Non Overlapping Intervals Minimum Removal in DSA Python - Interactive Practice

Practice - 5 Tasks

1fill in blank

easy

Complete the code to sort intervals by their end time.

DSA Python

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

Drag options to blanks, or click blank then click option'

A[1]

B[

C[0]

D[2]

Attempts:

3 left

💡 Hint

Common Mistakes

Sorting by start time instead of end time.

Using wrong index like 0 or 2.

We sort intervals by their end time, which is at index 1 in each interval list.

2fill in blank

medium

Complete the code to check if current interval overlaps with previous.

DSA Python

if intervals[i][0] < intervals[[1]][1]:

Drag options to blanks, or click blank then click option'

A0

Bi+1

Ci-1

Dlen(intervals)-1

Attempts:

3 left

💡 Hint

Common Mistakes

Using i+1 which is next interval, not previous.

Using 0 or last index incorrectly.

We compare the start of current interval with end of previous interval at i-1.

3fill in blank

hard

Fix the error in updating the end time after removing an interval.

DSA Python

if intervals[i][1] < intervals[prev][[1]]:
    prev = i

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A1

B0

C2

D-1

Attempts:

3 left

💡 Hint

Common Mistakes

Using index 0 which is start time.

Using invalid index like 2 or -1.

We compare end times at index 1, so update prev if current end is smaller.

4fill in blank

hard

Fill both blanks to complete the loop that counts removals.

DSA Python

removals = 0
prev = 0
for i in range(1, [1]):
    if intervals[i][0] < intervals[prev][1]:
        removals += 1
        if intervals[i][[2]] < intervals[prev][1]:
            prev = i
    else:
        prev = i

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Alen(intervals)

B0

C1

D2

Attempts:

3 left

💡 Hint

Common Mistakes

Using 0 or 2 for loop range end.

Comparing start time instead of end time.

Loop runs from 1 to length of intervals; compare end time at index 1.

5fill in blank

hard

Fill all three blanks to complete the function that returns minimum removals.

DSA Python

def eraseOverlapIntervals(intervals):
    if not intervals:
        return 0
    intervals.sort(key=lambda x: x[1])
    removals = 0
    prev = 0
    for i in range(1, [2]):
        if intervals[i][0] < intervals[prev][1]:
            removals += 1
            if intervals[i][[3]] < intervals[prev][1]:
                prev = i
        else:
            prev = i
    return removals

Drag options to blanks, or click blank then click option'

A[1]

Blen(intervals)

C1

D[0]

Attempts:

3 left

💡 Hint

Common Mistakes

Sorting by start time instead of end time.

Looping with wrong range.

Comparing wrong indices.

Sort by end time [1], loop till length, compare end time at index 1.

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