Remove Covered Intervals

📋

Remove Covered Intervals

Edge cases:

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IDE

def remove_covered_intervals(intervals: list[list[int]]) -> int:public int removeCoveredIntervals(int[][] intervals)int removeCoveredIntervals(vector<vector<int>>& intervals)function removeCoveredIntervals(intervals)

def remove_covered_intervals(intervals):
    # Write your solution here
    pass

class Solution {
    public int removeCoveredIntervals(int[][] intervals) {
        // Write your solution here
        return 0;
    }
}

#include <vector>
using namespace std;

int removeCoveredIntervals(vector<vector<int>>& intervals) {
    // Write your solution here
    return 0;
}

function removeCoveredIntervals(intervals) {
    // Write your solution here
}

Common Bugs to Avoid

Wrong: 3Failing to remove covered intervals due to incorrect sorting order or coverage condition.✅ Sort intervals by start ascending and end descending, then track max end to detect coverage.

Wrong: 0Incorrect handling of empty or single interval input, returning zero instead of correct count.✅ Add explicit checks for empty and single-element inputs to return correct counts.

Wrong: 3Not removing duplicates or treating identical intervals as uncovered.✅ Treat identical intervals as covered except one; ensure coverage condition includes equality.

Wrong: 4Greedy approach counting intervals without coverage check, leading to overcount.✅ Implement coverage detection by sorting and tracking max end to exclude covered intervals.

Wrong: TLEUsing brute force nested loops causing quadratic time complexity.✅ Optimize to O(n log n) by sorting intervals and performing a single pass coverage check.

✓

Test Cases

Try testing your solution on empty and single interval inputs to ensure base cases are handled.

Consider tricky cases where intervals have same start or are nested deeply to catch subtle bugs.

Optimize your solution to avoid nested loops; sorting plus single pass coverage detection is key.

t1_01basic

Input{"intervals":[[1,4],[2,8],[3,6]]}

Expected2

⏱ Performance - must finish in 2000ms

Interval [3,6] is covered by [2,8], so it is removed. Remaining intervals are [1,4] and [2,8].

💡 Sort intervals by start ascending and end descending to identify coverage.

💡 Check if current interval is fully contained within the previous interval.

💡 Count intervals that are not covered by comparing ends while iterating sorted list.

Why it failed: Incorrect count due to missing coverage check or wrong sorting order. Fix by sorting intervals by start ascending and end descending, then track max end to detect coverage.

✓ Correctly identifies covered intervals and counts remaining ones.

t1_02basic

Input{"intervals":[[1,4],[1,2],[3,4]]}

Expected1

⏱ Performance - must finish in 2000ms

Interval [3,4] is covered by [1,4], so it is removed. Remaining intervals are [1,2] and [1,4].

💡 Consider intervals with same start but different ends.

💡 Sorting by start ascending and end descending helps group intervals correctly.

💡 Track the maximum end seen so far to detect coverage.

Why it failed: Fails to handle intervals with same start but different ends; coverage detection misses intervals covered by earlier ones with same start. Fix sorting to order by start ascending and end descending.

✓ Handles intervals with same start correctly and counts uncovered intervals.

t2_01edge

Input{"intervals":[]}

Expected0

⏱ Performance - must finish in 2000ms

Empty input means no intervals remain.

💡 Check how your code handles empty input arrays.

💡 Ensure no assumptions about minimum length in loops.

💡 Return 0 immediately if intervals list is empty.

Why it failed: Code crashes or returns non-zero on empty input due to missing empty check. Fix by adding early return for empty intervals list.

✓ Correctly returns 0 for empty input.

t2_02edge

Input{"intervals":[[5,10]]}

Expected1

⏱ Performance - must finish in 2000ms

Single interval cannot be covered by any other, so it remains.

💡 Single element input should always return 1.

💡 Avoid loops that assume multiple intervals.

💡 Return length of intervals if only one interval present.

Why it failed: Fails single element input by returning 0 or error due to loop assumptions. Fix by handling single interval case explicitly or ensuring loops handle n=1.

