Remove Covered Intervals

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Steps

setup

Start with unsorted intervals

We begin with the original input intervals in their given order.

💡 Seeing the original order helps appreciate why sorting is necessary.

Line:

intervals = [[1,4],[3,6],[2,8]]

💡 Intervals are initially unordered; coverage detection is harder without sorting.

setup

Sort intervals by start ascending, end descending

Sort intervals so that starts increase and ends decrease for equal starts.

💡 Sorting this way ensures that if an interval covers another, it appears before it in the list.

Line:

intervals.sort(key=lambda x: (x[0], -x[1]))

💡 Sorted intervals make coverage detection a simple linear scan.

setup

Initialize count and max_end

Set count to 0 and max_end to 0 before starting iteration.

💡 Count tracks how many intervals are not covered; max_end tracks the largest end seen so far.

Line:

count = 0
max_end = 0

💡 Initial values prepare for coverage checks.

traverse

Check first interval [1,4]

Examine interval at index 0 with end=4; since 4 > max_end(0), count increments and max_end updates.

💡 First interval always counts as not covered because max_end starts at 0.

Line:

for _, end in intervals:
    if end > max_end:
        count += 1
        max_end = end

💡 Intervals with ends larger than max_end are not covered and extend coverage.

traverse

Check second interval [2,8]

Examine interval at index 1 with end=8; since 8 > max_end(4), count increments and max_end updates.

💡 A larger end means this interval is not covered and extends coverage.

Line:

if end > max_end:
    count += 1
    max_end = end

💡 max_end updates to the largest end to track coverage boundaries.

traverse

Check third interval [3,6]

Examine interval at index 2 with end=6; since 6 <= max_end(8), it is covered and not counted.

💡 If end is not greater than max_end, the interval is covered by a previous one.

Line:

if end > max_end:
    count += 1
    max_end = end

💡 Intervals with ends less or equal to max_end are covered and skipped.

reconstruct

Iteration complete, return count

All intervals processed; return the count of intervals not covered.

💡 The count now reflects how many intervals remain after removing covered ones.

Line:

return count

💡 The final count is the answer to the problem.

reconstruct

Final answer displayed

The algorithm outputs 2, meaning two intervals remain after removing covered ones.

💡 This confirms the correctness of the coverage detection logic.

Line:

print(remove_covered_intervals(intervals))

💡 The output matches the expected result.

prune

Summary: Covered interval identified

Interval [3,6] is covered by [2,8] and thus excluded from count.

💡 This step highlights the key coverage relationship found by the algorithm.

💡 Understanding coverage is about comparing ends after sorting by start.

reconstruct

Visualization complete

The algorithm has finished processing all intervals and the final count is confirmed.

💡 This final step wraps up the visualization and confirms understanding.

💡 You have now seen every key operation in the algorithm.

def remove_covered_intervals(intervals):  # STEP 1-2
    intervals.sort(key=lambda x: (x[0], -x[1]))  # STEP 2
    count = 0  # STEP 3
    max_end = 0  # STEP 3
    for _, end in intervals:  # STEP 4-6
        if end > max_end:  # STEP 4-6
            count += 1  # STEP 4-6
            max_end = end  # STEP 4-6
    return count  # STEP 7

if __name__ == '__main__':  # STEP 8
    intervals = [[1,4],[3,6],[2,8]]  # STEP 1
    print(remove_covered_intervals(intervals))  # Output: 2  # STEP 8

📊

Remove Covered Intervals - Watch the Algorithm Execute, Step by Step

Watching each step helps you understand how sorting by start ascending and end descending simplifies detecting covered intervals, and how updating max_end tracks coverage efficiently.

setup

[1,4]

0

[3,6]

1

[2,8]

2

Result: 0

sort

[1,4]

0

[2,8]

1

[3,6]

2

Result: 0

initialize

[1,4]

0

[2,8]

1

[3,6]

