Non Overlapping Intervals Minimum Removal in DSA Python - Cheat Sheet & Quick Revision

Recall & Review

beginner

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

To find the minimum number of intervals to remove from a list so that the remaining intervals do not overlap.

intermediate

Which greedy strategy is commonly used to solve the Non Overlapping Intervals Minimum Removal problem?

Sort intervals by their end time and iteratively select intervals that start after the last selected interval ends.

intermediate

Why do we sort intervals by their end time in this problem?

Sorting by end time helps to always pick the interval that leaves the most room for the rest, minimizing removals.

intermediate

What is the time complexity of the common solution to the Non Overlapping Intervals Minimum Removal problem?

O(n log n), where n is the number of intervals, due to sorting the intervals by their end time.

beginner

What does the output represent in the Non Overlapping Intervals Minimum Removal problem?

The minimum number of intervals that must be removed to make the rest of the intervals non-overlapping.

What is the first step in solving the Non Overlapping Intervals Minimum Removal problem?

A Sort intervals by their end time

B Sort intervals by their start time

C Remove intervals randomly

D Count overlapping intervals without sorting

Sorting intervals by their end time allows a greedy approach to select intervals that minimize overlap.

If intervals are [[1,3],[2,4],[3,5]], what is the minimum number of intervals to remove to avoid overlap?

A 0

B 1

C 2

D 3

Removing [2,4] leaves [1,3] and [3,5] which do not overlap.

Which data structure is most useful to keep track of the last selected interval's end time?

A Variable to store end time

B Stack

C Queue

D Hash map

A simple variable to store the end time of the last selected interval is enough for comparison.

What does it mean if two intervals overlap?

A Intervals have the same end time

B Intervals have the same start time

C Intervals are adjacent

D One interval starts before the other ends

Overlap means one interval starts before the other ends, causing a conflict.

What is the main reason to remove intervals in this problem?

A To find the longest interval

B To minimize the total length of intervals

C To maximize the number of intervals remaining without overlap

D To sort intervals by start time

Removing intervals helps to keep the maximum number of intervals that do not overlap.

Explain the greedy approach to solve the Non Overlapping Intervals Minimum Removal problem.

Think about choosing intervals that finish earliest to leave room for others.

Describe how to determine if two intervals overlap and why this matters in the problem.

Focus on interval start and end comparisons.

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