Insert Interval into Sorted List in DSA Python - Time & Space Complexity

Time Complexity: Insert Interval into Sorted List

O(n)

Understanding Time Complexity

We want to understand how the time needed to insert a new interval into a sorted list of intervals changes as the list grows.

Specifically, how does the number of steps grow when we add one interval to many sorted intervals?

Scenario Under Consideration

Analyze the time complexity of the following code snippet.

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

This code inserts a new interval into a sorted list of intervals, merging overlapping intervals.

Identify Repeating Operations

Identify the loops, recursion, array traversals that repeat.

• Primary operation: Three while loops that each traverse parts of the intervals list.
• How many times: Each loop moves forward through the list without going back, together covering all intervals once.

How Execution Grows With Input

As the number of intervals grows, the code checks each interval at most once to find where to insert and merge.

Input Size (n)	Approx. Operations
10	About 10 checks and merges
100	About 100 checks and merges
1000	About 1000 checks and merges

Pattern observation: The number of operations grows roughly in direct proportion to the number of intervals.

Final Time Complexity

Time Complexity: O(n)

This means the time to insert grows linearly with the number of intervals in the list.

Common Mistake

[X] Wrong: "Since the list is sorted, we can insert the interval in constant time without checking all intervals."

[OK] Correct: Even though the list is sorted, we must check intervals to find overlaps and merge them, which requires looking at many intervals in the worst case.

Interview Connect

Understanding this linear time complexity helps you explain how to efficiently handle interval merging, a common problem in coding interviews and real-world scheduling tasks.

Self-Check

"What if the intervals list was not sorted? How would the time complexity change?"

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