539. Minimum Time Difference
Explanation
To solve this problem, we can convert each time point to minutes from midnight, sort the minutes, and then find the minimum difference between consecutive times. Since the time wraps around (e.g., 23:59 is close to 00:00), we also need to consider the circular nature of time.
- Convert each time to minutes from midnight.
- Sort the minutes.
- Find the minimum difference between consecutive times.
- Also, consider the circular nature of time.
Time Complexity: O(n log n) - Sorting the time points \n Space Complexity: O(n) - Storing the minutes from midnight
import java.util.*;
class Solution {
public int findMinDifference(List<String> timePoints) {
List<Integer> minutes = new ArrayList<>();
for (String time : timePoints) {
String[] parts = time.split(":");
int hour = Integer.parseInt(parts[0]);
int minute = Integer.parseInt(parts[1]);
minutes.add(hour * 60 + minute);
}
Collections.sort(minutes);
int minDiff = Integer.MAX_VALUE;
for (int i = 1; i < minutes.size(); i++) {
minDiff = Math.min(minDiff, minutes.get(i) - minutes.get(i - 1));
}
minDiff = Math.min(minDiff, 24 * 60 + minutes.get(0) - minutes.get(minutes.size() - 1));
return minDiff;
}
}Code Editor (Testing phase)
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