# Solving Leetcode Interviews in Seconds with AI: Finding the Users Active Minutes


	# Introduction
	In this blog post, we will explore how to solve the LeetCode problem "1817" using AI. LeetCode is a popular platform for preparing for coding interviews, and with the help of AI tools like [Chatmagic](https://www.chatmagic.app), we can generate solutions quickly and efficiently - helping you pass the interviews and get the job offer without having to study for months.

	# Problem Statement
	> You are given the logs for users' actions on LeetCode, and an integer k. The logs are represented by a 2D integer array logs where each logs[i] = [IDi, timei] indicates that the user with IDi performed an action at the minute timei. Multiple users can perform actions simultaneously, and a single user can perform multiple actions in the same minute. The user active minutes (UAM) for a given user is defined as the number of unique minutes in which the user performed an action on LeetCode. A minute can only be counted once, even if multiple actions occur during it. You are to calculate a 1-indexed array answer of size k such that, for each j (1 <= j <= k), answer[j] is the number of users whose UAM equals j. Return the array answer as described above.   Example 1:  Input: logs = [[0,5],[1,2],[0,2],[0,5],[1,3]], k = 5 Output: [0,2,0,0,0] Explanation: The user with ID=0 performed actions at minutes 5, 2, and 5 again. Hence, they have a UAM of 2 (minute 5 is only counted once). The user with ID=1 performed actions at minutes 2 and 3. Hence, they have a UAM of 2. Since both users have a UAM of 2, answer[2] is 2, and the remaining answer[j] values are 0.  Example 2:  Input: logs = [[1,1],[2,2],[2,3]], k = 4 Output: [1,1,0,0] Explanation: The user with ID=1 performed a single action at minute 1. Hence, they have a UAM of 1. The user with ID=2 performed actions at minutes 2 and 3. Hence, they have a UAM of 2. There is one user with a UAM of 1 and one with a UAM of 2. Hence, answer[1] = 1, answer[2] = 1, and the remaining values are 0.    Constraints:  1 <= logs.length <= 104 0 <= IDi <= 109 1 <= timei <= 105 k is in the range [The maximum UAM for a user, 105].  

	# Explanation
	Here's the solution to the problem, incorporating efficiency and clarity:

*   **High-Level Approach:**
    *   Use a dictionary to store the unique active minutes for each user. The keys are user IDs, and the values are sets of active minutes.
    *   Iterate through the logs, updating the active minutes set for each user.
    *   Count the number of users with each UAM value and store it in the `answer` array.

*   **Complexity:**
    *   Runtime: O(N), where N is the number of logs.
    *   Space: O(U + K), where U is the number of unique users and K is the maximum possible UAM (given as input `k`).

	
	# Code
	```python
	def findingUsersActiveMinutes(logs, k):
    """
    Calculates the number of users with each UAM value.

    Args:
        logs: A list of logs, where each log is a list [user_id, timestamp].
        k: The maximum UAM to consider.

    Returns:
        A list of length k, where the i-th element is the number of users with UAM i.
    """

    user_active_minutes = {}  # {user_id: set(active_minutes)}
    for user_id, timestamp in logs:
        if user_id not in user_active_minutes:
            user_active_minutes[user_id] = set()
        user_active_minutes[user_id].add(timestamp)

    uam_counts = [0] * k  # Initialize the answer array
    for user_id in user_active_minutes:
        uam = len(user_active_minutes[user_id])
        if 1 <= uam <= k:
            uam_counts[uam - 1] += 1  # Adjust index to be 0-based

    return uam_counts
	```