Insert Interval into Sorted List in DSA C++ - 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 a new interval to the list?

Scenario Under Consideration

Analyze the time complexity of the following code snippet.

vector<vector<int>> insertInterval(vector<vector<int>> &intervals, vector<int> &newInterval) {
    vector<vector<int>> result;
    int i = 0, n = intervals.size();
    while (i < n && intervals[i][1] < newInterval[0]) {
        result.push_back(intervals[i++]);
    }
    while (i < n && intervals[i][0] <= newInterval[1]) {
        newInterval[0] = min(newInterval[0], intervals[i][0]);
        newInterval[1] = max(newInterval[1], intervals[i][1]);
        i++;
    }
    result.push_back(newInterval);
    while (i < n) {
        result.push_back(intervals[i++]);
    }
    return result;
}
    

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

Identify Repeating Operations

Identify the loops, recursion, array traversals that repeat.

• Primary operation: Three while loops that traverse the intervals list.
• How many times: Each loop runs over parts of the list without overlap, together covering all intervals once.

How Execution Grows With Input

The code goes through the list of intervals once, splitting the work into three parts but never revisiting intervals.

Input Size (n)	Approx. Operations
10	About 10 steps to check and insert intervals
100	About 100 steps, each interval checked once
1000	About 1000 steps, linear growth

Pattern observation: The number of steps grows directly with the number of intervals; doubling intervals roughly doubles the work.

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 there are three loops, the time complexity is O(3n) or O(n²)."

[OK] Correct: The loops run one after another, each covering different parts of the list without overlap, so total work is still proportional to n, not n squared.

Interview Connect

Understanding how to analyze linear scans like this is a key skill. It shows you can break down code into parts and see how they add up to total work.

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 merge overlapping intervals and keep the list sorted

B. To add the new interval at the end without merging

C. To remove all intervals that come before the new interval

D. To sort the intervals again after insertion without merging

Solution

1. Step 1: Understand the problem context

   We have a sorted list of intervals that do not overlap. When inserting a new interval, we must keep the list sorted and merge any overlapping intervals.

2. Step 2: Identify the correct approach

   Simply adding the new interval without merging can cause overlaps. Sorting again without merging breaks the non-overlapping property. Removing intervals before the new one is incorrect.

3. Final Answer:

   To merge overlapping intervals and keep the list sorted -> Option A

4. Quick Check:

   Insert Interval = Merge and keep sorted [OK]

Hint: Always merge overlapping intervals after insertion [OK]

Common Mistakes:

• Inserting without merging overlaps
• Sorting intervals again without merging
• Removing intervals unnecessarily

2. Which of the following C++ code snippets correctly declares a vector of intervals where each interval is a pair of integers?

easy

A. std::vector intervals;

B. std::vector intervals;

C. std::vector intervals;

D. std::vector> intervals;

Solution

1. Step 1: Understand vector and pair syntax

   In C++, a vector of pairs is declared as std::vector>. The pair must specify both types inside angle brackets.

2. Step 2: Check each option

   std::vector> intervals; correctly uses std::pair. std::vector intervals; is invalid syntax for vector. std::vector intervals; misses the second type in pair. std::vector intervals; misses template parameters for pair.

3. Final Answer:

   std::vector> intervals; -> Option D

4. Quick Check:

   Vector of pairs syntax = std::vector> intervals; [OK]

Hint: Use std::vector> for intervals [OK]

Common Mistakes:

• Missing template parameters in pair
• Incorrect vector declaration syntax
• Using incomplete pair types

3. Given the sorted intervals [(1, 3), (6, 9)] and a new interval (2, 5), what will be the resulting list after insertion and merging?

medium

A. [(1, 3), (2, 5), (6, 9)]

B. [(1, 5), (6, 9)]

C. [(2, 5), (6, 9)]

D. [(1, 9)]

Solution

1. Step 1: Identify overlapping intervals

   The new interval (2, 5) overlaps with (1, 3) because 2 <= 3. They should be merged into (1, 5).

2. Step 2: Merge and keep other intervals

   After merging (1, 3) and (2, 5) into (1, 5), the interval (6, 9) remains unchanged.

3. Final Answer:

   [(1, 5), (6, 9)] -> Option B

4. Quick Check:

   Merge overlapping intervals = [(1, 5), (6, 9)] [OK]

Hint: Merge intervals if new start ≤ existing end [OK]

Common Mistakes:

• Not merging overlapping intervals
• Inserting without sorting
• Merging all intervals incorrectly

4. Consider the following C++ code snippet for inserting an interval into a sorted list of intervals:

std::vector> insertInterval(std::vector>& intervals, std::pair newInterval) {
    std::vector> result;
    int i = 0;
    while (i < intervals.size() && intervals[i].second < newInterval.first) {
        result.push_back(intervals[i]);
        i++;
    }
    while (i < intervals.size() && intervals[i].first <= newInterval.second) {
        newInterval.first = std::min(newInterval.first, intervals[i].first);
        newInterval.second = std::max(newInterval.second, intervals[i].second);
        i++;
    }
    result.push_back(newInterval);
    while (i < intervals.size()) {
        result.push_back(intervals[i]);
        i++;
    }
    return result;
}

What is the bug in this code if any?

medium

A. No bug; the code correctly inserts and merges intervals

B. The first while loop condition should use <= instead of <

C. The second while loop should check intervals[i].second instead of intervals[i].first

D. The newInterval is not pushed back at the correct position

Solution

1. Step 1: Analyze the first while loop

   The first loop adds all intervals that end before the new interval starts, correctly skipping non-overlapping intervals before newInterval.

2. Step 2: Analyze the second while loop

   The second loop merges all intervals overlapping with newInterval by updating its start and end correctly.

3. Step 3: Analyze the final steps

   After merging, newInterval is added, then remaining intervals are appended. This maintains sorted order and merges properly.

4. Final Answer:

   No bug; the code correctly inserts and merges intervals -> Option A

5. Quick Check:

   Code logic matches insert and merge steps [OK]

Hint: Check loop conditions carefully for overlap and order [OK]

Common Mistakes:

• Changing loop conditions incorrectly
• Pushing newInterval too early or late
• Merging intervals incorrectly

5. You have a sorted list of intervals: [(1, 2), (3, 5), (6, 7), (8, 10), (12, 16)]. You want to insert a new interval (4, 8). What is the final list after insertion and merging?

hard

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

B. [(1, 2), (4, 10), (12, 16)]

C. [(1, 2), (3, 10), (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) because their ranges intersect.

2. Step 2: Merge all overlapping intervals

   Merge (3, 5), (6, 7), (8, 10), and (4, 8) into one interval. The merged interval starts at min(3,4,6,8) = 3 and ends at max(5,7,8,10) = 10.

3. Step 3: Keep non-overlapping intervals

   Intervals (1, 2) and (12, 16) remain unchanged.

4. Final Answer:

   [(1, 2), (3, 10), (12, 16)] -> Option C

5. Quick Check:

   Merge overlapping intervals = [(1, 2), (3, 10), (12, 16)] [OK]

Hint: Merge all intervals overlapping new interval into one [OK]

Common Mistakes:

• Not merging all overlapping intervals
• Inserting new interval without merging
• Incorrect interval boundaries after merge