You are given a string `s` consisting of lowercase English letters, and an integer `k`.

First, **convert** `s` into an integer by replacing each letter with its position in the alphabet (i.e., replace `'a'` with `1`, `'b'` with `2`, ..., `'z'` with `26`). Then, **transform** the integer by replacing it with the **sum of its digits**. Repeat the **transform** operation `k` **times** in total.

For example, if `s = "zbax "` and `k = 2`, then the resulting integer would be `8` by the following operations:

*   **Convert**: `"zbax " ➝ "(26)(2)(1)(24) " ➝ "262124 " ➝ 262124`
*   **Transform #1**: `262124 ➝ 2 + 6 + 2 + 1 + 2 + 4 ➝ 17`
*   **Transform #2**: `17 ➝ 1 + 7 ➝ 8`

Return _the resulting integer after performing the operations described above_.

**Example 1:**

**Input:** s =  "iiii ", k = 1
**Output:** 36
**Explanation:** The operations are as follows:
- Convert:  "iiii " ➝  "(9)(9)(9)(9) " ➝  "9999 " ➝ 9999
- Transform #1: 9999 ➝ 9 + 9 + 9 + 9 ➝ 36
Thus the resulting integer is 36.

**Example 2:**

**Input:** s =  "leetcode ", k = 2
**Output:** 6
**Explanation:** The operations are as follows:
- Convert:  "leetcode " ➝  "(12)(5)(5)(20)(3)(15)(4)(5) " ➝  "12552031545 " ➝ 12552031545
- Transform #1: 12552031545 ➝ 1 + 2 + 5 + 5 + 2 + 0 + 3 + 1 + 5 + 4 + 5 ➝ 33
- Transform #2: 33 ➝ 3 + 3 ➝ 6
Thus the resulting integer is 6.

**Example 3:**

**Input:** s =  "zbax ", k = 2
**Output:** 8

**Constraints:**

*   `1 <= s.length <= 100`
*   `1 <= k <= 10`
*   `s` consists of lowercase English letters.


from collections import defaultdict


def finding_users_active_minutes(logs, k):
    user_minutes = defaultdict(set)
    for log in logs:
        user_minutes[log[0]].add(log[1])
    result = [0] * k
    for minutes in user_minutes.values():
        if len(minutes) <= k:
            result[len(minutes) - 1] += 1
    return result
