Why Insert Interval into Sorted List in DSA Python?

The Big Idea

What if you could add new events to your schedule without worrying about overlaps or order?

The Scenario

Imagine you have a list of booked meeting times sorted by start time. You want to add a new meeting without messing up the order or overlapping times.

The Problem

Manually checking where to put the new meeting and merging overlapping times is slow and easy to mess up. You might forget to merge intervals or place the new one in the wrong spot.

The Solution

Using the Insert Interval method, you can quickly find the right place to add the new meeting and merge any overlaps automatically, keeping the list sorted and clean.

Before vs After

✗ Before

intervals = [[1,3],[6,9]]
new_interval = [2,5]
# Manually check and merge intervals one by one

✓ After

def insert_interval(intervals, new_interval):
    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
    return result

What It Enables

This method lets you keep a sorted list of intervals updated efficiently, perfect for scheduling and timeline management.

Real Life Example

Adding a new appointment to your calendar without overlapping or losing existing bookings.

Key Takeaways

Manually inserting intervals is error-prone and slow.

Insert Interval method merges overlaps and keeps order automatically.

Useful for managing schedules, timelines, and bookings.

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