Insert Interval into Sorted List in DSA Python - Interactive Practice

Practice - 5 Tasks

1fill in blank

easy

Complete the code to insert the new interval into the list.

DSA Python

def insert_interval(intervals, new_interval):
    result = []
    i = 0
    while i < len(intervals) and intervals[i][1] < [1][0]:
        result.append(intervals[i])
        i += 1
    return result

Drag options to blanks, or click blank then click option'

Anew_interval

Bintervals

Cresult

Di

Attempts:

3 left

💡 Hint

Common Mistakes

Comparing with intervals instead of new_interval

Using wrong index in comparison

We compare the end of intervals[i] with the start of new_interval to find where to insert.

2fill in blank

medium

Complete the code to merge overlapping intervals with the new interval.

DSA Python

while i < len(intervals) and intervals[i][0] <= [1][1]:
    new_interval[0] = min(new_interval[0], intervals[i][0])
    new_interval[1] = max(new_interval[1], intervals[i][1])
    i += 1

Drag options to blanks, or click blank then click option'

Aintervals[i]

Bnew_interval

Cresult

Di

Attempts:

3 left

💡 Hint

Common Mistakes

Using wrong variable for comparison

Not updating new_interval boundaries correctly

We merge intervals overlapping with new_interval by updating new_interval boundaries.

3fill in blank

hard

Fix the error in adding the merged new interval to the result list.

DSA Python

result.append([1])

Drag options to blanks, or click blank then click option'

Anew_interval

Bintervals[i]

Cintervals

Dresult

Attempts:

3 left

💡 Hint

Common Mistakes

Appending intervals[i] which may be out of range

Appending the whole intervals list

After merging, we add the updated new_interval to the result list.

4fill in blank

hard

Fill both blanks to add remaining intervals after the new interval.

DSA Python

while [1] < len(intervals):
    result.[2](intervals[i])
    i += 1

Drag options to blanks, or click blank then click option'

Ai

Bappend

Cintervals

Dinsert

Attempts:

3 left

💡 Hint

Common Mistakes

Using wrong variable for index

Using insert instead of append

We continue from index i and append remaining intervals to result.

5fill in blank

hard

Fill all three blanks to complete the insert_interval function.

DSA Python

def insert_interval(intervals, new_interval):
    result = []
    i = 0
    while i < len(intervals) and intervals[i][1] < [1][0]:
        result.append(intervals[i])
        i += 1
    while i < len(intervals) and intervals[i][0] <= [2][1]:
        [3][0] = min([3][0], intervals[i][0])
        new_interval[1] = max(new_interval[1], intervals[i][1])
        i += 1
    result.append(new_interval)
    while i < len(intervals):
        result.append(intervals[i])
        i += 1
    return result

Drag options to blanks, or click blank then click option'

Anew_interval

Bintervals

Dresult

Attempts:

3 left

💡 Hint

Common Mistakes

Using intervals instead of new_interval in comparisons

Not updating new_interval boundaries correctly

We use new_interval to compare and update boundaries during merging.

Practice

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

easy

A. To add the new interval and merge any overlapping intervals to keep the list sorted and non-overlapping

B. To replace the first interval with the new interval regardless of overlap

C. To insert the new interval at the end without checking for overlaps

D. To delete all intervals that come after the new interval

Solution

1. Step 1: Understand the problem requirement

   The goal is to insert a new interval into a sorted list of intervals without overlaps.

2. Step 2: Merge overlapping intervals after insertion

   After inserting, merge any intervals that overlap to maintain the sorted, non-overlapping property.

3. Final Answer:

   To add the new interval and merge any overlapping intervals to keep the list sorted and non-overlapping -> Option A

4. Quick Check:

   Insert and merge intervals = To add the new interval and merge any overlapping intervals to keep the list sorted and non-overlapping [OK]

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

Common Mistakes:

• Ignoring merging after insertion
• Inserting without maintaining order
• Replacing intervals incorrectly

2. Which of the following Python code snippets correctly inserts a new interval into a sorted list of intervals without merging?

easy

A. intervals.remove(new_interval)

B. intervals.append(new_interval)

C. intervals.sort()

D. intervals.insert(2, new_interval)

Solution

1. Step 1: Identify correct insertion method

   Using insert(index, element) places the new interval at the correct position without losing order.

