Why Insert Interval into Sorted List in DSA Rust?

The Big Idea

What if you could add new time slots perfectly 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 confusing. You might forget to merge intervals or place the new one in the wrong spot, causing scheduling errors.

The Solution

Using the "Insert Interval into Sorted List" method, you can quickly find the right place for the new meeting and merge any overlaps automatically. This keeps the list sorted and clean without mistakes.

Before vs After

✗ Before

let mut intervals = vec![(1,3), (6,9)];
// Manually find place and merge
intervals.push((2,5));
// Need to sort and merge manually

✓ After

let mut intervals = vec![(1,3), (6,9)];
let new_interval = (2,5);
insert_interval(&mut intervals, new_interval);

What It Enables

This lets you manage schedules or ranges efficiently, ensuring no overlaps and always keeping the list sorted.

Real Life Example

Adding a new booked time to a calendar app without breaking existing appointments or causing conflicts.

Key Takeaways

Manual insertion and merging is error-prone and slow.

Insert Interval method automates placement and merging.

Keeps intervals sorted and non-overlapping easily.

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