Why Insert Interval into Sorted List in DSA Java?

The Big Idea

What if you could add new time slots perfectly without worrying about overlaps or order?

The Scenario

Imagine you have a list of time slots booked for meetings, sorted by start time. Now, you want to add a new meeting without messing up the order or overlapping existing meetings.

The Problem

Manually checking each time slot to find where the new meeting fits is slow and confusing. You might forget to merge overlapping meetings or place the new one in the wrong spot, causing double bookings or disorder.

The Solution

Using the Insert Interval method, you can quickly find the right place for the new meeting, merge any overlaps automatically, and keep the list sorted and clean without extra effort.

Before vs After

✗ Before

for (int i = 0; i < intervals.length; i++) {
  if (newInterval.start < intervals[i].start) {
    // insert here manually
  }
  // check overlaps manually
}

✓ After

List<Interval> result = new ArrayList<>();
for (Interval interval : intervals) {
  if (interval.end < newInterval.start) result.add(interval);
  else if (interval.start > newInterval.end) {
    result.add(newInterval);
    newInterval = interval;
  } else {
    newInterval.start = Math.min(newInterval.start, interval.start);
    newInterval.end = Math.max(newInterval.end, interval.end);
  }
}
result.add(newInterval);

What It Enables

This method lets you manage schedules or ranges efficiently, ensuring no overlaps and keeping everything in order automatically.

Real Life Example

Adding a new appointment to a doctor's daily schedule without causing conflicts or disorder.

Key Takeaways

Manual insertion is error-prone and slow.

Insert Interval merges overlaps and keeps order automatically.

Useful for managing time slots, ranges, or any sorted intervals.

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