leetcode/problems/src/array/MinimumTimeDifference.java at master · KindleBooks66/leetcode · GitHub

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package array;

import java.util.*;

import java.util.stream.Collectors;

/**

* Created by gouthamvidyapradhan on 30/07/2019 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.

* <p>Solution: O(N log N) convert each time value of the form hh:mm to minutes and sort the array.

* For every pair (i, j) where j = i + 1 (also for the case where i = 0 and j = N - 1) check the

* minute difference and return the minimum time difference as the answer.

public class MinimumTimeDifference {

public static void main(String[] args) {

List<String> list = Arrays.asList("23:59", "00:00");

System.out.println(new MinimumTimeDifference().findMinDifference(list));

}

public int findMinDifference(List<String> timePoints) {

List<Integer> timeInMinutes =

timePoints

.stream()

.map(

t -> {

String[] strings = t.split(":");

return Integer.parseInt(strings[0]) * 60 + Integer.parseInt(strings[1]);

})

.sorted(Integer::compareTo)

.collect(Collectors.toList());

int min = Integer.MAX_VALUE;

for (int i = 1, l = timeInMinutes.size(); i < l; i++) {

int prev = timeInMinutes.get(i - 1);

int curr = timeInMinutes.get(i);

min = Math.min(min, curr - prev);

min = Math.min(min, ((24 * 60) - curr) + prev);

}

int prev = timeInMinutes.get(0);

int curr = timeInMinutes.get(timeInMinutes.size() - 1);

min = Math.min(min, curr - prev);

min = Math.min(min, ((24 * 60) - curr) + prev);

return min;

}

leetcode/problems/src/array/MinimumTimeDifference.java at master · KindleBooks66/leetcode · GitHub

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