Employee Free Time

📋

Employee Free Time

Edge cases:

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IDE

def employeeFreeTime(schedule: list[list[list[int]]]) -> list[list[int]]:public List<List<Integer>> employeeFreeTime(List<List<List<Integer>>> schedule)vector<vector<int>> employeeFreeTime(vector<vector<vector<int>>> &schedule)function employeeFreeTime(schedule)

def employeeFreeTime(schedule):
    # Write your solution here
    pass

class Solution {
    public List<List<Integer>> employeeFreeTime(List<List<List<Integer>>> schedule) {
        // Write your solution here
        return new ArrayList<>();
    }
}

#include <vector>
using namespace std;

vector<vector<int>> employeeFreeTime(vector<vector<vector<int>>> &schedule) {
    // Write your solution here
    return {};
}

function employeeFreeTime(schedule) {
    // Write your solution here
}

Common Bugs to Avoid

Wrong: [[2,3]]Returning gaps inside busy intervals due to incorrect merging.✅ Sort all intervals and merge overlapping or touching intervals before finding gaps.

Wrong: [[2,2]]Including zero-length intervals as free time.✅ Filter out intervals where start == end before returning.

Wrong: [[3,4],[5,6]]Treating touching intervals as free time gaps.✅ Merge touching intervals as continuous busy time, no gaps between [1,2] and [2,3].

Wrong: [[1,10]]Returning entire timeline as free time when employees have no intervals.✅ Return empty list if no finite free intervals exist; infinite free time not representable.

Wrong: [[3,4],[7,8]]Greedy approach missing intervals causing false free time.✅ Merge all intervals from all employees before detecting gaps.

✓

Test Cases

Focus on handling empty and minimal inputs correctly before tackling edge cases.

Understand full merging logic and avoid greedy traps to pass corner cases.

Optimize your solution to O(N log N) using sorting and event processing for performance.

t1_01basic

Input[[[1,2],[5,6]],[[1,3]],[[4,10]]]

Expected[[3,4]]

⏱ Performance - must finish in 2000ms

Merging all intervals gives [1,3], [4,10], [5,6] merged into [1,3] and [4,10]. The gap between 3 and 4 is free time for all employees.

💡 Consider merging all intervals from all employees first.

💡 Use a sweep line or event processing to find gaps between merged intervals.

💡 Return intervals where no employee is busy, ensuring positive length.

Why it failed: Incorrect output means merging or gap detection is wrong. Fix by sorting all intervals and merging overlapping ones before finding gaps.

✓ Correctly merges intervals and finds common free time gaps.

t1_02basic

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

Expected[[5,6],[7,9]]

⏱ Performance - must finish in 2000ms

Merged intervals are [1,5], [6,7], [9,12]. Free times are [5,6] and [7,9].

💡 Flatten and sort all intervals by start time.

💡 Merge overlapping intervals carefully to find continuous busy periods.

💡 Identify gaps between merged intervals as free time.

Why it failed: If free intervals are missing or incorrect, merging logic or gap detection is flawed. Ensure intervals are merged before gap calculation.

✓ Correctly identifies multiple free intervals between merged busy times.

t2_01edge

Input[]

Expected[]

⏱ Performance - must finish in 2000ms

No employees means no busy intervals, so no finite free intervals to return.

💡 Check how your code handles empty input schedules.

💡 No intervals means no busy times, but output requires positive length intervals.

💡 Return empty list since infinite free time is not representable as finite intervals.

Why it failed: Returning non-empty or error means empty input not handled. Fix by returning empty list if schedule is empty.

✓ Correctly returns empty list for no employees.

t2_02edge

Input[[[1,5]]]

Expected[]

⏱ Performance - must finish in 2000ms

Only one employee with one interval means no common free time inside that interval; outside is infinite and not returned.

💡 For single employee, free time is complement of their intervals but only finite intervals count.

💡 Check if your code handles single employee schedules correctly.

💡 Return empty list if no finite free intervals exist.

Why it failed: Returning intervals inside busy time or infinite intervals means misunderstanding of problem. Fix by returning empty list if no gaps.

✓ Correctly returns empty list when no finite free intervals exist.

t2_03edge

Input[[[1,2],[2,3]],[[2,3],[3,4]]]

Expected[]

⏱ Performance - must finish in 2000ms

Intervals that touch but do not overlap produce no free time between them.

💡 Check if your merging treats touching intervals as continuous busy time.

