Non Overlapping Intervals Minimum Removal in DSA Python - Execution Trace

Concept Flow - Non Overlapping Intervals Minimum Removal

Sort intervals by end time

↓

Initialize count=0, prev_end = -∞

↓

For each interval

↓

Check if interval start >= prev_end?

No→Remove interval, count++

Yes↓

Update prev_end to current interval end

↓

Repeat for all intervals

↓

Return count of removals

Sort intervals by their end time, then iterate to count how many intervals must be removed to avoid overlaps.

Execution Sample

DSA Python

intervals = [[1,2],[2,3],[1,3]]
intervals.sort(key=lambda x: x[1])
count = 0
prev_end = float('-inf')
for start, end in intervals:
    if start >= prev_end:
        prev_end = end
    else:
        count += 1
print(count)

This code counts the minimum number of intervals to remove so that no intervals overlap.

Execution Table

Step	Operation	Current Interval	prev_end	Overlap?	Action	Removal Count	Intervals Removed
1	Sort intervals	[1,2],[2,3],[1,3]	-∞	N/A	Intervals sorted by end time	0	[]
2	Check interval	[1,2]	-∞	No	Update prev_end to 2	0	[]
3	Check interval	[2,3]	2	No	Update prev_end to 3	0	[]
4	Check interval	[1,3]	3	Yes	Remove interval	1	[[1,3]]
5	End	N/A	3	N/A	Return removal count	1	[[1,3]]

💡 All intervals processed; minimum removals = 1

Variable Tracker

Variable	Start	After Step 2	After Step 3	After Step 4	Final
prev_end	-∞	2	3	3	3
count	0	0	0	1	1
intervals_removed	[]	[]	[]	[[1,3]]	[[1,3]]

Key Moments - 3 Insights

Why do we sort intervals by their end time instead of start time?

Sorting by end time ensures we always keep the interval that finishes earliest, leaving more room for following intervals. This is shown in Step 1 of the execution_table where intervals are sorted by their end times.

Why do we update prev_end only when intervals do not overlap?

Updating prev_end to the current interval's end time only when no overlap occurs ensures we track the end of the last kept interval. This is clear in Steps 2 and 3 where prev_end updates, but in Step 4 it stays the same because the interval overlaps and is removed.

How do we know when to remove an interval?

If the current interval's start is less than prev_end, it overlaps with the last kept interval and must be removed. This is shown in Step 4 where overlap is detected and removal count increases.

Visual Quiz - 3 Questions

Look at the execution_table, what is the value of prev_end after Step 3?

A 2

B 3

C -∞

D 1

💡 Hint

Check the 'prev_end' column in the execution_table row for Step 3.

At which step does the removal count increase?

A Step 4

B Step 3

C Step 2

D Step 5

💡 Hint

Look at the 'Removal Count' column in the execution_table to see when it changes from 0 to 1.

If the intervals were [[1,2],[2,3],[3,4]], how would the removal count change?

A It would be 1

B It would be 2

C It would be 0

D It would be 3

💡 Hint

Check the logic in the code and variable_tracker for how intervals without overlap affect removal count.

Concept Snapshot

Non Overlapping Intervals Minimum Removal:
- Sort intervals by end time
- Initialize count=0, prev_end=-∞
- For each interval:
  - If start >= prev_end, update prev_end
  - Else, increment removal count
- Return removal count
This greedy approach keeps max intervals without overlap.

Full Transcript

To find the minimum number of intervals to remove so that no intervals overlap, we first sort the intervals by their end time. We then keep track of the end time of the last interval we accepted (prev_end). For each interval, if its start time is greater than or equal to prev_end, we accept it and update prev_end. Otherwise, it overlaps and we remove it, increasing our removal count. This method ensures we keep the maximum number of intervals without overlap by always choosing the interval that ends earliest. The execution table shows each step, the current interval checked, whether it overlaps, and the removal count. The variable tracker shows how prev_end and count change over time. Key moments clarify why sorting by end time is important, when to update prev_end, and when to remove intervals. The visual quiz tests understanding of these steps and outcomes.

Practice

1. What is the main goal of the Non Overlapping Intervals Minimum Removal problem?

easy

A. To find the minimum number of intervals to remove so that the rest do not overlap

B. To find the maximum number of overlapping intervals

C. To sort intervals by their start time

D. To merge all overlapping intervals into one

Solution

1. Step 1: Understand the problem goal

   The problem asks to remove the fewest intervals so that no two intervals overlap.

