human
I found an interesting problem on https://leetcode.com/problems/subarrays-with-k-different-integers/:
Given an array A of positive integers, call a (contiguous, not necessarily distinct) subarray of A good if the number of different integers in that subarray is exactly K.
(For example, [1,2,3,1,2] has 3 different integers: 1, 2, and 3.)
Return the number of good subarrays of A.
 
Example 1:
Input: A = [1,2,1,2,3], K = 2
Output: 7
Explanation: Subarrays formed with exactly 2 different integers: [1,2], [2,1], [1,2], [2,3], [1,2,1], [2,1,2], [1,2,1,2].

Example 2:
Input: A = [1,2,1,3,4], K = 3
Output: 3
Explanation: Subarrays formed with exactly 3 different integers: [1,2,1,3], [2,1,3], [1,3,4].

 
Note:

1 <= A.length <= 20000
1 <= A[i] <= A.length
1 <= K <= A.length

I tried it in Python, but could not do it. Can you solve it?
assistant
#Here is the Python program you requested.

class Solution:
    
    def _identify_limit(self, A, current_limit, K,  char_count_after_limit):
        char_count_after_limit = {k: v for k, v in char_count_after_limit.items()}
        unique_chars = {i for i in char_count_after_limit if char_count_after_limit[i] > 0}
        # print(unique_chars)
        while len(unique_chars) == K : 
            # print(\"WHILE\", char_count_after_limit)
            temp_char = A[current_limit]
            char_count_after_limit[temp_char] = char_count_after_limit[temp_char] - 1
            if char_count_after_limit[temp_char] == 0: 
                unique_chars.remove(temp_char)
            current_limit += 1
            
        char_count_after_limit[A[current_limit - 1]] = char_count_after_limit[A[current_limit - 1]] + 1
        return  current_limit - 1, char_count_after_limit
    
    def subarraysWithKDistinct(self, A: List[int], K: int) -> int:
        # char_info = {item : [] for item in set(A)} 
        # unique_char_count = {}
        # unique_char = set()
        # start_idx = 0 
        # end_idx = 0
        # num_substr = 0
        # current_subarray = []
        # current_valid_count = 0 
        # # In any valid subarray what is the index upto which left index could be moved without making subarray bad 
        # # Whenever end_idx is moved to right keeping subarray good compute limit_idx again 
        # limit_idx = 0
        # while end_idx < len(A) + 1 : 
        #     if len(unique_char) == K : 
        #         # temp_unique_char = {i for i in unique_char}
        #         # temp_char_count = {k:v for k, v in unique_char_count.items()}
        #         # temp_start_idx = start_idx 
        #         # while len(temp_unique_char) == K:
        #         #     temp_char = A[temp_start_idx]
        #         #     temp_char_count[temp_char] = temp_char_count[temp_char] - 1
        #         #     if temp_char_count[temp_char] == 0: 
        #         #         temp_unique_char.remove(temp_char)
        #         #     temp_start_idx += 1
        # ...