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, 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

    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