Merge Overlapping Intervals in DSA Java - Practice Problems & Challenges

Challenge - 5 Problems

❓ Predict Output

intermediate

Output of merging intervals with overlapping pairs

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

DSA Java

import java.util.*;

public class Main {
    public static List<int[]> mergeIntervals(int[][] intervals) {
        if (intervals.length == 0) return new ArrayList<>();
        Arrays.sort(intervals, (a, b) -> Integer.compare(a[0], b[0]));
        List<int[]> merged = new ArrayList<>();
        int[] current = intervals[0];
        for (int i = 1; i < intervals.length; i++) {
            if (intervals[i][0] <= current[1]) {
                current[1] = Math.max(current[1], intervals[i][1]);
            } else {
                merged.add(current);
                current = intervals[i];
            }
        }
        merged.add(current);
        return merged;
    }

    public static void main(String[] args) {
        int[][] intervals = {{1,3},{2,6},{8,10},{15,18}};
        List<int[]> result = mergeIntervals(intervals);
        for (int[] interval : result) {
            System.out.print("[" + interval[0] + "," + interval[1] + "] ");
        }
    }
}

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

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

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

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

💡 Hint

Think about how intervals that overlap are combined into one.

The code sorts intervals by start time, then merges overlapping intervals by extending the end time. Intervals [1,3] and [2,6] overlap, so they merge into [1,6]. Others remain separate.

❓ Predict Output

intermediate

Result of merging intervals with nested overlaps

What is the output of this Java code merging nested overlapping intervals?

DSA Java

import java.util.*;

public class Main {
    public static List<int[]> mergeIntervals(int[][] intervals) {
        Arrays.sort(intervals, Comparator.comparingInt(a -> a[0]));
        List<int[]> merged = new ArrayList<>();
        int[] current = intervals[0];
        for (int i = 1; i < intervals.length; i++) {
            if (intervals[i][0] <= current[1]) {
                current[1] = Math.max(current[1], intervals[i][1]);
            } else {
                merged.add(current);
                current = intervals[i];
            }
        }
        merged.add(current);
        return merged;
    }

    public static void main(String[] args) {
        int[][] intervals = {{1,10},{2,5},{6,9},{11,15}};
        List<int[]> result = mergeIntervals(intervals);
        for (int[] interval : result) {
            System.out.print("[" + interval[0] + "," + interval[1] + "] ");
        }
    }
}

A [1,15]

B [1,5] [6,9] [11,15]

C [1,10] [11,15]

D [2,5] [6,9] [11,15]

💡 Hint

Nested intervals merge into the largest covering interval.

Intervals [1,10], [2,5], and [6,9] overlap and merge into [1,10]. Interval [11,15] is separate.

🔧 Debug

advanced

Identify the error in merging intervals code

What error does this Java code produce when merging intervals?

DSA Java

import java.util.*;

public class Main {
    public static List<int[]> mergeIntervals(int[][] intervals) {
        Arrays.sort(intervals, (a, b) -> a[0] - b[0]);
        List<int[]> merged = new ArrayList<>();
        int[] current = intervals[0];
        for (int i = 1; i < intervals.length; i++) {
            if (intervals[i][0] <= current[1]) {
                current[1] = Math.max(current[1], intervals[i][1]);
            } else {
                merged.add(current);
                current = intervals[i];
            }
        }
        merged.add(current);
        return merged;
    }

    public static void main(String[] args) {
        int[][] intervals = {{1,4},{4,5}};
        List<int[]> result = mergeIntervals(intervals);
        for (int[] interval : result) {
            System.out.print("[" + interval[0] + "," + interval[1] + "] ");
        }
    }
}

A ArrayIndexOutOfBoundsException

B NullPointerException

C No error, outputs: [1,5]

D ClassCastException

💡 Hint

Check the loop boundary condition carefully.

The for loop uses i <= intervals.length which causes an index out of bounds when i equals intervals.length.

🚀 Application

advanced

Number of merged intervals after merging

Given intervals {{1,4},{2,3},{5,7},{6,8},{9,10}}, how many intervals remain after merging overlapping ones?

A 3

B 2

C 5

D 4

💡 Hint

Merge intervals that overlap or touch.

Intervals [1,4] and [2,3] merge into [1,4]. Intervals [5,7] and [6,8] merge into [5,8]. Interval [9,10] remains separate. Total 3 intervals.

🧠 Conceptual

expert

Correct order of steps to merge intervals

What is the correct order of steps to merge overlapping intervals?

A 2,1,3,4

B 1,2,3,4

C 1,3,2,4

D 3,1,2,4

💡 Hint

Think about what must happen first before merging.

You must first sort intervals, then initialize merged list, then iterate and merge or add intervals.

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