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

Time Complexity: Insert Interval into Sorted List

O(n)

Understanding Time Complexity

We want to understand how long it takes to add a new time interval into a list of sorted intervals.

Specifically, how the time grows as the list gets bigger.

Scenario Under Consideration

Analyze the time complexity of the following code snippet.

fn insert_interval(intervals: &mut Vec<(i32, i32)>, new_interval: (i32, i32)) {
    let mut result = Vec::new();
    let mut i = 0;
    let n = intervals.len();

    while i < n && intervals[i].1 < new_interval.0 {
        result.push(intervals[i]);
        i += 1;
    }

    let mut merged = new_interval;
    while i < n && intervals[i].0 <= merged.1 {
        merged.0 = std::cmp::min(merged.0, intervals[i].0);
        merged.1 = std::cmp::max(merged.1, intervals[i].1);
        i += 1;
    }
    result.push(merged);

    while i < n {
        result.push(intervals[i]);
        i += 1;
    }

    *intervals = result;
}
    

This code inserts a new interval into a sorted list of intervals, merging any 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 runs over a portion of the list without overlap, together covering all intervals once.

How Execution Grows With Input

As the number of intervals grows, the code checks each interval once in order.

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 operations grow roughly in direct proportion to the number of intervals.

Final Time Complexity

Time Complexity: O(n)

This means the time to insert and merge intervals 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(n^2)."

[OK] Correct: The loops run one after another without overlapping intervals, so together they only pass through the list once, keeping the time linear, not quadratic.

Interview Connect

Understanding this linear time complexity helps you explain how to efficiently handle sorted data and merging tasks, a common real-world problem.

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 intervals?

easy

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

B. To replace all intervals with the new interval

C. To sort the intervals in descending order

D. To add the new interval and merge any overlapping intervals

Solution

1. Step 1: Understand the problem context

   We have a sorted list of intervals and want to insert a new interval without breaking the order.

2. Step 2: Identify the required operation

   We must insert the new interval and merge any intervals that overlap with it to keep the list sorted and non-overlapping.

3. Final Answer:

   To add the new interval and merge any overlapping intervals -> Option D

4. Quick Check:

   Insert and merge intervals = To add the new interval and merge any overlapping intervals [OK]

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

Common Mistakes:

• Replacing all intervals instead of merging
• Sorting intervals in descending order
• Removing intervals before the new one

2. Which Rust syntax correctly declares a vector of intervals where each interval is a tuple of two integers?

easy

A. let intervals: Vec<(i32, i32)> = vec![(1, 3), (6, 9)];

B. let intervals = [(1, 3), (6, 9)];

C. let intervals: Vec<[i32; 2]> = vec![[1, 3], [6, 9]];

D. let intervals: Vec<(int, int)> = vec![(1, 3), (6, 9)];

Solution

1. Step 1: Identify correct Rust tuple and vector syntax

   Rust uses (i32, i32) for tuple of two integers and Vec for vector type.

2. Step 2: Check each option

   let intervals: Vec<(i32, i32)> = vec![(1, 3), (6, 9)]; uses Vec<(i32, i32)> with vec! macro and tuples correctly. let intervals = [(1, 3), (6, 9)]; is an array, not a vector. let intervals: Vec<[i32; 2]> = vec![[1, 3], [6, 9]]; uses arrays inside vector, which is valid but different type. let intervals: Vec<(int, int)> = vec![(1, 3), (6, 9)]; uses 'int' which is invalid in Rust.

3. Final Answer:

   let intervals: Vec<(i32, i32)> = vec![(1, 3), (6, 9)]; -> Option A

4. Quick Check:

   Correct Rust tuple vector syntax = let intervals: Vec<(i32, i32)> = vec![(1, 3), (6, 9)]; [OK]

Hint: Use Vec<(i32, i32)> and vec! macro for interval list [OK]

Common Mistakes:

• Using 'int' instead of 'i32'
• Confusing arrays and tuples
• Not using vec! macro for vectors

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

medium

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

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

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

D. [(1, 9)]

Solution

1. Step 1: Insert new interval and find overlaps

   The new interval (2, 5) overlaps with (1, 3) because 2 <= 3.

2. Step 2: Merge overlapping intervals

   Merge (1, 3) and (2, 5) into (1, 5). The interval (6, 9) remains unchanged.

3. Final Answer:

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

4. Quick Check:

   Merged 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 this Rust code snippet for inserting an interval into a sorted list:

fn insert_interval(intervals: &mut Vec<(i32, i32)>, new_interval: (i32, i32)) {
    let mut i = 0;
    while i < intervals.len() {
        if intervals[i].1 < new_interval.0 {
            i += 1;
        } else if intervals[i].0 > new_interval.1 {
            intervals.insert(i, new_interval);
            return;
        } else {
            intervals[i].0 = intervals[i].0.min(new_interval.0);
            intervals[i].1 = intervals[i].1.max(new_interval.1);
            intervals.remove(i + 1);
        }
    }
    intervals.push(new_interval);
}

What is the main bug in this code?

medium

A. It never increments index i inside the else block causing infinite loop

B. It inserts the new interval at the wrong position

C. It removes the wrong interval when merging overlapping intervals

D. It does not handle empty intervals vector

Solution

1. Step 1: Examine the else branch for merging

   The code updates intervals[i] by merging the new interval but never increments the index i inside the else block.

2. Step 2: Identify infinite loop cause

   Because i is not incremented after merging and removing intervals[i+1], the loop can get stuck processing the same index repeatedly, causing an infinite loop.

3. Final Answer:

   It never increments index i inside the else block causing infinite loop -> Option A

4. Quick Check:

   Missing i increment in else block = infinite loop [OK]

Hint: Always increment index after merging to avoid infinite loop [OK]

Common Mistakes:

• Forgetting to increment index after removal
• Assuming insert always works without checks
• Ignoring empty vector edge case

5. You have a sorted list of intervals vec![(1, 2), (3, 5), (6, 7), (8, 10), (12, 16)]. Insert the new interval (4, 8) and return the updated list after merging all overlapping intervals.

hard

A. [(1, 2), (3, 16)]

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

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

D. [(1, 2), (3, 5), (6, 8), (8, 10), (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 (3, 10). The intervals (1, 2) and (12, 16) remain unchanged.

3. Final Answer:

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

4. Quick Check:

   Merged intervals after insertion = [(1, 2), (3, 10), (12, 16)] [OK]

Hint: Merge all intervals overlapping new interval's range [OK]

Common Mistakes:

• Merging only partially overlapping intervals
• Not merging intervals that touch at boundaries
• Replacing intervals incorrectly