✓ Correctly returns 1 for single interval input.

t2_03edge

Input{"intervals":[[1,5],[1,5],[1,5]]}

Expected1

⏱ Performance - must finish in 2000ms

All intervals are identical; only one remains after removing covered duplicates.

💡 Identical intervals cover each other.

💡 Sorting by start ascending and end descending groups identical intervals together.

💡 Count only unique intervals or skip covered duplicates.

Why it failed: Fails to remove duplicates or treats identical intervals as uncovered. Fix by ensuring coverage check treats identical intervals as covered except one.

✓ Correctly removes duplicates and returns 1.

t2_04edge

Input{"intervals":[[1,2],[3,4],[5,6],[10,20]]}

Expected4

⏱ Performance - must finish in 2000ms

No intervals cover any other; all remain.

💡 Check intervals with no coverage at all.

💡 Sorting by start ascending and end descending keeps order consistent.

💡 Ensure coverage detection does not remove intervals incorrectly.

Why it failed: Incorrectly removes intervals when no coverage exists due to faulty coverage condition. Fix coverage check to require full containment.

✓ Correctly returns count of all intervals when none are covered.

t3_01corner

Input{"intervals":[[1,10],[2,3],[4,5],[6,7],[8,9]]}

Expected1

⏱ Performance - must finish in 2000ms

All smaller intervals are covered by [1,10], so only one remains.

💡 Beware of greedy approaches that count intervals without checking full coverage.

💡 Sort intervals and track max end to detect coverage correctly.

💡 Update max end only when current interval extends coverage.

Why it failed: Greedy approach counting intervals without coverage check returns wrong count. Fix by sorting and tracking max end to detect coverage properly.

✓ Correctly identifies all covered intervals and returns 1.

t3_02corner

Input{"intervals":[[1,4],[2,8],[3,6]]}

Expected2

⏱ Performance - must finish in 2000ms

Interval [3,6] is covered by [2,8], so it is removed. Remaining intervals are [1,4] and [2,8].

💡 Check that coverage is inclusive: c ≤ a and b ≤ d.

💡 Sorting by start ascending and end descending helps detect coverage.

💡 Track max end and compare current interval's end to detect coverage.

Why it failed: Incorrect coverage detection due to off-by-one or strict inequality errors. Fix by using inclusive coverage condition c ≤ a and b ≤ d.

✓ Correctly applies coverage condition and counts uncovered intervals.

t3_03corner

Input{"intervals":[[1,10],[2,6],[3,5],[5,7]]}

Expected1

⏱ Performance - must finish in 2000ms

All intervals except [1,10] are covered by it, so only one remains.

💡 Sorting descending by start can cause errors; sort ascending by start.

💡 Check coverage by comparing current interval with max end seen so far.

💡 Update max end only when current interval extends coverage.

Why it failed: Sorting intervals in descending order causes coverage detection to fail. Fix by sorting intervals by start ascending and end descending.

✓ Correctly handles intervals regardless of input order.

t4_01performance

Input{"intervals":[[0,1000],[1,1001],[2,1002],[3,1003],[4,1004],[5,1005],[6,1006],[7,1007],[8,1008],[9,1009],[10,1010],[11,1011],[12,1012],[13,1013],[14,1014],[15,1015],[16,1016],[17,1017],[18,1018],[19,1019],[20,1020],[21,1021],[22,1022],[23,1023],[24,1024],[25,1025],[26,1026],[27,1027],[28,1028],[29,1029],[30,1030],[31,1031],[32,1032],[33,1033],[34,1034],[35,1035],[36,1036],[37,1037],[38,1038],[39,1039],[40,1040],[41,1041],[42,1042],[43,1043],[44,1044],[45,1045],[46,1046],[47,1047],[48,1048],[49,1049],[50,1050],[51,1051],[52,1052],[53,1053],[54,1054],[55,1055],[56,1056],[57,1057],[58,1058],[59,1059],[60,1060],[61,1061],[62,1062],[63,1063],[64,1064],[65,1065],[66,1066],[67,1067],[68,1068],[69,1069],[70,1070],[71,1071],[72,1072],[73,1073],[74,1074],[75,1075],[76,1076],[77,1077],[78,1078],[79,1079],[80,1080],[81,1081],[82,1082],[83,1083],[84,1084],[85,1085],[86,1086],[87,1087],[88,1088],[89,1089],[90,1090],[91,1091],[92,1092],[93,1093],[94,1094],[95,1095],[96,1096],[97,1097],[98,1098],[99,1099]]}

Expectednull

⏱ Performance - must finish in 2000ms

n=100 intervals with length 1000 each, O(n log n) sorting + O(n) coverage check must complete within 2s.

