Subarrays in an Array using the most optimal sliding window two pointer approach.
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[chandramoulyv]

Wonderful explanation sir. I am able to understand the logic easily.

[sailendrachettri]

Brute force approach was able to pass 1007 test cases :)
Thank you for better approach.

[AJ-jqhm]

This felt pretty easy, especially after the question asked a few days ago called "Count Good Subarrays". At first I used a set to count distinct elements, and a hashmap to traverse the window.. But since the value of nums[i] was pretty small, it was just easier and faster to take 2 arrays of size 2001.

class Solution {
public int countCompleteSubarrays(int[] arr) {
int[] set = new int[2001], temp = new int[2001];;
int dist = 0, left = 0, ans = 0, count = 0;
for(int i : arr){
set[i]++;
dist += set[i] == 1 ? 1 : 0;
}
for(int i = 0; i < arr.length; ++i){
temp[arr[i]]++;
count += temp[arr[i]] == 1 ? 1 : 0;
while(left <= i && count == dist){
ans += arr.length - i;
temp[arr[left]]--;
if(temp[arr[left]] == 0)
count--;
left++;
}
}
return ans;
}
}

[Anikait-hd]

the new leetcode ui is such a disappointment according to me . What is your thought on it?