Insert Interval into Sorted List in DSA Python - Build from Scratch

Insert Interval into Sorted List

📖 Scenario: Imagine you have a calendar with booked time slots represented as intervals. You want to add a new meeting time and keep the calendar sorted and merged so no overlapping meetings exist.

🎯 Goal: Build a program that inserts a new interval into a sorted list of non-overlapping intervals and merges any overlapping intervals.

📋 What You'll Learn

Create a list called intervals with sorted, non-overlapping intervals as lists of two integers.

Create a list called new_interval representing the interval to insert.

Write code to insert new_interval into intervals and merge overlapping intervals.

Print the final list of merged intervals.

💡 Why This Matters

🌍 Real World

Calendars, booking systems, and timeline management often require inserting and merging time intervals to avoid conflicts.

💼 Career

Understanding interval merging is useful for software engineers working on scheduling apps, event planners, and resource allocation tools.

1

Create the initial sorted list of intervals

Create a list called intervals with these exact intervals: [1, 3], [6, 9].

DSA Python

# Your code here

Hint

Use a list of lists to represent intervals.

2

Create the new interval to insert

Create a list called new_interval with the exact values [2, 5].

DSA Python

intervals = [[1, 3], [6, 9]]
# Your code here

Hint

Just assign the list [2, 5] to new_interval.

3

Insert and merge the new interval

Write code to insert new_interval into intervals and merge overlapping intervals. Use a list called merged to store the result. Iterate over intervals + [new_interval] sorted by start time. Merge intervals if they overlap.

DSA Python

intervals = [[1, 3], [6, 9]]
new_interval = [2, 5]
# Your code here

Hint

Sort all intervals including the new one, then merge overlapping ones by comparing the end of the last merged interval with the start of the current.

4

Print the merged intervals

Print the list merged to show the final merged intervals after insertion.

DSA Python

intervals = [[1, 3], [6, 9]]
new_interval = [2, 5]
merged = []
for interval in sorted(intervals + [new_interval], key=lambda x: x[0]):
    if not merged or merged[-1][1] < interval[0]:
        merged.append(interval)
    else:
        merged[-1][1] = max(merged[-1][1], interval[1])
# Your code here

Hint

Use print(merged) to display the final intervals.

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