2

Result: 0

compare

i

[1,4]

0

[2,8]

1

[3,6]

2

Result: 1

compare

[1,4]

0

i

[2,8]

1

[3,6]

2

Result: 2

compare

[1,4]

0

[2,8]

1

i

[3,6]

2

Result: 2

return

[1,4]

0

[2,8]

1

[3,6]

2

Result: 2

output

[1,4]

0

[2,8]

1

[3,6]

2

Result: 2

prune

[1,4]

0

[2,8]

1

covered

[3,6]

2

Result: 2

done

[1,4]

0

[2,8]

1

[3,6]

2

Result: 2

Key Takeaways

✓ Sorting intervals by start ascending and end descending simplifies coverage detection.

Without sorting, it is difficult to detect coverage efficiently; sorting orders intervals so coverage checks become a simple linear scan.

✓ Tracking the maximum end seen so far lets us identify covered intervals in one pass.

This insight shows how a single variable can represent coverage boundaries, avoiding nested loops.

✓ Intervals with ends less or equal to max_end are covered and excluded from the count.

This concrete example from the trace ([3,6] covered by [2,8]) clarifies how coverage is detected by comparing ends.

Practice

1. You are given a list of meeting time intervals consisting of start and end times. Your task is to find the minimum number of conference rooms required so that all meetings can be held without overlap. Which of the following approaches guarantees an optimal solution with efficient time complexity?

easy

A. Sort intervals by start time and use a min-heap to track the earliest ending meeting to reuse rooms efficiently.

B. Use a brute force approach checking every pair of intervals for overlap and counting maximum overlaps.

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

D. Sort intervals by end time and greedily assign rooms without tracking ongoing meetings.

Solution

1. Step 1: Understand the problem requires tracking overlapping intervals to find minimum rooms.

   Brute force checks all pairs but is inefficient; greedy by end time alone doesn't track concurrent overlaps.

2. Step 2: Recognize that sorting by start time and using a min-heap to track earliest ending meeting allows reusing rooms optimally.

   This approach efficiently manages room allocation by freeing rooms as meetings end.

3. Final Answer:

   Option A -> Option A

4. Quick Check:

   Min-heap approach is standard optimal solution for this problem [OK]

Hint: Min-heap tracks earliest end to reuse rooms efficiently [OK]

Common Mistakes:

• Assuming greedy by end time alone suffices
• Thinking brute force is efficient enough
• Confusing DP for interval scheduling with room allocation

2. Consider the following Python code implementing the min-heap approach to find the minimum number of meeting rooms. Given the input intervals = [[0,30],[5,10],[15,20]], what is the value of the heap after processing the second interval (i=1)?

easy

A. [10]

B. [30, 10]

C. [30, 20]

D. [5, 10]

Solution

1. Step 1: Trace heap after first interval [0,30].

   Heap contains [30] after pushing end time of first meeting.

2. Step 2: Process second interval [5,10].

   Since 5 < 30 (heap[0]), no pop occurs; push 10. Heap now contains [10, 30] (min-heap property).

3. Final Answer:

   Option B -> Option B

4. Quick Check:

   Heap stores end times; after second interval, both 30 and 10 are in heap [OK]

Hint: Heap stores end times; no pop if start < earliest end [OK]

Common Mistakes:

• Popping when start < earliest end
• Confusing heap contents order
• Forgetting to push current interval's end

3. You are given a list of time intervals representing booked meeting rooms. Your task is to combine all overlapping intervals into the minimum number of non-overlapping intervals. Which algorithmic approach guarantees an optimal and efficient solution for this problem?

easy

A. Use a greedy approach that always merges the earliest starting interval with the next one without sorting.

B. Sort intervals by start time and then merge overlapping intervals in a single pass.

C. Use dynamic programming to find the maximum number of non-overlapping intervals and then merge accordingly.

D. Check every pair of intervals with nested loops to merge overlaps until no overlaps remain.

Solution

1. Step 1: Understand the problem requires merging overlapping intervals efficiently.

   Sorting intervals by their start time ensures that overlapping intervals are adjacent, enabling a single pass merge.

