Merge Overlapping Intervals in DSA Java - Interactive Practice

Practice - 5 Tasks

1fill in blank

easy

Complete the code to sort intervals by their start time.

DSA Java

Arrays.sort(intervals, (a, b) -> a[[1]] - b[[1]]);

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

A0

B2

C1

Dlength

Attempts:

3 left

💡 Hint

Common Mistakes

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

Using 'length' or 2 causes index out of bounds errors.

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

2fill in blank

medium

Complete the code to check if two intervals overlap.

DSA Java

if (intervals[i][0] <= merged.get(merged.size() - 1)[[1]]) {

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A1

B0

C2

Dlength

Attempts:

3 left

💡 Hint

Common Mistakes

Comparing start with start instead of end misses overlaps.

Using index 0 instead of 1 causes wrong comparisons.

We check if the current interval's start is less than or equal to the last merged interval's end, which is at index 1.

3fill in blank

hard

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

DSA Java

merged.get(merged.size() - 1)[1] = Math.max(merged.get(merged.size() - 1)[1], intervals[i][[1]]);

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

A0

B2

C1

Dlength

Attempts:

3 left

💡 Hint

Common Mistakes

Using index 0 updates start time incorrectly.

Using index 2 or length causes index errors.

We update the end time at index 1 by taking the maximum of current end and new interval's end at index 1.

4fill in blank

hard

Fill both blanks to add a new interval to merged list and convert list to array.

DSA Java

merged.add(new int[] {intervals[i][[1]], intervals[i][[2]]});
return merged.toArray(new int[[3]][]);

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

A0

B1

Cmerged.size()

Dintervals.length

Attempts:

3 left

💡 Hint

Common Mistakes

Swapping start and end indices.

Using intervals.length instead of merged.size() for array size.

We add the current interval's start (index 0) and end (index 1). Then convert merged list to array with size merged.size().

5fill in blank

hard

Fill all three blanks to complete the merge intervals method.

DSA Java

public int[][] merge(int[][] intervals) {
    if (intervals.length == 0) return new int[0][];
    Arrays.sort(intervals, (a, b) -> a[[1]] - b[[1]]);
    List<int[]> merged = new ArrayList<>();
    merged.add(intervals[0]);
    for (int i = 1; i < intervals.length; i++) {
        if (intervals[i][0] <= merged.get(merged.size() - 1)[[2]]) {
            merged.get(merged.size() - 1)[1] = Math.max(merged.get(merged.size() - 1)[1], intervals[i][1]);
        } else {
            merged.add(intervals[i]);
        }
    }
    return merged.toArray(new int[[3]][]);
}

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

A0

B1

Cmerged.size()

Dintervals.length

Attempts:

3 left

💡 Hint

Common Mistakes

Using wrong indices for start or end.

Using intervals.length instead of merged.size() for array size.

Sort by start (index 0), check overlap with last merged end (index 1), and convert list to array with merged.size().

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