Why Intervals Are a Common Problem Pattern in DSA Python - Test Your Knowledge

Practice - 5 Tasks

1fill in blank

easy

Complete the code to merge overlapping intervals.

DSA Python

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

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

A0

B-1

C2

D1

Attempts:

3 left

💡 Hint

Common Mistakes

Sorting by the end time instead of start time.

Using an invalid index like 2 or -1.

We sort intervals by their start time, which is at index 0.

2fill in blank

medium

Complete the code to check if 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

Comparing the end time of current interval instead of start time.

Using wrong index for start time.

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 start time index instead of end time.

Using negative index incorrectly.

The end time is at index 1, so we take the max of last merged end and current interval end.

4fill in blank

hard

Fill both blanks to create a list of merged intervals.

DSA Python

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

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A0

Blen(intervals)

C1

Dintervals

Attempts:

3 left

💡 Hint

Common Mistakes

Starting merged list with wrong index.

Looping with incorrect range.

Start merged list with first interval at index 0, then loop from 1 to length of intervals.

5fill in blank

hard

Fill all three blanks to complete the dictionary comprehension that counts interval overlaps by start time.

DSA Python

overlap_count = { [1]: sum(1 for interval in intervals if interval[[2]] == start) for start in [3] }

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Astart

B0

C[interval[0] for interval in intervals]

Dinterval

Attempts:

3 left

💡 Hint

Common Mistakes

Using wrong variable names in comprehension.

Using wrong index for start time.

We create a dictionary with keys as start times, counting how many intervals start at each time.

Practice

1. Why are intervals commonly used in problems involving scheduling or merging tasks?

easy

A. Because intervals represent time ranges which help manage overlapping tasks

B. Because intervals are always sorted by default in data structures

C. Because intervals only contain single points, making them simple to handle

D. Because intervals do not require any sorting or comparison

Solution

1. Step 1: Understand what intervals represent

   Intervals represent ranges, often time ranges, which can overlap or be separate.

2. Step 2: Connect intervals to scheduling problems

   Scheduling tasks often need to check if time ranges overlap to avoid conflicts or merge tasks.

3. Final Answer:

   Because intervals represent time ranges which help manage overlapping tasks -> Option A

4. Quick Check:

   Intervals = time ranges for tasks [OK]

Hint: Intervals model time ranges, key for overlaps and merges [OK]

Common Mistakes:

• Thinking intervals are single points
• Assuming intervals are always sorted
• Ignoring overlap checks

2. Which of the following is the correct Python syntax to represent an interval as a tuple from 3 to 7?

easy

A. interval = {3, 7}

B. interval = [3, 7]

C. interval = (3, 7)

D. interval = 3..7

Solution

1. Step 1: Recall Python tuple syntax

   Tuples use parentheses and commas, e.g., (3, 7) to represent fixed pairs.

2. Step 2: Check each option

   A creates a set, B a list, D invalid syntax; only C = (3, 7) uses parentheses for tuple correctly.

3. Final Answer:

   interval = (3, 7) -> Option C

4. Quick Check:

   Tuple syntax = (start, end) [OK]

Hint: Use parentheses and comma for tuples [OK]

Common Mistakes:

• Using square brackets for tuples
• Using curly braces which create sets
• Using invalid range syntax

3. What is the output of the following code?

intervals = [(1, 3), (2, 4), (5, 7)]
merged = []
for start, end in sorted(intervals):
    if not merged or merged[-1][1] < start:
        merged.append((start, end))
    else:
        merged[-1] = (merged[-1][0], max(merged[-1][1], end))
print(merged)

medium

A. [(1, 3), (2, 4), (5, 7)]

B. [(2, 4), (5, 7)]

C. [(1, 7)]

D. [(1, 4), (5, 7)]

Solution

1. Step 1: Sort intervals and iterate

   Sorted intervals are [(1,3), (2,4), (5,7)]. Start with empty merged list.

2. Step 2: Merge overlapping intervals

   First interval (1,3) added. Second (2,4) overlaps with (1,3), merged to (1,4). Third (5,7) does not overlap, added as is.

3. Final Answer:

   [(1, 4), (5, 7)] -> Option D

4. Quick Check:

   Merged intervals = [(1,4), (5,7)] [OK]

Hint: Sort intervals, merge if overlapping, else add new [OK]

Common Mistakes:

• Not sorting intervals before merging
• Overwriting intervals incorrectly
• Appending duplicates instead of merging

4. Identify the error in this code snippet that merges intervals:

intervals = [(1, 5), (3, 7), (8, 10)]
merged = []
for start, end in intervals:
    if merged[-1][1] < start:
        merged.append((start, end))
    else:
        merged[-1] = (merged[-1][0], max(merged[-1][1], end))
print(merged)

medium

A. Accessing merged[-1] before any elements are added causes an error

B. Intervals are not sorted before merging

C. Using max() function incorrectly

D. Appending intervals inside else block instead of if block

Solution

1. Step 1: Check initial merged list usage

   merged is empty initially, so merged[-1] access causes IndexError on first iteration.

2. Step 2: Confirm error cause

   Code tries to check merged[-1][1] before any append, which is invalid.

3. Final Answer:

   Accessing merged[-1] before any elements are added causes an error -> Option A

4. Quick Check:

   Empty list access = error [OK]

Hint: Check list is not empty before accessing last element [OK]

Common Mistakes:

• Ignoring empty list before access
• Not sorting intervals before merging
• Misplacing append statements

5. Given a list of intervals representing booked meeting times, how would you find the minimum number of meeting rooms required?

Example: intervals = [(0, 30), (5, 10), (15, 20)]

hard

A. Sort intervals by end time and count how many end before next start

B. Sort start and end times separately, then use two pointers to count overlaps

C. Merge all intervals and count the total length

D. Count intervals with the same start time only

Solution

1. Step 1: Separate and sort start and end times

   Extract start times [0,5,15] and end times [10,20,30], sort both lists.

2. Step 2: Use two pointers to track overlaps

   Move through start and end lists; increment room count when a meeting starts before another ends.

3. Final Answer:

   Sort start and end times separately, then use two pointers to count overlaps -> Option B

4. Quick Check:

   Two-pointer overlap count = rooms needed [OK]

Hint: Sort starts and ends, count overlaps with two pointers [OK]

Common Mistakes:

• Merging intervals instead of counting overlaps
• Ignoring end times in counting rooms
• Counting only intervals with same start