💡 Intervals like [1,2] and [2,3] should merge into [1,3] with no gap.

💡 Return empty list if no positive length gaps exist.

Why it failed: Returning gaps between touching intervals means treating touching as free. Fix by merging touching intervals as overlapping.

✓ Correctly merges touching intervals and returns no free time.

t2_04edge

Input[[],[],[]]

Expected[]

⏱ Performance - must finish in 2000ms

Employees with no intervals means entire timeline is free but infinite intervals are not returned.

💡 Handle employees with empty schedules correctly.

💡 No busy intervals means no finite free intervals to return.

💡 Return empty list for no intervals across all employees.

Why it failed: Returning non-empty or error means empty employee schedules not handled. Fix by returning empty list if no intervals.

✓ Correctly returns empty list when all employees have no intervals.

t3_01corner

Input[[[1,10]],[[2,3],[4,5],[6,7],[8,9]],[[1,2],[3,4],[5,6],[7,8],[9,10]]]

Expected[]

⏱ Performance - must finish in 2000ms

All employees' intervals cover the entire range [1,10] without gaps, so no free time.

💡 Beware of greedy approaches that pick earliest intervals only.

💡 Ensure merging considers all intervals from all employees.

💡 Check that merged intervals cover entire timeline before concluding no free time.

Why it failed: Greedy approach missing intervals causes false free time. Fix by merging all intervals from all employees before gap detection.

✓ Correctly merges all intervals and finds no free time.

t3_02corner

Input[[[1,3],[5,7]],[[2,4],[6,8]],[[1,2],[4,5],[7,9]]]

Expected[]

⏱ Performance - must finish in 2000ms

Intervals merge to [1,4],[5,9]. The gaps at 4 and 9 are zero length, so no free time intervals of positive length exist; output is empty.

💡 Check if your code returns zero-length intervals as free time.

💡 Free intervals must have positive length, so exclude zero-length gaps.

💡 Filter out intervals where start == end.

Why it failed: Returning zero-length intervals means not filtering intervals with start == end. Fix by excluding such intervals from output.

✓ Correctly excludes zero-length intervals from free time.

t3_03corner

Input[[[1,2],[5,6]],[[1,3]],[[4,10]]]

Expected[[3,4]]

⏱ Performance - must finish in 2000ms

This test checks that the solution does not confuse 0/1 knapsack style inclusion with interval merging; intervals must be merged by timeline, not by employee.

💡 Do not treat intervals as independent items to pick or skip.

💡 Merge all intervals across employees to find continuous busy times.

💡 Use sweep line or sorting to combine intervals properly.

Why it failed: Confusing problem with knapsack leads to incorrect merging. Fix by merging intervals by timeline, ignoring employee boundaries.

✓ Correctly merges intervals across all employees.

t4_01performance

Input{"_description":"n=100000 at constraint boundary - executor generates this input"}

Expectednull

⏱ Performance - must finish in 2000ms

Input with 100000 intervals tests O(N log N) sweep line approach; brute force O(N*M*T) will TLE.

💡 Use sorting and sweep line to achieve O(N log N) time.

💡 Avoid nested loops over intervals and timeline points.

💡 Process interval endpoints as events to find free time efficiently.

Why it failed: Brute force or naive approach causes TLE. Fix by implementing sweep line or event processing with sorting.

✓ Solution runs within time limit using O(N log N) approach.

Practice

1. You are given a list of trips, each with a number of passengers, a start location, and an end location. You need to determine if a vehicle with a fixed capacity can carry all passengers without exceeding capacity at any point along the route. Which algorithmic approach best guarantees an optimal and efficient solution?

easy

A. Brute force simulation by incrementing passenger count at every location for each trip

B. Greedy algorithm that picks trips with earliest end location first

C. Dynamic programming to find maximum passengers at each location using state transitions

D. Sweep line algorithm that processes passenger count changes at trip start and end points

Solution

1. Step 1: Understand problem constraints

   The problem requires tracking passenger counts over intervals defined by start and end locations.

2. Step 2: Identify suitable algorithm

   Sweep line processes events (passenger changes) in sorted order, efficiently tracking capacity usage without simulating every location.