2. Step 2: Understand other options

   append adds at end without order, sort only sorts but doesn't insert, remove deletes an element.

3. Final Answer:

   intervals.insert(2, new_interval) -> Option D

4. Quick Check:

   Use insert() to add at specific position [OK]

Hint: Use insert(index, value) to place interval correctly [OK]

Common Mistakes:

• Using append without sorting
• Using remove instead of insert
• Relying on sort without insertion

3. What is the output of the following code?

intervals = [[1,3],[6,9]]
new_interval = [2,5]

# Insert new_interval and merge
result = []
i = 0
while i < len(intervals) and intervals[i][1] < new_interval[0]:
    result.append(intervals[i])
    i += 1
while i < len(intervals) and intervals[i][0] <= new_interval[1]:
    new_interval[0] = min(new_interval[0], intervals[i][0])
    new_interval[1] = max(new_interval[1], intervals[i][1])
    i += 1
result.append(new_interval)
while i < len(intervals):
    result.append(intervals[i])
    i += 1
print(result)

medium

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

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

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

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

Solution

1. Step 1: Add intervals before new_interval

   The first while loop checks if intervals[i][1] < new_interval[0]. For i=0, 3 < 2 is false, so no intervals added before. result = [].

2. Step 2: Merge overlapping intervals

   Intervals overlapping with new_interval [2,5] are [1,3] and [2,5] merged to [1,5].

3. Step 3: Add remaining intervals

   Remaining interval [6,9] is added after merged interval.

4. Final Answer:

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

5. Quick Check:

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

Hint: Merge intervals overlapping new_interval boundaries [OK]

Common Mistakes:

• Not merging overlapping intervals
• Appending new_interval without merging
• Incorrect loop conditions

4. The following code tries to insert and merge a new interval but produces incorrect output. What is the bug?

def insert_interval(intervals, new_interval):
    result = []
    for interval in intervals:
        if interval[1] < new_interval[0]:
            result.append(interval)
        elif interval[0] > new_interval[1]:
            result.append(new_interval)
        else:
            new_interval[0] = min(new_interval[0], interval[0])
            new_interval[1] = max(new_interval[1], interval[1])
    result.append(new_interval)
    return result

print(insert_interval([[1,3],[6,9]], [2,5]))

medium

A. Appending new_interval inside loop causes duplicates

B. Not updating new_interval correctly during merge

C. Missing return statement

D. Incorrect comparison operators in conditions

Solution

1. Step 1: Analyze loop behavior

   Appending new_interval inside the loop when interval[0] > new_interval[1] causes multiple appends.

2. Step 2: Understand consequences

   This leads to duplicates and incorrect order in the result list.

3. Final Answer:

   Appending new_interval inside loop causes duplicates -> Option A

4. Quick Check:

   Append new_interval once after loop [OK]

Hint: Append merged interval only once after loop ends [OK]

Common Mistakes:

• Appending inside loop multiple times
• Not merging intervals properly
• Confusing interval boundaries

5. Given a sorted list of intervals [[1,2],[3,5],[6,7],[8,10],[12,16]] and a new interval [4,8], what is the final list after inserting and merging?

hard

A. [[1,2],[3,5],[6,7],[8,10],[4,8],[12,16]]

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

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

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

Solution

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

   Intervals [3,5], [6,7], and [8,10] overlap with [4,8].

2. Step 2: Merge overlapping intervals

   Merge [3,5], [6,7], [8,10], and [4,8] into [3,10].

3. Step 3: Combine with non-overlapping intervals

   Keep [1,2] and [12,16] as is, final list is [[1,2],[3,10],[12,16]].

4. Final Answer:

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

5. Quick Check:

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

Hint: Merge all intervals overlapping new interval boundaries [OK]

Common Mistakes:

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