💡 Use sorting with O(n log n) complexity.

💡 Avoid nested loops to prevent O(n^2) time.

💡 Track max end in a single pass after sorting.

Why it failed: Brute force O(n^2) approach causes TLE on n=100. Fix by implementing sorting + single pass coverage detection with O(n log n) complexity.

✓ Efficient O(n log n) solution passes performance test.

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. What is the time complexity of the optimal car pooling solution using the difference array and prefix sum approach, given n trips and maximum location L?

medium

A. O(n * L) because we must simulate passenger count at every location for each trip

B. O(n + L) because we process each trip once and then sweep through the difference array once

C. O(n log n) due to sorting trips by start location before processing

D. O(L) because the difference array size dominates and trips are processed in constant time

Solution

1. Step 1: Analyze trip processing

   Each of the n trips updates two positions in the difference array -> O(n)

2. Step 2: Analyze prefix sum sweep

   We iterate over the difference array of size L once -> O(L)

3. Final Answer:

   Option B -> Option B

4. Quick Check:

   Sum of O(n) + O(L) operations [OK]

Hint: Difference array updates O(n), prefix sum O(L) [OK]

Common Mistakes:

• Confusing brute force O(n*L) with optimal
• Assuming sorting is required
• Ignoring prefix sum sweep cost

3. The following code attempts to count intervals containing each point using a line sweep. Identify the bug that causes incorrect counts for points exactly at interval ends.

medium

A. Line adding interval end event should use end + 1 instead of end.

B. Sorting key should sort points before interval starts.

C. Active count should be incremented after processing points, not before.

D. Result array initialization size is incorrect.

Solution

1. Step 1: Identify event creation for interval ends

   Interval end events use 'end' instead of 'end + 1', so points exactly at interval end are processed after the interval ends.

2. Step 2: Understand effect on counting

   Because end events are processed at 'end', points at 'end' see active count after decrement, missing that interval.

3. Final Answer:

   Option A -> Option A

4. Quick Check:

   Changing to end + 1 fixes counting for points at interval ends [OK]

Hint: Interval end event must be at end+1 to include points at end [OK]

Common Mistakes:

• Using end instead of end+1
• Misordering events
• Incorrect active count update

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. The following code attempts to find the maximum number of non-overlapping intervals. Which line contains a subtle bug that can cause incorrect results on some inputs?

def max_non_overlapping_intervals(intervals):
    intervals.sort(key=lambda x: x[0])
    removals = 0
    last_end = intervals[0][1]
    for i in range(1, len(intervals)):
        if intervals[i][0] <= last_end:
            removals += 1
            last_end = min(last_end, intervals[i][1])
        else:
            last_end = intervals[i][1]
    return len(intervals) - removals

medium

A. Line 9: Updating last_end to intervals[i][1] when no overlap

B. Line 2: Sorting intervals by start time instead of end time

C. Line 7: Updating last_end to min(last_end, intervals[i][1]) inside overlap condition

D. Line 5: Using '<=' instead of '<' in the overlap condition

Solution

1. Step 1: Understand overlap condition

   Intervals that start exactly at last_end do not overlap and should be allowed.

2. Step 2: Identify incorrect operator

   Using '<=' causes intervals starting exactly at last_end to be considered overlapping, reducing count incorrectly.

3. Final Answer:

   Option D -> Option D

4. Quick Check:

   Change '<=' to '<' fixes the bug [OK]

Hint: Overlap check must use strict less than, not less or equal [OK]

Common Mistakes:

• Confusing inclusive vs exclusive overlap conditions
• Sorting by start time is allowed but less optimal
• Incorrect last_end updates