LeetCode in Kotlin
1736. Latest Time by Replacing Hidden Digits
Easy
You are given a string time in the form of hh:mm, where some of the digits in the string are hidden (represented by ?).
The valid times are those inclusively between 00:00 and 23:59.
Return the latest valid time you can get from time by replacing the hidden digits.
Example 1:
Input: time = “2?:?0”
Output: “23:50”
Explanation: The latest hour beginning with the digit ‘2’ is 23 and the latest minute ending with the digit ‘0’ is 50.
Example 2:
Input: time = “0?:3?”
Output: “09:39”
Example 3:
Input: time = “1?:22”
Output: “19:22”
Constraints:
timeis in the formathh:mm.- It is guaranteed that you can produce a valid time from the given string.
Solution
class Solution {
fun maximumTime(time: String): String {
val sb: StringBuilder = StringBuilder()
val strs: Array<String> = time.split(":").dropLastWhile({ it.isEmpty() }).toTypedArray()
val hour: String = strs[0]
val min: String = strs[1]
if (hour[0] == '?') {
if (hour[1] == '?') {
sb.append("23")
} else if (hour[1] > '3') {
sb.append("1")
sb.append(hour[1])
} else {
sb.append("2")
sb.append(hour[1])
}
} else if (hour[0] == '0' || hour[0] == '1') {
if (hour[1] == '?') {
sb.append(hour[0])
sb.append("9")
} else {
sb.append(hour)
}
} else if (hour[0] == '2') {
if (hour[1] == '?') {
sb.append("23")
} else {
sb.append(hour)
}
}
sb.append(":")
if (min[0] == '?') {
if (min[1] == '?') {
sb.append("59")
} else {
sb.append("5")
sb.append(min[1])
}
return sb.toString()
}
sb.append(min[0])
if (min[1] == '?') {
sb.append("9")
} else {
sb.append(min[1])
}
return sb.toString()
}
}