LeetCode in Kotlin
949. Largest Time for Given Digits
Medium
Given an array arr of 4 digits, find the latest 24-hour time that can be made using each digit exactly once.
24-hour times are formatted as "HH:MM", where HH is between 00 and 23, and MM is between 00 and 59. The earliest 24-hour time is 00:00, and the latest is 23:59.
Return the latest 24-hour time in "HH:MM" format. If no valid time can be made, return an empty string.
Example 1:
Input: arr = [1,2,3,4]
Output: “23:41”
Explanation: The valid 24-hour times are “12:34”, “12:43”, “13:24”, “13:42”, “14:23”, “14:32”, “21:34”, “21:43”, “23:14”, and “23:41”. Of these times, “23:41” is the latest.
Example 2:
Input: arr = [5,5,5,5]
Output: “”
Explanation: There are no valid 24-hour times as “55:55” is not valid.
Constraints:
arr.length == 40 <= arr[i] <= 9
Solution
class Solution {
fun largestTimeFromDigits(arr: IntArray): String {
val buf = StringBuilder()
var maxHour: String? = ""
for (i in 0..3) {
for (j in 0..3) {
if (i != j) {
val hour = getTime(arr, i, j, 23, false)
val min = getTime(arr, i, j, 59, true)
if (hour != null && min != null && hour > maxHour!!) {
buf.setLength(0)
buf.append(hour).append(':').append(min)
maxHour = hour
}
}
}
}
return buf.toString()
}
private fun getTime(arr: IntArray, i: Int, j: Int, limit: Int, exclude: Boolean): String? {
var n1 = -1
var n2 = -1
for (k in 0..3) {
if (exclude && k != i && k != j || !exclude && (k == i || k == j)) {
if (n1 == -1) {
n1 = arr[k]
} else {
n2 = arr[k]
}
}
}
var s1: String? = n1.toString() + n2
var s2: String? = n2.toString() + n1
var v1 = s1!!.toInt()
if (v1 > limit) {
v1 = -1
s1 = null
}
var v2 = s2!!.toInt()
if (v2 > limit) {
v2 = -1
s2 = null
}
return if (v1 >= v2) s1 else s2
}
}