LeetCode 759: Employee Free Time Solution
Master LeetCode problem 759 (Employee Free Time), a hard challenge, with our optimized solutions in Java, C++, and Python. Explore detailed explanations, test your code in our interactive editor, and prepare for coding interviews.
759. Employee Free Time
Problem Explanation
Explanation
To find the free time between the schedules of multiple employees, we can first merge all the intervals into a single list and then sort them based on their start times. We then iterate through the sorted list to find the gaps between the intervals, which represent the free time.
Algorithm:
- Merge all the intervals into a single list.
- Sort the merged list based on the start times.
- Iterate through the sorted list to find the free time gaps between intervals.
Time Complexity
The time complexity of this approach is O(n log n), where n is the total number of intervals.
Space Complexity
The space complexity is O(n) to store the merged intervals.
Solution Code
import java.util.*;
class Solution {
public List<Interval> employeeFreeTime(List<List<Interval>> schedule) {
List<Interval> mergedIntervals = new ArrayList<>();
for (List<Interval> employeeSchedule : schedule) {
mergedIntervals.addAll(employeeSchedule);
}
Collections.sort(mergedIntervals, (a, b) -> a.start - b.start);
List<Interval> result = new ArrayList<>();
Interval prev = mergedIntervals.get(0);
for (Interval curr : mergedIntervals) {
if (prev.end < curr.start) {
result.add(new Interval(prev.end, curr.start));
prev = curr;
} else {
prev = prev.end < curr.end ? curr : prev;
}
}
return result;
}
}Try It Yourself
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