LeetCode-in-Java
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
public class Solution {
public String largestTimeFromDigits(int[] arr) {
StringBuilder buf = new StringBuilder();
String maxHour = "";
for (int i = 0; i < 4; i++) {
for (int j = 0; j < 4; j++) {
if (i != j) {
String hour = getTime(arr, i, j, 23, false);
String min = getTime(arr, i, j, 59, true);
if (hour != null && min != null && hour.compareTo(maxHour) > 0) {
buf.setLength(0);
buf.append(hour).append(':').append(min);
maxHour = hour;
}
}
}
}
return buf.toString();
}
private String getTime(int[] arr, int i, int j, int limit, boolean exclude) {
int n1 = -1;
int n2 = -1;
for (int k = 0; k < 4; k++) {
if ((exclude && k != i && k != j) || (!exclude && (k == i || k == j))) {
if (n1 == -1) {
n1 = arr[k];
} else {
n2 = arr[k];
}
}
}
String s1 = String.valueOf(n1) + n2;
String s2 = String.valueOf(n2) + n1;
int v1 = Integer.parseInt(s1);
if (v1 > limit) {
v1 = -1;
s1 = null;
}
int v2 = Integer.parseInt(s2);
if (v2 > limit) {
v2 = -1;
s2 = null;
}
return v1 >= v2 ? s1 : s2;
}
}