Insert Interval into Sorted List in DSA Java - Practice Problems & Challenges

Challenge - 5 Problems

❓ Predict Output

intermediate

Output after inserting an interval that merges with existing intervals

What is the output of the intervals list after inserting [4, 8] into [[1, 3], [6, 9]]?

DSA Java

import java.util.*;

public class Solution {
    public static List<int[]> insertInterval(List<int[]> intervals, int[] newInterval) {
        List<int[]> result = new ArrayList<>();
        int i = 0;
        while (i < intervals.size() && intervals.get(i)[1] < newInterval[0]) {
            result.add(intervals.get(i));
            i++;
        }
        while (i < intervals.size() && intervals.get(i)[0] <= newInterval[1]) {
            newInterval[0] = Math.min(newInterval[0], intervals.get(i)[0]);
            newInterval[1] = Math.max(newInterval[1], intervals.get(i)[1]);
            i++;
        }
        result.add(newInterval);
        while (i < intervals.size()) {
            result.add(intervals.get(i));
            i++;
        }
        return result;
    }

    public static void main(String[] args) {
        List<int[]> intervals = new ArrayList<>();
        intervals.add(new int[]{1, 3});
        intervals.add(new int[]{6, 9});
        int[] newInterval = {4, 8};
        List<int[]> result = insertInterval(intervals, newInterval);
        for (int[] interval : result) {
            System.out.print("[" + interval[0] + ", " + interval[1] + "] ");
        }
    }
}

A [[1, 3], [4, 9]]

B [[1, 3], [4, 8], [6, 9]]

C [[1, 8], [6, 9]]

D [[1, 9]]

💡 Hint

Think about how intervals that overlap with the new interval are merged.

The new interval [4, 8] overlaps with [6, 9], so they merge into [4, 9]. The first interval [1, 3] does not overlap and remains unchanged.

❓ Predict Output

intermediate

Output after inserting an interval that does not overlap

What is the output of the intervals list after inserting [10, 12] into [[1, 5], [6, 9]]?

DSA Java

import java.util.*;

public class Solution {
    public static List<int[]> insertInterval(List<int[]> intervals, int[] newInterval) {
        List<int[]> result = new ArrayList<>();
        int i = 0;
        while (i < intervals.size() && intervals.get(i)[1] < newInterval[0]) {
            result.add(intervals.get(i));
            i++;
        }
        while (i < intervals.size() && intervals.get(i)[0] <= newInterval[1]) {
            newInterval[0] = Math.min(newInterval[0], intervals.get(i)[0]);
            newInterval[1] = Math.max(newInterval[1], intervals.get(i)[1]);
            i++;
        }
        result.add(newInterval);
        while (i < intervals.size()) {
            result.add(intervals.get(i));
            i++;
        }
        return result;
    }

    public static void main(String[] args) {
        List<int[]> intervals = new ArrayList<>();
        intervals.add(new int[]{1, 5});
        intervals.add(new int[]{6, 9});
        int[] newInterval = {10, 12};
        List<int[]> result = insertInterval(intervals, newInterval);
        for (int[] interval : result) {
            System.out.print("[" + interval[0] + ", " + interval[1] + "] ");
        }
    }
}

A [[1, 5], [6, 12]]

B [[1, 12]]

C [[1, 5], [6, 9], [10, 12]]

D [[10, 12], [1, 5], [6, 9]]

💡 Hint

New interval does not overlap, so it should be added at the end.

Since [10, 12] does not overlap with any existing intervals, it is simply added at the end.

