Non Overlapping Intervals Minimum Removal in DSA Python - Practice Problems & Challenges

Challenge - 5 Problems

❓ Predict Output

intermediate

Output of minimum removals for overlapping intervals

What is the output of the following code that calculates the minimum number of intervals to remove to avoid overlaps?

DSA Python

def eraseOverlapIntervals(intervals):
    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
    return count

intervals = [[1,3],[2,4],[3,5],[7,8]]
print(eraseOverlapIntervals(intervals))

A 3

B 2

C 1

D 0

💡 Hint

Sort intervals by their end time and count how many overlap.

The code sorts intervals by their end time. It then iterates and counts intervals that overlap with the last chosen interval. Here, only one interval needs removal to avoid overlap.

❓ Predict Output

intermediate

Result of minimum removals with nested intervals

What does the function return for this set of intervals?

DSA Python

def eraseOverlapIntervals(intervals):
    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
    return count

intervals = [[1,10],[2,3],[3,4],[4,5]]
print(eraseOverlapIntervals(intervals))

A 0

B 3

C 2

D 1

💡 Hint

Check how many intervals overlap with the earliest finishing intervals.

After sorting by end time to [[2,3],[3,4],[4,5],[1,10]], the code keeps [2,3], [3,4], [4,5] (non-overlapping as start >= previous end), skips [1,10], removes 1 interval.

🔧 Debug

advanced

Identify the error in interval removal code

What error will this code produce when run?

DSA Python

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

intervals = [[1,2],[2,3],[1,3]]
print(eraseOverlapIntervals(intervals))

A Outputs 1

B Raises an error

C Outputs 0

D Outputs 2

💡 Hint

Check if sorting by start time affects overlap counting.

The code sorts by start time instead of end time, but still counts overlaps correctly. It outputs 1 because one interval overlaps.

🧠 Conceptual

advanced

Why sort intervals by end time in minimum removal problem?

Why is sorting intervals by their end time the best approach for minimizing removals to avoid overlaps?

A It sorts intervals randomly to check all possibilities.

B It allows choosing intervals that finish earliest, maximizing non-overlapping intervals.

C It sorts intervals by length to remove the longest intervals first.

D It groups intervals by their start time to remove earliest starting intervals.

💡 Hint

Think about how earliest finishing intervals help fit more intervals without overlap.

Sorting by end time helps pick intervals that free up space earliest, allowing more intervals to fit without overlap, minimizing removals.

❓ Predict Output

expert

Output of minimum removals for complex overlapping intervals

What is the output of the code for this complex set of intervals?

DSA Python

def eraseOverlapIntervals(intervals):
    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
    return count

intervals = [[0,30],[5,10],[15,20],[25,35],[27,40],[40,50]]
print(eraseOverlapIntervals(intervals))

A 2

B 4

C 3

D 1

💡 Hint

Sort by end time and count overlaps carefully.

After sorting by end time to [[5,10],[15,20],[0,30],[25,35],[27,40],[40,50]], keeps [5,10], [15,20], [25,35], [40,50]; removes [0,30] and [27,40], totaling 2 removals.

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