LeetCode in Kotlin
539. Minimum Time Difference
Medium
Given a list of 24-hour clock time points in “HH:MM” format, return the minimum minutes difference between any two time-points in the list.
Example 1:
Input: timePoints = [“23:59”,”00:00”]
Output: 1
Example 2:
Input: timePoints = [“00:00”,”23:59”,”00:00”]
Output: 0
Constraints:
2 <= timePoints.length <= 2 * 104timePoints[i]is in the format “HH:MM”.
Solution
class Solution {
fun findMinDifference(timePoints: List<String>): Int {
return if (timePoints.size < 300) {
smallInputSize(timePoints)
} else largeInputSize(timePoints)
}
private fun largeInputSize(timePoints: List<String>): Int {
val times = BooleanArray(60 * 24)
for (time in timePoints) {
val hours = time.substring(0, 2).toInt()
val minutes = time.substring(3, 5).toInt()
if (times[hours * 60 + minutes]) {
return 0
}
times[hours * 60 + minutes] = true
}
var prev = -1
var min = 60 * 24
for (i in 0 until times.size + times.size / 2) {
if (i < times.size) {
if (times[i] && prev == -1) {
prev = i
} else if (times[i]) {
min = Math.min(min, i - prev)
prev = i
}
} else {
if (times[i - times.size] && prev == -1) {
prev = i
} else if (times[i - times.size]) {
min = Math.min(min, i - prev)
prev = i
}
}
}
return min
}
private fun smallInputSize(timePoints: List<String>): Int {
val times = IntArray(timePoints.size)
var j = 0
for (time in timePoints) {
val hours = time.substring(0, 2).toInt()
val minutes = time.substring(3, 5).toInt()
times[j++] = hours * 60 + minutes
}
times.sort()
var min = 60 * 24
for (i in 1..times.size) {
min = if (i == times.size) {
Math.min(min, times[0] + 60 * 24 - times[times.size - 1])
} else {
Math.min(min, times[i] - times[i - 1])
}
if (min == 0) {
return 0
}
}
return min
}
}