2. Step 2: Differentiate from related problems

   Unlike merging intervals or counting overlaps, here we minimize removals to avoid overlaps.

3. Final Answer:

   To find the minimum number of intervals to remove so that the rest do not overlap -> Option A

4. Quick Check:

   Minimum removals = non-overlapping intervals [OK]

Hint: Focus on removing intervals to avoid overlaps, not merging [OK]

Common Mistakes:

• Confusing with merging intervals problem
• Trying to maximize overlaps instead of minimizing removals
• Sorting intervals incorrectly

2. Which of the following Python code snippets correctly sorts a list of intervals by their end time?

easy

A. intervals.sort(key=lambda x: x[0])

B. intervals.sort(reverse=True)

C. intervals.sort(key=lambda x: x[1])

D. intervals.sort(key=lambda x: x[1], reverse=True)

Solution

1. Step 1: Identify sorting by end time

   Sorting by end time means using the second element of each interval, which is index 1.

2. Step 2: Check the correct syntax

   Using intervals.sort(key=lambda x: x[1]) sorts intervals ascending by end time.

3. Final Answer:

   intervals.sort(key=lambda x: x[1]) -> Option C

4. Quick Check:

   Sort by end time = key=lambda x: x[1] [OK]

Hint: Sort intervals by end time using key=lambda x: x[1] [OK]

Common Mistakes:

• Sorting by start time instead of end time
• Using reverse=True when not needed
• Incorrect lambda syntax

3. What is the output of the following code?

intervals = [[1,3],[2,4],[3,5]]
intervals.sort(key=lambda x: x[1])
count = 0
end = float('-inf')
for interval in intervals:
    if interval[0] >= end:
        end = interval[1]
    else:
        count += 1
print(count)

medium

A. 2

B. 1

C. 0

D. 3

Solution

1. Step 1: Sort intervals by end time

   Sorted intervals: [[1,3],[2,4],[3,5]] (already sorted by end time)

2. Step 2: Iterate and count overlaps

   Start with end = -inf
   Interval [1,3]: 1 >= -inf true, update end=3
   Interval [2,4]: 2 >= 3 false, count=1
   Interval [3,5]: 3 >= 3 true, update end=5

3. Final Answer:

   1 -> Option B

4. Quick Check:

   One interval removed to avoid overlap [OK]

Hint: Count intervals that start before current end to find removals [OK]

Common Mistakes:

• Not updating end correctly
• Counting non-overlapping intervals instead of removals
• Sorting by start time instead of end time

4. Find the error in this code snippet that tries to solve the minimum removal of overlapping intervals:

def eraseOverlapIntervals(intervals):
    intervals.sort(key=lambda x: x[0])
    count = 0
    end = float('-inf')
    for interval in intervals:
        if interval[0] >= end:
            end = interval[1]
        else:
            count += 1
    return count

medium

A. The function should return intervals, not count

B. Using float('-inf') for end causes errors

C. The count variable should be initialized to 1

D. Sorting by start time instead of end time causes incorrect results

Solution

1. Step 1: Check sorting key

   The code sorts intervals by start time, but to minimize removals, sorting by end time is needed.

2. Step 2: Understand impact of sorting

   Sorting by start time can cause wrong overlap checks and incorrect count of removals.

3. Final Answer:

   Sorting by start time instead of end time causes incorrect results -> Option D

4. Quick Check:

   Sort by end time to minimize removals [OK]

Hint: Always sort intervals by end time for this problem [OK]

Common Mistakes:

• Sorting by start time
• Incorrect initialization of end
• Returning wrong value

5. Given intervals [[1,2],[2,3],[3,4],[1,3]], what is the minimum number of intervals to remove to make the rest non-overlapping?

hard

A. 1

B. 2

C. 3

D. 0

Solution

1. Step 1: Sort intervals by end time

   Sorted intervals: [[1,2],[2,3],[1,3],[3,4]]

2. Step 2: Select intervals greedily and count removals

   Start end = -inf
   [1,2]: 1 >= -inf true, end=2
   [2,3]: 2 >= 2 true, end=3
   [1,3]: 1 >= 3 false, count=1 (remove this)
   [3,4]: 3 >= 3 true, end=4

3. Final Answer:

   1 -> Option A

4. Quick Check:

   One interval removed to avoid overlap [OK]

Hint: Sort by end time, greedily pick intervals, count removals [OK]

Common Mistakes:

• Not sorting by end time
• Counting non-overlapping intervals instead of removals
• Removing wrong intervals