LeetCode in Kotlin
2933. High-Access Employees
Medium
You are given a 2D 0-indexed array of strings, access_times, with size n. For each i where 0 <= i <= n - 1, access_times[i][0] represents the name of an employee, and access_times[i][1] represents the access time of that employee. All entries in access_times are within the same day.
The access time is represented as four digits using a 24-hour time format, for example, "0800" or "2250".
An employee is said to be high-access if he has accessed the system three or more times within a one-hour period.
Times with exactly one hour of difference are not considered part of the same one-hour period. For example, "0815" and "0915" are not part of the same one-hour period.
Access times at the start and end of the day are not counted within the same one-hour period. For example, "0005" and "2350" are not part of the same one-hour period.
Return a list that contains the names of high-access employees with any order you want.
Example 1:
Input: access_times = [[“a”,”0549”],[“b”,”0457”],[“a”,”0532”],[“a”,”0621”],[“b”,”0540”]]
Output: [“a”]
Explanation: “a” has three access times in the one-hour period of [05:32, 06:31] which are 05:32, 05:49, and 06:21. But “b” does not have more than two access times at all. So the answer is [“a”].
Example 2:
Input: access_times = [[“d”,”0002”],[“c”,”0808”],[“c”,”0829”],[“e”,”0215”],[“d”,”1508”],[“d”,”1444”],[“d”,”1410”],[“c”,”0809”]]
Output: [“c”,”d”]
Explanation: “c” has three access times in the one-hour period of [08:08, 09:07] which are 08:08, 08:09, and 08:29. “d” has also three access times in the one-hour period of [14:10, 15:09] which are 14:10, 14:44, and 15:08. However, “e” has just one access time, so it can not be in the answer and the final answer is [“c”,”d”].
Example 3:
Input: access_times = [[“cd”,”1025”],[“ab”,”1025”],[“cd”,”1046”],[“cd”,”1055”],[“ab”,”1124”],[“ab”,”1120”]]
Output: [“ab”,”cd”]
Explanation: “ab” has three access times in the one-hour period of [10:25, 11:24] which are 10:25, 11:20, and 11:24. “cd” has also three access times in the one-hour period of [10:25, 11:24] which are 10:25, 10:46, and 10:55. So the answer is [“ab”,”cd”].
Constraints:
1 <= access_times.length <= 100access_times[i].length == 21 <= access_times[i][0].length <= 10access_times[i][0]consists only of English small letters.access_times[i][1].length == 4access_times[i][1]is in 24-hour time format.access_times[i][1]consists only of'0'to'9'.
Solution
class Solution {
private fun isPossible(a: Int, b: Int): Boolean {
val hb = b / 100
val ha = a / 100
var mind = b % 100
val mina = a % 100
if (hb == 23 && ha == 0) {
return false
}
if (hb - ha > 1) {
return false
}
if (hb - ha == 1) {
mind += 60
}
return mind - mina < 60
}
private fun isHighAccess(list: List<Int>): Boolean {
if (list.size < 3) {
return false
}
var i = 0
var j = 1
var k = 2
while (k < list.size) {
val a = list[i++]
val b = list[j++]
val c = list[k++]
if (isPossible(a, c) && isPossible(b, c) && isPossible(a, b)) {
return true
}
}
return false
}
private fun stringToInt(str: String): Int {
var i = 1000
var `val` = 0
for (ch in str.toCharArray()) {
val n = ch.code - '0'.code
`val` += i * n
i = i / 10
}
return `val`
}
fun findHighAccessEmployees(accessTimes: List<List<String>>): List<String> {
val map = HashMap<String, MutableList<Int>>()
for (list in accessTimes) {
val temp = map.getOrDefault(list[0], ArrayList())
val `val` = stringToInt(list[1])
temp.add(`val`)
map[list[0]] = temp
}
val ans: MutableList<String> = ArrayList()
for ((key, temp) in map) {
temp.sort()
if (isHighAccess(temp)) {
ans.add(key)
}
}
return ans
}
}