2. Step 2: Evaluate each option's approach.

   Use a greedy approach that always merges the earliest starting interval with the next one without sorting. fails because without sorting, intervals may be merged incorrectly or missed. Use dynamic programming to find the maximum number of non-overlapping intervals and then merge accordingly. is unrelated as DP is not needed here. Check every pair of intervals with nested loops to merge overlaps until no overlaps remain. is brute force with O(n²) time, inefficient for large inputs. Sort intervals by start time and then merge overlapping intervals in a single pass. correctly sorts and merges in O(n log n) time.

3. Final Answer:

   Option B -> Option B

4. Quick Check:

   Sorting + one pass merge is the classic optimal pattern for merging intervals. [OK]

Hint: Sort intervals first, then merge in one pass [OK]

Common Mistakes:

• Trying to merge without sorting
• Using nested loops leading to O(n²) time

4. Suppose intervals can be reused multiple times (i.e., you can merge overlapping intervals repeatedly, possibly reusing intervals). Which modification to the merge intervals algorithm correctly handles this scenario?

hard

A. Use the same sorting + one pass merge approach; repeated intervals will be merged naturally.

B. Sort intervals and repeatedly apply the merge pass until no changes occur, which may take multiple passes.

C. Use a brute force nested loop approach to merge intervals until no overlaps remain.

D. Use a graph-based approach to find connected components of overlapping intervals and merge each component.

Solution

1. Step 1: Understand that reusing intervals means overlaps can form complex connected groups.

   Simple one-pass merge assumes intervals appear once and sorted order suffices.

2. Step 2: Recognize that overlapping intervals form connected components in a graph.

   Building a graph where intervals are nodes and edges represent overlaps allows merging all connected intervals correctly.

3. Step 3: Evaluate options.

   Use the same sorting + one pass merge approach; repeated intervals will be merged naturally. fails because one pass merge assumes no reuse. Sort intervals and repeatedly apply the merge pass until no changes occur, which may take multiple passes. is inefficient and may not terminate quickly. Use a brute force nested loop approach to merge intervals until no overlaps remain. is brute force and inefficient. Use a graph-based approach to find connected components of overlapping intervals and merge each component. correctly models the problem and merges all connected intervals.

4. Final Answer:

   Option D -> Option D

5. Quick Check:

   Graph connected components capture all overlapping intervals including reuse. [OK]

Hint: Model reuse as graph connected components [OK]

Common Mistakes:

• Assuming one pass merge works with reuse
• Using repeated merges without termination guarantee

5. Suppose balloons can be reused multiple times after bursting (i.e., an arrow can burst a balloon multiple times if shot at different points). How does this affect the algorithm to find the minimum number of arrows needed?

hard

A. The original greedy algorithm still applies without changes.

B. We must use dynamic programming to track multiple bursts per balloon.

C. The problem reduces to counting unique balloons since reuse removes overlap constraints.

D. We can shoot arrows only at balloon start points to minimize arrows.

Solution

1. Step 1: Understand reuse effect

   If balloons can be burst multiple times independently, overlapping intervals no longer reduce arrow count.

2. Step 2: Simplify problem

   Each balloon requires at least one arrow, so minimum arrows equal number of unique balloons.

3. Step 3: Algorithm implication

   Overlap and interval sorting become irrelevant; counting balloons suffices.

4. Final Answer:

   Option C -> Option C

5. Quick Check:

   Reuse breaks overlap optimization, so count balloons directly [OK]

Hint: Reuse removes overlap benefit; count balloons directly [OK]

Common Mistakes:

• Assuming greedy still works unchanged
• Trying to track multiple bursts with DP unnecessarily
• Ignoring that reuse breaks interval overlap logic