Merge Overlapping Intervals in DSA Python - Interactive Practice

Practice - 5 Tasks

1fill in blank

easy

Complete the code to sort the intervals by their start time.

DSA Python

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

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A1

B0

C2

D-1

Attempts:

3 left

💡 Hint

Common Mistakes

Using index 1 instead of 0 causes sorting by end time, which is incorrect.

Using negative index like -1 sorts by end but is less clear.

We sort intervals by their start time, which is at index 0 in each interval list.

2fill in blank

medium

Complete the code to check if the current interval overlaps with the last merged interval.

DSA Python

if intervals[i][[1]] <= merged[-1][1]:

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A1

B-1

C2

D0

Attempts:

3 left

💡 Hint

Common Mistakes

Checking wrong index for start time.

Using '<' instead of '<=' misses touching intervals.

We compare the start of the current interval (index 0) with the end of the last merged interval.

3fill in blank

hard

Fix the error in updating the end time of the last merged interval.

DSA Python

merged[-1][1] = max(merged[-1][1], intervals[i][[1]])

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A1

B0

C2

D-1

Attempts:

3 left

💡 Hint

Common Mistakes

Using index 0 (start) instead of 1 (end) causes wrong merging.

Using negative index -1 can be confusing here.

We update the end time using the end of the current interval, which is at index 1.

4fill in blank

hard

Fill both blanks to append the first interval and loop through the rest.

DSA Python

merged = [intervals[[1]]]
for i in range([2], len(intervals)):
    # merging logic here

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A0

B1

C2

Dlen(intervals)

Attempts:

3 left

💡 Hint

Common Mistakes

Starting loop from 0 causes duplicate processing of first interval.

Appending wrong interval index to merged.

We start merged with the first interval at index 0 and loop from index 1 to the end.

5fill in blank

hard

Fill all three blanks to complete the merge intervals function.

DSA Python

def merge(intervals):
    if not intervals:
        return []
    intervals.sort(key=lambda x: x[[1]])
    merged = [intervals[[2]]]
    for i in range([3], len(intervals)):
        if intervals[i][0] <= merged[-1][1]:
            merged[-1][1] = max(merged[-1][1], intervals[i][1])
        else:
            merged.append(intervals[i])
    return merged

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

A0

C1

Attempts:

3 left

💡 Hint

Common Mistakes

Mixing up start and end indices.

Looping from 0 instead of 1 causing duplicates.

Sort by start index 0, start merged with intervals[0], loop from 1 to end.