Merge Overlapping Intervals in DSA Java - Time & Space Complexity

Time Complexity: Merge Overlapping Intervals

O(n log n)

Understanding Time Complexity

We want to know how the time to merge overlapping intervals grows as we add more intervals.

How does the number of intervals affect the work done to combine them?

Scenario Under Consideration

Analyze the time complexity of the following code snippet.

import java.util.*;

public List<int[]> mergeIntervals(List<int[]> intervals) {
    Collections.sort(intervals, (a, b) -> a[0] - b[0]);
    List<int[]> merged = new ArrayList<>();
    for (int[] interval : intervals) {
        if (merged.isEmpty() || merged.get(merged.size() - 1)[1] < interval[0]) {
            merged.add(interval);
        } else {
            merged.get(merged.size() - 1)[1] = Math.max(merged.get(merged.size() - 1)[1], interval[1]);
        }
    }
    return merged;
}
    

This code sorts intervals by start time, then merges overlapping ones into a new list.

Identify Repeating Operations

• Primary operation: Sorting the list of intervals and then looping through each interval once.
• How many times: Sorting takes time based on all intervals, then the loop runs once per interval.

How Execution Grows With Input

As the number of intervals grows, sorting takes more time, and merging checks each interval once.

Input Size (n)	Approx. Operations
10	About 10 log 10 + 10 = 40 operations
100	About 100 log 100 + 100 = 700 operations
1000	About 1000 log 1000 + 1000 = 10,000 operations

Pattern observation: The time grows a bit faster than the number of intervals because of sorting, but the merging step grows linearly.

Final Time Complexity

Time Complexity: O(n log n)

This means the time to merge intervals grows a little faster than the number of intervals because sorting takes extra work.

Common Mistake

[X] Wrong: "Merging intervals is just a simple loop, so it must be O(n)."

[OK] Correct: Sorting the intervals first takes extra time, which dominates the total time, making it O(n log n).

Interview Connect

Understanding this time complexity helps you explain why sorting is important before merging and shows you can analyze combined steps clearly.

Self-Check

"What if the intervals were already sorted? How would the time complexity change?"

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