539. Minimum Time Difference

Description

Given a list of 24-hour clock time points in “Hour:Minutes” format, find the minimum minutes difference between any two time points in the list. Example 1: Input: [“23:59”,“00:00”] Output: 1 Note: The number of time points in the given list is at least 2 and won’t exceed 20000. The input time is legal and ranges from 00:00 to 23:59.

Problem URL

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Solution

给一个string的list，找到这个list中表示的时间中的任意两个时间的最小差值，可能跨天。

We use a Integer list to store the value of coverted time from String to 1440(because a day has 1440 minutes). Then sort the list from small to big. Calculate the minimum difference, and a corner case, from time[0] to tomorrow’s time[time.size() - 1]. Return the smaller one.

Code

class Solution { public int findMinDifference(List<String> timePoints) { int min = Integer.MAX_VALUE; List<Integer> time = new ArrayList<>(); for (int i = 0; i < timePoints.size(); i++){ int hour = Integer.valueOf(timePoints.get(i).substring(0,2)); int minute = Integer.valueOf(timePoints.get(i).substring(3,5)); time.add(hour * 60 + minute); } Collections.sort(time, (Integer a, Integer b) -> a - b); for (int i = 1; i < time.size(); i++){ min = Math.min(min, time.get(i) - time.get(i - 1)); } int cornerCase = time.get(0) + (1440 - time.get(time.size() - 1)); return Math.min(cornerCase, min); } }

Time Complexity: O(n) Space Complexity: O(n)

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Review

We could also use Bucket sort, using a boolean int[] with size 1440 to document wether a time is showed in timepoints list. Then iteratively traverse the bucket, find minimum difference and calculate corner case.