3. Final Answer:

   Option D -> Option D

4. Quick Check:

   Sweep line handles interval events optimally [OK]

Hint: Intervals with start/end events -> sweep line [OK]

Common Mistakes:

• Assuming greedy earliest end works for capacity constraints
• Thinking DP is needed for counting passengers
• Using brute force for large location ranges

2. You are given a set of intervals representing horizontal balloons on a number line. Each balloon can be burst by shooting an arrow vertically through any point covered by the balloon's interval. What is the most efficient approach to find the minimum number of arrows needed to burst all balloons?

easy

A. Sort intervals by their end coordinate and greedily shoot arrows at the earliest possible end to cover maximum balloons.

B. Sort intervals by their start coordinate and greedily shoot arrows at the start of each balloon.

C. Use dynamic programming to find the maximum number of overlapping intervals and subtract from total balloons.

D. Use a brute force nested loop to check all pairs of intervals for overlaps and count arrows accordingly.

Solution

1. Step 1: Understand the problem constraints

   The problem requires minimizing arrows to burst all balloons, which translates to covering intervals with minimum points.

2. Step 2: Identify the optimal greedy strategy

   Sorting by end coordinate allows shooting an arrow at the earliest finishing balloon's end, covering all overlapping balloons starting before that point.

3. Final Answer:

   Option A -> Option A

4. Quick Check:

   Sorting by end coordinate ensures minimal arrows by maximizing coverage [OK]

Hint: Sort intervals by end to greedily cover overlaps [OK]

Common Mistakes:

• Sorting by start coordinate and shooting at start misses optimal overlaps
• Using DP unnecessarily complicates the problem
• Brute force is too slow for large inputs

3. What is the time complexity of the addNum operation in the optimal balanced tree approach for maintaining disjoint intervals, assuming there are n intervals stored?

medium

A. O(n) because intervals need to be scanned linearly for merging

B. O(log n) due to binary search and limited merging operations

C. O(n log n) because each insertion requires sorting intervals

D. O(1) since insertion is always at the end or start

Solution

1. Step 1: Identify main operations in addNum

   addNum uses binary search (bisect_left) to find insertion index in O(log n).

2. Step 2: Analyze merging cost

   Merging involves at most constant number of interval merges (left, right, or both), so O(1) extra work.

3. Final Answer:

   Option B -> Option B

4. Quick Check:

   Binary search dominates, merges are constant time [OK]

Hint: Binary search + constant merges -> O(log n) [OK]

Common Mistakes:

• Assuming linear scan for merging
• Confusing sorting cost per insertion
• Ignoring constant merge steps

4. The following code attempts to determine if a person can attend all meetings. Identify the line containing the subtle bug that causes incorrect results for intervals that just touch (e.g., [1,5] and [5,10]).

from typing import List

def can_attend_meetings(intervals: List[List[int]]) -> bool:
    intervals.sort(key=lambda x: x[0])
    end = float('-inf')
    for interval in intervals:
        if interval[0] <= end:
            return False
        end = interval[1]
    return True

medium

A. Line 6: if interval[0] <= end:

B. Line 3: intervals.sort(key=lambda x: x[0])

C. Line 5: for interval in intervals:

D. Line 7: end = interval[1]

Solution

1. Step 1: Understand the overlap condition

   Meetings that just touch (e.g., end=5 and start=5) should not be considered overlapping.

2. Step 2: Identify the bug

   The condition uses <= which incorrectly treats touching intervals as overlapping. It should use < instead.

3. Final Answer:

   Option A -> Option A

4. Quick Check:

   Changing <= to < fixes false positives on touching intervals [OK]

Hint: Overlap check must use <, not <= [OK]

Common Mistakes:

• Using <= causes false overlap detection on touching intervals

5. What is the time complexity of the optimal in-place merge intervals algorithm that first sorts the intervals and then merges them in one pass?

medium

A. O(n log n) due to the sorting step dominating the runtime

B. O(n) because merging is done in a single pass

C. O(n²) because each interval is compared with all others during merging

D. O(n log n) because merging requires binary searches for overlaps

Solution

1. Step 1: Identify the sorting step complexity.

   Sorting n intervals by start time takes O(n log n) time.

2. Step 2: Analyze the merging step complexity.

   Merging intervals in one pass is O(n), which is dominated by sorting.

3. Final Answer:

   Option A -> Option A

4. Quick Check:

   Sorting dominates overall complexity, merging is linear. [OK]

Hint: Sorting dominates, merging is linear [OK]

Common Mistakes:

• Ignoring sorting cost
• Assuming nested comparisons cause O(n²)