Insert Interval into Sorted List in DSA Java - Cheat Sheet & Quick Revision

Recall & Review

beginner

What is the main goal when inserting an interval into a sorted list of intervals?

To add the new interval into the correct position in the list and merge any overlapping intervals so the list remains sorted and non-overlapping.

beginner

How do you know if two intervals overlap?

Two intervals overlap if the start of one interval is less than or equal to the end of the other interval, and vice versa.

beginner

What is the first step when inserting a new interval into a sorted list?

Find all intervals that come before the new interval without overlapping and add them to the result as they are.

intermediate

What do you do when you find intervals that overlap with the new interval?

Merge them by updating the start to the minimum start and the end to the maximum end of all overlapping intervals including the new one.

beginner

After merging overlapping intervals, what is the final step?

Add the merged interval to the result list and then add all remaining intervals that come after it without overlap.

What condition indicates two intervals [a, b] and [c, d] overlap?

A b < c

B a > d

C a == d

D b >= c and a <= d

Two intervals overlap if the end of the first is at least the start of the second and the start of the first is at most the end of the second.

When inserting a new interval, what do you do with intervals that end before the new interval starts?

A Add them to the result as they are

B Merge them with the new interval

C Skip them

D Remove them

Intervals that end before the new interval starts do not overlap and should be added to the result unchanged.

What happens if the new interval does not overlap with any existing intervals?

A It is merged with the first interval

B It is ignored

C It is added in the correct sorted position

D It replaces all intervals

If there is no overlap, the new interval is simply inserted in the correct sorted position.

What is the time complexity of inserting an interval into a sorted list of n intervals?

A O(log n)

B O(n)

C O(n^2)

D O(1)

You may need to scan through the list once to find overlapping intervals, so the time complexity is O(n).

Which of these is NOT a step in inserting an interval into a sorted list?

A Sort the entire list again after insertion

B Merge overlapping intervals with the new interval

C Add all intervals before the new interval without overlap

D Add all intervals after the new interval without overlap

The list is already sorted, so you do not need to sort it again after insertion.

Explain step-by-step how to insert a new interval into a sorted list of intervals.

Think about how to keep the list sorted and non-overlapping.

Describe how to detect and merge overlapping intervals when inserting a new interval.

Focus on comparing interval boundaries.

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