Insert Interval into Sorted List in DSA Rust - Build from Scratch

Insert Interval into Sorted List

📖 Scenario: You are managing a schedule of booked time slots represented as intervals. Each interval is a pair of start and end times. The schedule is sorted by start times and intervals do not overlap.Now, you want to add a new time slot to this schedule. If the new time slot overlaps with existing ones, you need to merge them to keep the schedule clean and sorted.

🎯 Goal: Build a Rust program that inserts a new interval into a sorted list of non-overlapping intervals and merges any overlapping intervals.

📋 What You'll Learn

Create a vector called intervals with these exact intervals: [1, 3], [6, 9]

Create a tuple called new_interval with the exact interval: [2, 5]

Write a function called insert_interval that takes intervals and new_interval and returns a new vector with the new interval inserted and merged

Print the resulting vector of intervals after insertion and merging

💡 Why This Matters

🌍 Real World

Managing calendar events, booking systems, or time schedules where overlapping time slots need to be merged.

💼 Career

Useful for software developers working on scheduling apps, calendar integrations, or any system that manages time intervals.

1

Create the initial intervals list

Create a vector called intervals with the exact intervals [1, 3] and [6, 9].

DSA Rust

# Create the intervals vector with [1, 3] and [6, 9]
// Your code here

Hint

Use vec![(1, 3), (6, 9)] to create the vector of tuples.

2

Create the new interval to insert

Create a tuple called new_interval with the exact interval [2, 5].

DSA Rust

fn main() {
    let intervals = vec![(1, 3), (6, 9)];
    // Create the new_interval vector with [2, 5]
    // Your code here
}

Hint

Use a tuple (2, 5) to represent the new interval.

3

Write the insert_interval function

Write a function called insert_interval that takes intervals: Vec<(i32, i32)> and new_interval: (i32, i32) and returns a Vec<(i32, i32)> with the new interval inserted and merged. Use while loops and conditions to merge overlapping intervals.

DSA Rust

fn main() {
    let intervals = vec![(1, 3), (6, 9)];
    let new_interval = (2, 5);

    // Write the insert_interval function below
    // Your code here
}

Hint

Use two while loops to add intervals before and merge overlapping intervals, then add remaining intervals.

4

Print the merged intervals

Print the merged vector of intervals using println!("{:?}", merged).

DSA Rust

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

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

    while i < n && intervals[i].0 <= end {
        start = start.min(intervals[i].0);
        end = end.max(intervals[i].1);
        i += 1;
    }

    result.push((start, end));

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

    result
}

fn main() {
    let intervals = vec![(1, 3), (6, 9)];
    let new_interval = (2, 5);
    let merged = insert_interval(intervals, new_interval);

    // Print the merged intervals
    // Your code here
}

Hint

Use println!("{:?}", merged) to print the vector of tuples.

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