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Formatted question description: https://leetcode.ca/all/681.html

# 681. Next Closest Time

Medium

## Description

Given a time represented in the format “HH:MM”, form the next closest time by reusing the current digits. There is no limit on how many times a digit can be reused.

You may assume the given input string is always valid. For example, “01:34”, “12:09” are all valid. “1:34”, “12:9” are all invalid.

Example 1:

Input: “19:34”

Output: “19:39”

Explanation: The next closest time choosing from digits 1, 9, 3, 4, is 19:39, which occurs 5 minutes later. It is not 19:33, because this occurs 23 hours and 59 minutes later.

Example 2:

Input: “23:59”

Output: “22:22”

Explanation: The next closest time choosing from digits 2, 3, 5, 9, is 22:22. It may be assumed that the returned time is next day’s time since it is smaller than the input time numerically.

## Solution

Simulate all possible times minute by minute. If a time can be formed by reusing the current digit, return the time.

When returning the time, use String.format to convert it into the required format “HH:MM”.

• class Solution {
public String nextClosestTime(String time) {
String hourStr = time.substring(0, 2), minuteStr = time.substring(3);
Set<Integer> set = new HashSet<Integer>();
for (int i = 0; i < 2; i++) {
}
int[] timeArray = {Integer.parseInt(hourStr), Integer.parseInt(minuteStr)};
for (int i = 1; i <= 1440; i++) {
next(timeArray);
if (isReuse(timeArray, set)) {
String nextTime = String.format("%02d:%02d", timeArray, timeArray);
return nextTime;
}
}
return time;
}

public void next(int[] timeArray) {
timeArray++;
if (timeArray >= 60) {
timeArray %= 60;
timeArray++;
}
if (timeArray >= 24)
timeArray %= 24;
}

public boolean isReuse(int[] timeArray, Set<Integer> set) {
int hour = timeArray, minute = timeArray;
if (!set.contains(hour / 10))
return false;
else if (!set.contains(hour % 10))
return false;
else if (!set.contains(minute / 10))
return false;
else if (!set.contains(minute % 10))
return false;
else
return true;
}
}

• class Solution:
def nextClosestTime(self, time: str) -> str:
def check(t):
h, m = int(t[:2]), int(t[2:])
return 0 <= h < 24 and 0 <= m < 60

def dfs(curr):
if len(curr) == 4:
if not check(curr):
return
nonlocal ans, d
p = int(curr[:2]) * 60 + int(curr[2:])
if t < p < t + d:
d = p - t
ans = curr[:2] + ':' + curr[2:]
return
for c in s:
dfs(curr + c)

s = {c for c in time if c != ':'}
t = int(time[:2]) * 60 + int(time[3:])
d = inf
ans = None
dfs('')
if ans is None:
mi = min(int(c) for c in s)
ans = f'{mi}{mi}:{mi}{mi}'
return ans