Why Intervals Are a Common Problem Pattern in DSA Python - Challenge Your Understanding

Challenge - 5 Problems

❓ Predict Output

intermediate

Output of Merging Overlapping Intervals

What is the output of the following Python code that merges overlapping intervals?

DSA Python

def merge_intervals(intervals):
    intervals.sort(key=lambda x: x[0])
    merged = []
    for interval in intervals:
        if not merged or merged[-1][1] < interval[0]:
            merged.append(interval)
        else:
            merged[-1][1] = max(merged[-1][1], interval[1])
    return merged

intervals = [[1,3],[2,6],[8,10],[15,18]]
print(merge_intervals(intervals))

A [[1, 6], [8, 10], [15, 18]]

B [[1, 3], [2, 6], [8, 10], [15, 18]]

C [[1, 6], [2, 6], [8, 10], [15, 18]]

D [[1, 3], [8, 10], [15, 18]]

💡 Hint

Think about how overlapping intervals are combined into one.

The code sorts intervals by start time, then merges overlapping ones by updating the end time of the last merged interval.

🧠 Conceptual

intermediate

Why Sorting is Important in Interval Problems

Why is sorting intervals by their start time a common first step in interval problems?

A It helps to process intervals in order to easily detect overlaps and merge them.

B It reduces the number of intervals by removing duplicates automatically.

C It changes the intervals to a different data structure for faster access.

D It ensures intervals are all the same length for uniform processing.

💡 Hint

Think about how intervals relate to each other when ordered by start time.

Sorting by start time allows us to check each interval against the previous one to detect overlaps efficiently.

🔧 Debug

advanced

Identify the Error in Interval Overlap Check

What error does this code raise when checking if two intervals overlap?

DSA Python

def is_overlapping(a, b):
    return a[1] > b[0] and b[1] > a[0]

print(is_overlapping([1,3], [3,5]))

A Raises a TypeError due to wrong comparison operators.

B Returns True, which is incorrect because intervals do not overlap if they only touch at the boundary.

C Returns False, but intervals touching at the boundary should be considered overlapping.

D Raises an IndexError because intervals have only one element.

💡 Hint

Check how the code treats intervals that just touch at the edges.

The code returns False for intervals that touch at the boundary (like [1,3] and [3,5]) because it uses > instead of >=, so touching intervals are not considered overlapping.

🚀 Application

advanced

Find Maximum Number of Overlapping Intervals

Given intervals, what is the maximum number of intervals overlapping at any point?

DSA Python

intervals = [[1,4],[2,5],[7,9],[3,6]]

# What is the maximum overlap count?

A 2

B 3

C 4

D 1

💡 Hint

Visualize intervals on a timeline and count overlaps at each point.

Intervals [1,4], [2,5], and [3,6] overlap between 3 and 4, so maximum overlap is 3.

❓ Predict Output

expert

Output of Interval Scheduling Maximum Set

What is the output of the code that finds the maximum number of non-overlapping intervals?

DSA Python

def max_non_overlapping(intervals):
    intervals.sort(key=lambda x: x[1])
    count = 0
    end = float('-inf')
    for interval in intervals:
        if interval[0] >= end:
            count += 1
            end = interval[1]
    return count

intervals = [[1,2],[2,3],[3,4],[1,3]]
print(max_non_overlapping(intervals))

A 1

B 4

C 2

D 3

💡 Hint

Think about choosing intervals that end earliest to maximize count.

The code selects intervals greedily by earliest end time, resulting in 3 non-overlapping intervals: [1,2], [2,3], and [3,4].