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])

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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]:

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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.