LeetCode in Kotlin

3433. Count Mentions Per User

Medium

You are given an integer numberOfUsers representing the total number of users and an array events of size n x 3.

Each events[i] can be either of the following two types:

1. Message Event: ["MESSAGE", "timestampi", "mentions_stringi"]
   • This event indicates that a set of users was mentioned in a message at timestampi.
   • The mentions_stringi string can contain one of the following tokens:
     • id<number>: where <number> is an integer in range [0,numberOfUsers - 1]. There can be multiple ids separated by a single whitespace and may contain duplicates. This can mention even the offline users.
     • ALL: mentions all users.
     • HERE: mentions all online users.
2. Offline Event: ["OFFLINE", "timestampi", "idi"]
   • This event indicates that the user idi had become offline at timestampi for 60 time units. The user will automatically be online again at time timestampi + 60.

Return an array mentions where mentions[i] represents the number of mentions the user with id i has across all MESSAGE events.

All users are initially online, and if a user goes offline or comes back online, their status change is processed before handling any message event that occurs at the same timestamp.

Note that a user can be mentioned multiple times in a single message event, and each mention should be counted separately.

Example 1:

Input: numberOfUsers = 2, events = [[“MESSAGE”,”10”,”id1 id0”],[“OFFLINE”,”11”,”0”],[“MESSAGE”,”71”,”HERE”]]

Output: [2,2]

Explanation:

Initially, all users are online.

At timestamp 10, id1 and id0 are mentioned. mentions = [1,1]

At timestamp 11, id0 goes offline.

At timestamp 71, id0 comes back online and "HERE" is mentioned. mentions = [2,2]

Example 2:

Input: numberOfUsers = 2, events = [[“MESSAGE”,”10”,”id1 id0”],[“OFFLINE”,”11”,”0”],[“MESSAGE”,”12”,”ALL”]]

Output: [2,2]

Explanation:

Initially, all users are online.

At timestamp 10, id1 and id0 are mentioned. mentions = [1,1]

At timestamp 11, id0 goes offline.

At timestamp 12, "ALL" is mentioned. This includes offline users, so both id0 and id1 are mentioned. mentions = [2,2]

Example 3:

Input: numberOfUsers = 2, events = [[“OFFLINE”,”10”,”0”],[“MESSAGE”,”12”,”HERE”]]

Output: [0,1]

Explanation:

Initially, all users are online.

At timestamp 10, id0 goes offline.

At timestamp 12, "HERE" is mentioned. Because id0 is still offline, they will not be mentioned. mentions = [0,1]

Constraints:

• 1 <= numberOfUsers <= 100
• 1 <= events.length <= 100
• events[i].length == 3
• events[i][0] will be one of MESSAGE or OFFLINE.
• 1 <= int(events[i][1]) <= 105
• The number of id<number> mentions in any "MESSAGE" event is between 1 and 100.
• 0 <= <number> <= numberOfUsers - 1
• It is guaranteed that the user id referenced in the OFFLINE event is online at the time the event occurs.

Solution

class Solution {
    fun countMentions(numberOfUsers: Int, events: List<List<String>>): IntArray {
        val ans = IntArray(numberOfUsers)
        val l: MutableList<Int?> = ArrayList<Int?>()
        var c = 0
        for (i in events.indices) {
            val s = events[i][0]
            val ss = events[i][2]
            if (s == "MESSAGE") {
                if (ss == "ALL" || ss == "HERE") {
                    c++
                    if (ss == "HERE") {
                        l.add(events[i][1].toInt())
                    }
                } else {
                    val sss: Array<String?> = ss.split(" ".toRegex()).dropLastWhile { it.isEmpty() }.toTypedArray()
                    for (j in sss.indices) {
                        val jj = sss[j]!!.substring(2, sss[j]!!.length).toInt()
                        ans[jj]++
                    }
                }
            }
        }
        for (i in events.indices) {
            if (events[i][0] == "OFFLINE") {
                val id = events[i][2].toInt()
                val a = events[i][1].toInt() + 60
                for (j in l.indices) {
                    if (l[j]!! >= a - 60 && l[j]!! < a) {
                        ans[id]--
                    }
                }
            }
        }
        for (i in 0..<numberOfUsers) {
            ans[i] += c
        }
        return ans
    }
}

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