Insert Interval into Sorted List in DSA Java - Interactive Practice

Practice - 5 Tasks

1fill in blank

easy

Complete the code to declare the method that inserts an interval into a sorted list.

DSA Java

public List<Interval> [1](List<Interval> intervals, Interval newInterval) { }

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AinsertSorted

BaddInterval

CinsertInterval

DaddSorted

Attempts:

3 left

💡 Hint

Common Mistakes

Using a method name that does not clearly indicate insertion.

Choosing a method name that implies adding but not necessarily inserting in order.

The method name is 'insertInterval' as per the problem statement.

2fill in blank

medium

Complete the code to initialize the result list that will hold merged intervals.

DSA Java

List<Interval> result = new [1]<>();

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AArrayList

BLinkedList

CHashSet

DTreeSet

Attempts:

3 left

💡 Hint

Common Mistakes

Using a Set which does not maintain order or allows duplicates.

Using LinkedList unnecessarily when ArrayList suffices.

ArrayList is the appropriate list implementation to hold intervals in order.

3fill in blank

hard

Fix the error in the condition to check if the current interval ends before the new interval starts.

DSA Java

if (intervals.get(i).end [1] newInterval.start) { result.add(intervals.get(i)); i++; }

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A>=

B<

C<=

D>

Attempts:

3 left

💡 Hint

Common Mistakes

Using '>=' which includes touching intervals as non-overlapping incorrectly.

Using '>' which misses intervals that end exactly at newInterval.start.

The condition should check if the current interval ends before the new interval starts, so '<' is correct.

4fill in blank

hard

Fill both blanks to merge overlapping intervals correctly.

DSA Java

newInterval.start = Math.min(newInterval.start, intervals.get(i).[1]);
newInterval.end = Math.max(newInterval.end, intervals.get(i).[2]);

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Astart

Bend

Clength

Dvalue

Attempts:

3 left

💡 Hint

Common Mistakes

Using incorrect fields like length or value which do not exist.

Swapping start and end in min/max functions.

To merge intervals, update newInterval start to min of starts and end to max of ends.

5fill in blank

hard

Fill all three blanks to add the merged interval and remaining intervals to the result list.

DSA Java

result.[1](newInterval);
while (i [2] intervals.size()) {
    result.[3](intervals.get(i));
    i++;
}

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Aadd

B<

Dremove

Attempts:

3 left

💡 Hint

Common Mistakes

Using remove instead of add which deletes elements.

Using '>=' instead of '<' causing loop to skip or error.

Use add to insert intervals and '<' to loop while i is less than intervals size.

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