🔧 Debug

advanced

Identify the error in interval insertion code

What error will this code produce when inserting [2, 5] into [[1, 3], [6, 9]]? Code snippet: List intervals = new ArrayList<>(); intervals.add(new int[]{1, 3}); intervals.add(new int[]{6, 9}); int[] newInterval = {2, 5}; List result = insertInterval(intervals, newInterval); public static List insertInterval(List intervals, int[] newInterval) { List result = new ArrayList<>(); int i = 0; while (i < intervals.size() && intervals.get(i)[1] <= newInterval[0]) { result.add(intervals.get(i)); i++; } while (i < intervals.size() && intervals.get(i)[0] < newInterval[1]) { newInterval[0] = Math.min(newInterval[0], intervals.get(i)[0]); newInterval[1] = Math.max(newInterval[1], intervals.get(i)[1]); i++; } result.add(newInterval); while (i < intervals.size()) { result.add(intervals.get(i)); i++; } return result; }

A The code produces incorrect merging resulting in [[1, 3], [6, 9], [2, 5]]

B The code runs correctly and outputs [[1, 5], [6, 9]]

C The code causes an infinite loop

D The code throws IndexOutOfBoundsException

💡 Hint

Check the comparison operators in the while loops carefully.

The conditions use <= and < incorrectly, causing the first interval [1,3] to be added before merging, and the new interval added at the end, resulting in wrong output.

🧠 Conceptual

advanced

Understanding interval insertion complexity

What is the time complexity of inserting a new interval into a sorted list of n non-overlapping intervals using the standard merge approach?

A O(log n) because binary search finds the position quickly

B O(n) because we may need to scan all intervals once

C O(n log n) because sorting is required after insertion

D O(1) because insertion is at a fixed position

💡 Hint

Consider how many intervals you might need to check or merge.

Even though the list is sorted, we may need to scan all intervals to merge overlapping ones, so the complexity is linear O(n).

🚀 Application

expert

Result after multiple insertions with merges

Given an initially empty list, insert intervals in this order: [1, 2], [3, 5], [4, 7], [6, 8]. What is the final list of intervals?

A [[1, 7], [6, 8]]

B [[1, 8]]

C [[1, 2], [3, 5], [4, 7], [6, 8]]

D [[1, 2], [3, 8]]

💡 Hint

Merge intervals as you insert if they overlap.

After inserting [1, 2], list is [[1, 2]]. Insert [3, 5], list is [[1, 2], [3, 5]]. Insert [4, 7], merges with [3, 5] to become [3, 7]. Insert [6, 8], merges with [3, 7] to become [3, 8]. Final list: [[1, 2], [3, 8]].

Practice

1. What is the main goal when inserting a new interval into a sorted list of non-overlapping intervals?

easy

A. Merge the new interval with overlapping intervals and keep the list sorted

B. Add the new interval at the end without checking overlaps

C. Remove all intervals that come after the new interval

D. Sort the list again after adding the new interval anywhere

Solution

1. Step 1: Understand the problem context

   The list contains sorted, non-overlapping intervals. When inserting a new interval, we must keep the list sorted and merge any overlapping intervals.

2. Step 2: Identify correct insertion behavior

   Simply adding the new interval without merging or sorting breaks the list's properties. Proper merging keeps intervals non-overlapping and sorted.

3. Final Answer:

   Merge the new interval with overlapping intervals and keep the list sorted -> Option A

4. Quick Check:

   Insert interval = Merge and sort [OK]

Hint: Always merge overlapping intervals to keep list sorted [OK]

Common Mistakes:

• Inserting without merging overlaps
• Appending interval without sorting
• Removing intervals unnecessarily

2. Which of the following Java method signatures correctly represents a function to insert an interval into a sorted list of intervals?

easy

A. public void insertInterval(int intervals[], int newInterval[])

B. public List insertInterval(List intervals, int[] newInterval)

C. public int insertInterval(List<int> intervals, int newInterval)

D. public List<int> insertInterval(int[] intervals, int[] newInterval)

Solution

1. Step 1: Analyze parameter types for intervals

   Intervals are pairs of integers, so List is appropriate to hold multiple intervals. The new interval is an int[] of size 2.

2. Step 2: Check return type and parameters

   The method returns a List after insertion. public List insertInterval(List intervals, int[] newInterval) matches this signature correctly.

