Why Merge Overlapping Intervals in DSA Java?

The Big Idea

What if you could instantly combine all overlapping times without missing a single clash?

The Scenario

Imagine you have a list of booked meeting times in a day, but some meetings overlap. You want to find the total free time, but first, you need to combine all overlapping meetings into one.

The Problem

Manually checking each meeting against all others to merge overlaps is slow and confusing. It's easy to miss overlaps or merge incorrectly, especially when many meetings overlap in complex ways.

The Solution

Using the "Merge Overlapping Intervals" method, you sort the meetings by start time and then combine overlapping ones in a single pass. This makes merging simple, fast, and error-free.

Before vs After

✗ Before

List<Interval> merged = new ArrayList<>();
for (Interval i : intervals) {
  // manually check and merge overlaps
}

✓ After

Collections.sort(intervals, (a, b) -> a.start - b.start);
for (Interval interval : intervals) {
  if (merged.isEmpty() || merged.get(merged.size() - 1).end < interval.start) {
    merged.add(interval);
  } else {
    merged.get(merged.size() - 1).end = Math.max(merged.get(merged.size() - 1).end, interval.end);
  }
}

What It Enables

This lets you quickly find combined time ranges, making scheduling, booking, and resource planning easy and reliable.

Real Life Example

When booking rooms in an office, merging overlapping reservations helps show when rooms are truly free or busy without confusion.

Key Takeaways

Manually merging intervals is slow and error-prone.

Sorting intervals first simplifies merging overlaps.

One pass after sorting efficiently merges all overlaps.

Practice

1. What is the main purpose of merging overlapping intervals in a list of intervals?

easy

A. To combine intervals that overlap into a single continuous interval

B. To sort intervals by their length

C. To remove all intervals that have the same start point

D. To find the interval with the maximum end value

Solution

1. Step 1: Understand overlapping intervals

   Overlapping intervals share some common range, so merging combines them into one continuous interval.

2. Step 2: Purpose of merging

   Merging reduces multiple overlapping intervals into fewer intervals without overlaps, simplifying the list.

3. Final Answer:

   To combine intervals that overlap into a single continuous interval -> Option A

4. Quick Check:

   Merging = Combine overlaps [OK]

Hint: Merging means joining overlapping intervals into one [OK]

Common Mistakes:

• Thinking merging sorts intervals by length
• Confusing merging with removing duplicates
• Assuming merging finds max end only

2. Which of the following Java code snippets correctly sorts a list of intervals by their start time before merging?

easy

A. Collections.sort(intervals, (a, b) -> a.start - b.start);

B. intervals.sort((a, b) -> b.start - a.start);

C. Collections.sort(intervals, (a, b) -> a.end - b.end);

D. intervals.sort((a, b) -> a.length() - b.length());

Solution

1. Step 1: Identify sorting by start time ascending

   We need to sort intervals by their start value in ascending order for merging.

2. Step 2: Check each option

   Collections.sort(intervals, (a, b) -> a.start - b.start); sorts by start ascending correctly. intervals.sort((a, b) -> b.start - a.start); sorts descending. Collections.sort(intervals, (a, b) -> a.end - b.end); sorts by end, not start. intervals.sort((a, b) -> a.length() - b.length()); sorts by length, which is incorrect.

3. Final Answer:

   Collections.sort(intervals, (a, b) -> a.start - b.start); -> Option A

4. Quick Check:

   Sort by start ascending = Collections.sort(intervals, (a, b) -> a.start - b.start); [OK]

Hint: Sort intervals by start ascending before merging [OK]

Common Mistakes:

• Sorting by end time instead of start time
• Sorting in descending order
• Sorting by interval length

3. What is the output of the following Java code that merges overlapping intervals?

List<Interval> intervals = List.of(new Interval(1,3), new Interval(2,6), new Interval(8,10), new Interval(15,18));
List<Interval> merged = mergeIntervals(intervals);
for (Interval i : merged) {
    System.out.print("[" + i.start + "," + i.end + "] ");
}

Assume mergeIntervals merges intervals correctly.

medium

A. [1,3] [2,6] [8,10] [15,18]

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

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

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

Solution

1. Step 1: Identify overlapping intervals

   Intervals (1,3) and (2,6) overlap, so they merge into (1,6). Others do not overlap.

2. Step 2: List merged intervals

   Merged list is (1,6), (8,10), (15,18).

3. Final Answer:

   [1,6] [8,10] [15,18] -> Option C

4. Quick Check:

   Merged intervals = [1,6] [8,10] [15,18] [OK]

Hint: Merge overlapping intervals by extending end time [OK]

Common Mistakes:

• Not merging overlapping intervals
• Printing original intervals without merge
• Merging incorrectly leaving duplicates

4. Identify the bug in this Java method that merges intervals:

public List<Interval> mergeIntervals(List<Interval> intervals) {
    Collections.sort(intervals, (a, b) -> a.start - b.start);
    List<Interval> merged = new ArrayList<>();
    for (Interval interval : intervals) {
        if (merged.isEmpty() || merged.get(merged.size() - 1).end < interval.start) {
            merged.add(interval);
        } else {
            merged.get(merged.size() - 1).end = interval.end;
        }
    }
    return merged;
}

medium

A. It sorts intervals by end instead of start

B. It does not update the end to the max of overlapping intervals

C. It adds intervals even if they overlap

D. It returns the original list instead of merged list

Solution

1. Step 1: Check merging logic

   When intervals overlap, the code sets end to interval.end directly, which may be smaller than current end.

2. Step 2: Correct approach

   It should update end to max of current end and interval.end to cover full overlap.

3. Final Answer:

   It does not update the end to the max of overlapping intervals -> Option B

4. Quick Check:

   Update end = max(end, interval.end) needed [OK]

Hint: Update merged end to max of overlapping ends [OK]

Common Mistakes:

• Setting end to interval.end without max check
• Sorting by wrong field
• Adding overlapping intervals separately

5. Given a list of intervals where some intervals are nested inside others, how does the merge overlapping intervals algorithm handle nested intervals like [(1,10), (2,5), (6,9)]? What is the final merged list?

hard

A. [(2,5), (6,9)]

B. [(1,5), (6,9), (10,10)]

C. [(1,5), (6,10)]

D. [(1,10)]

Solution

1. Step 1: Analyze nested intervals

   Intervals (2,5) and (6,9) are fully inside (1,10), so they overlap completely.

2. Step 2: Merge nested intervals

   Merging results in a single interval (1,10) covering all nested intervals.

3. Final Answer:

   [(1,10)] -> Option D

4. Quick Check:

   Nested intervals merge into outer interval [OK]

Hint: Nested intervals merge into the largest containing interval [OK]

Common Mistakes:

• Keeping nested intervals separate
• Splitting outer interval incorrectly
• Ignoring full overlap cases