Next Closest Time

Similar Problems:

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.

Github: code.dennyzhang.com

Credits To: leetcode.com

Leave me comments, if you have better ways to solve.

## Blog link: https://code.dennyzhang.com/next-closest-time ## Basic Ideas: Only pow(4, 4) combinations ## Get all of them, and rule out the invalid onces ## Convert the time string to seconds ## Compare the absolute diff with rotated enforced ## ## Assumption: "11:11" -> "11:11" ## Complexity: Time O(1), Space O(1) class Solution: def nextClosestTime(self, time): """ :type time: str :rtype: str """ import sys ch_set = set(time) ch_set.remove(':') if len(ch_set) == 1: return time l = [""] for i in range(4): l2 = [] for ch in ch_set: for item in l: if i == 1: l2.append("%s%s:" % (item, ch)) else: l2.append("%s%s" % (item, ch)) l = l2 minutes = self.getMinutes(time) min_diff, index = sys.maxsize, None total_minutes = 24*60 for i in range(len(l)): t = l[i] # check whether t is valid if int(t[0])*10+int(t[1]) >= 24: continue if int(t[3])*10+int(t[4]) >= 60: continue diff = (self.getMinutes(t)-minutes+total_minutes) % total_minutes if diff == 0: continue if min_diff > diff: min_diff, index = diff, i return l[index] def getMinutes(self, time): return (int(time[0])*10 + int(time[1]))*60+(int(time[3])*10 + int(time[4]))

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