3. Final Answer:

   public List insertInterval(List intervals, int[] newInterval) -> Option B

4. Quick Check:

   Correct method signature uses List and int[] [OK]

Hint: Use List for intervals and int[] for new interval [OK]

Common Mistakes:

• Using int[] instead of List for intervals
• Wrong return type like void or int
• Using List<int> instead of List

3. Given the intervals [[1,3],[6,9]] and newInterval = [2,5], what is the resulting list after inserting the new interval?

medium

A. [[1,3],[6,9],[2,5]]

B. [[1,3],[2,5],[6,9]]

C. [[1,5],[6,9]]

D. [[2,5],[6,9]]

Solution

1. Step 1: Identify overlapping intervals

   The new interval [2,5] overlaps with [1,3] because 2 ≤ 3. So they merge into [1,5].

2. Step 2: Merge and keep other intervals

   After merging, the list becomes [[1,5],[6,9]] as [6,9] does not overlap.

3. Final Answer:

   [[1,5],[6,9]] -> Option C

4. Quick Check:

   Merged intervals = [[1,5],[6,9]] [OK]

Hint: Merge intervals if new start ≤ current end [OK]

Common Mistakes:

• Not merging overlapping intervals
• Appending new interval without sorting
• Incorrectly splitting intervals

4. Consider this Java code snippet for inserting an interval:

List insertInterval(List intervals, int[] newInterval) {
  List result = new ArrayList<>();
  for (int[] interval : intervals) {
    if (interval[1] < newInterval[0]) {
      result.add(interval);
    } else if (interval[0] > newInterval[1]) {
      result.add(newInterval);
      newInterval = interval;  // Fix to avoid multiple additions
    } else {
      newInterval[0] = Math.min(newInterval[0], interval[0]);
      newInterval[1] = Math.max(newInterval[1], interval[1]);
    }
  }
  result.add(newInterval);
  return result;
}

What is the bug in this code?

medium

A. It adds newInterval multiple times if intervals after it exist

B. It never adds intervals that come before newInterval

C. It modifies the input intervals list directly

D. It does not merge overlapping intervals correctly

Solution

1. Step 1: Analyze insertion of newInterval in else-if block

   When interval[0] > newInterval[1], it adds newInterval to result, but continues looping, causing newInterval to be added again later.

2. Step 2: Identify repeated addition of newInterval

   Because newInterval is added again after the loop, it can be added multiple times, causing duplicates.

3. Final Answer:

   It adds newInterval multiple times if intervals after it exist -> Option A

4. Quick Check:

   Repeated addition of newInterval causes duplicates [OK]

Hint: Add newInterval once after loop to avoid duplicates [OK]

Common Mistakes:

• Adding newInterval inside loop multiple times
• Not updating newInterval correctly
• Forgetting to add remaining intervals

5. You have intervals [[1,2],[3,5],[6,7],[8,10],[12,16]] and newInterval = [4,8]. After inserting and merging, what is the final list?

hard

A. [[1,2],[3,8],[12,16]]

B. [[1,2],[3,5],[6,8],[8,10],[12,16]]

C. [[1,2],[4,8],[12,16]]

D. [[1,2],[3,10],[12,16]]

Solution

1. Step 1: Identify intervals overlapping with [4,8]

   Intervals [3,5], [6,7], and [8,10] overlap with [4,8] because their ranges intersect.

2. Step 2: Merge overlapping intervals

   Merge [3,5], [6,7], [8,10], and [4,8] into one interval [3,10]. The others remain unchanged.

3. Step 3: Construct final list

   The final list is [[1,2],[3,10],[12,16]].

4. Final Answer:

   [[1,2],[3,10],[12,16]] -> Option D

5. Quick Check:

   Merged intervals = [[1,2],[3,10],[12,16]] [OK]

Hint: Merge all overlapping intervals into one continuous interval [OK]

Common Mistakes:

• Not merging all overlapping intervals
• Leaving duplicates of overlapping intervals
• Incorrectly splitting merged intervals