LeetCode in Kotlin
3114. Latest Time You Can Obtain After Replacing Characters
Easy
You are given a string s representing a 12-hour format time where some of the digits (possibly none) are replaced with a "?".
12-hour times are formatted as "HH:MM", where HH is between 00 and 11, and MM is between 00 and 59. The earliest 12-hour time is 00:00, and the latest is 11:59.
You have to replace all the "?" characters in s with digits such that the time we obtain by the resulting string is a valid 12-hour format time and is the latest possible.
Return the resulting string.
Example 1:
Input: s = “1?:?4”
Output: “11:54”
Explanation: The latest 12-hour format time we can achieve by replacing "?" characters is "11:54".
Example 2:
Input: s = “0?:5?”
Output: “09:59”
Explanation: The latest 12-hour format time we can achieve by replacing "?" characters is "09:59".
Constraints:
s.length == 5s[2]is equal to the character":".- All characters except
s[2]are digits or"?"characters. - The input is generated such that there is at least one time between
"00:00"and"11:59"that you can obtain after replacing the"?"characters.
Solution
class Solution {
fun findLatestTime(s: String): String {
val nm = StringBuilder()
if (s[0] == '?' && s[1] == '?') {
nm.append("11")
} else if (s[0] != '?' && s[1] == '?') {
nm.append(s[0])
if (s[0] == '1') {
nm.append("1")
} else {
nm.append("9")
}
} else if (s[0] == '?' && s[1] != '?') {
if (s[1] in '2'..'9') {
nm.append("0")
} else {
nm.append("1")
}
nm.append(s[1])
} else {
nm.append(s[0])
nm.append(s[1])
}
nm.append(":")
if (s[3] == '?' && s[4] == '?') {
nm.append("59")
} else if (s[3] != '?' && s[4] == '?') {
nm.append(s[3])
nm.append("9")
} else if (s[3] == '?' && s[4] != '?') {
nm.append("5")
nm.append(s[4])
} else {
nm.append(s[3])
nm.append(s[4])
}
return nm.toString()
}
}