What is the brute force approach and complexity for Employee Free Time?

A gap can only start where some interval ends; test each end against all intervals with time complexity O(N²) and space complexity O(N).

What is the optimal approach and complexity for Employee Free Time?

Merge the k sorted lists with a min-heap; report a gap whenever the next start passes lastEnd with optimal time complexity O(N log k) and space complexity O(k).

How do you identify the pattern for Employee Free Time?

Look for key signals in the prompt: "all employees are free" indicates Nobody is busy → only the union of busy times matters, not who is busy.; "each employee's list is sorted" indicates k sorted lists → k-way merge with a min-heap instead of a full sort.; "finite, positive-length intervals" indicates Skip the endless time before the first and after the last interval, and skip start == lastEnd..

Employee Free Time

Hard·LeetCode #759 Premium Pro OnlyPattern Transfer

LeetCode

Problem Statement

You get the work schedule of k employees. schedule[e] is the list of times employee e is busy. Each employee's list is sorted and its intervals don't overlap.

Return every stretch of time when all employees are free at once, as a sorted list of intervals. Only return gaps of positive, finite length: skip the endless time before everyone's first interval and after everyone's last one, and skip a "gap" like [5,5] where one interval ends exactly when another starts.

About the signature: on LeetCode the input is vector<vector<Interval>>, where Interval is a small class LeetCode gives you with two fields, start and end. So schedule[e][i].start is when employee e's i-th busy interval begins. The answer is a vector<Interval>. Our code shows Interval as a plain struct only so it is readable on its own.

Input

schedule: a list of k lists. schedule[e] holds employee e's busy intervals, sorted by start.

Output

The common free intervals, sorted by start. It can be empty.

Example 1

Input: schedule = [[[1,2],[5,6]], [[1,3]], [[4,10]]]

Output: [[3,4]]

Together, people are busy during [1,3] and [4,10]. The only finite gap is [3,4]. Time before 1 and after 10 is endless, so it is left out.

Example 2

Input: schedule = [[[1,3],[6,7]], [[2,4]], [[2,5],[9,12]]]

Output: [[5,6],[7,9]]

Busy time joined together: [1,5], [6,7], [9,12]. The gaps between them are [5,6] and [7,9].

Constraints

• 1 ≤ schedule.length ≤ 50
• 1 ≤ schedule[i].length ≤ 50
• 0 ≤ schedule[i][j].start < schedule[i][j].end ≤ 10^8

Pattern Recognition Clues

Recognising these key signals in the problem description immediately identifies the pattern:

“all employees are free”

→ Nobody is busy → only the union of busy times matters, not who is busy.

“each employee's list is sorted”

→ k sorted lists → k-way merge with a min-heap instead of a full sort.

“finite, positive-length intervals”

→ Skip the endless time before the first and after the last interval, and skip start == lastEnd.

Approach Complexity Ladder

Brute Force — check every possible gap startO(N²)O(N)

A gap can only start where some interval ends; test each end against all intervals

Better — flatten, sort, merge, report gapsO(N log N)O(N)

Sort all intervals by start; walk with lastEnd and report every gap

Optimal — k-way merge with a min-heapO(N log k)O(k)

Merge the k sorted lists with a min-heap; report a gap whenever the next start passes lastEnd

Brute Force Intuition

Put every busy interval into one list. A free gap can only begin at some interval's end. For each distinct end e, check all intervals: if nobody is busy at e (no interval has start ≤ e < end), the gap runs from e to the next start after e. If nothing starts after e, the gap is endless, so skip it.

Recognition rule for this sub-pattern:

Two (or k) interval lists in time order — sort first if needed. Output their overlap, then advance whichever ends first; with k lists, use a min-heap.

Remember this · Intervals

Trigger

Input is [start, end] pairs; overlap, merge, insert, meeting rooms, free time, minimum removals.

Think

After sorting, do I only ever need to compare against the most recent interval?

Key question

Sort by start or by end — which one makes the greedy choice obvious?

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