Insert Interval into Sorted List in DSA Rust - Practice Problems & Challenges

Challenge - 5 Problems

❓ Predict Output

intermediate

Output of inserting an interval that merges with one existing interval

What is the output list of intervals after inserting [4, 8] into the sorted list [[1, 3], [6, 9]]?

DSA Rust

fn insert_interval(intervals: Vec<[i32; 2]>, new_interval: [i32; 2]) -> Vec<[i32; 2]> {
    let mut result = Vec::new();
    let mut i = 0;
    let n = intervals.len();
    let mut new_start = new_interval[0];
    let mut new_end = new_interval[1];

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

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

    result.push([new_start, new_end]);

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

    result
}

fn main() {
    let intervals = vec![[1, 3], [6, 9]];
    let new_interval = [4, 8];
    let result = insert_interval(intervals, new_interval);
    print!("{:?}", result);
}

A [[1, 3], [4, 9]]

B [[1, 3], [4, 8], [6, 9]]

C [[4, 8], [6, 9]]

D [[1, 9]]

💡 Hint

Think about how overlapping intervals merge into one.

The new interval [4, 8] overlaps with [6, 9], so they merge into [4, 9]. The first interval [1, 3] does not overlap and stays as is.

❓ Predict Output

intermediate

Output after inserting an interval that does not overlap

What is the output list of intervals after inserting [10, 11] into the sorted list [[1, 2], [3, 5], [6, 7]]?

DSA Rust

fn insert_interval(intervals: Vec<[i32; 2]>, new_interval: [i32; 2]) -> Vec<[i32; 2]> {
    let mut result = Vec::new();
    let mut i = 0;
    let n = intervals.len();
    let mut new_start = new_interval[0];
    let mut new_end = new_interval[1];

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

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

    result.push([new_start, new_end]);

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

    result
}

fn main() {
    let intervals = vec![[1, 2], [3, 5], [6, 7]];
    let new_interval = [10, 11];
    let result = insert_interval(intervals, new_interval);
    print!("{:?}", result);
}

A [[10, 11], [1, 2], [3, 5], [6, 7]]

B [[1, 2], [3, 5], [6, 7]]

C [[1, 2], [3, 5], [6, 7], [10, 11]]

D [[1, 2], [3, 5], [6, 7], [10, 10]]

💡 Hint

The new interval does not overlap and should be added at the end.

Since [10, 11] does not overlap with any existing intervals, it is simply appended at the end.

🔧 Debug

advanced

Identify the error in interval insertion code

What error does the following Rust code produce when inserting [2, 5] into [[1, 3], [6, 9]]? fn insert_interval(intervals: Vec<[i32; 2]>, new_interval: [i32; 2]) -> Vec<[i32; 2]> { let mut result = Vec::new(); let mut i = 0; let n = intervals.len(); let mut new_start = new_interval[0]; let mut new_end = new_interval[1]; while i < n && intervals[i][1] <= new_start { result.push(intervals[i]); i += 1; } while i < n && intervals[i][0] < new_end { new_start = new_start.min(intervals[i][0]); new_end = new_end.max(intervals[i][1]); i += 1; } result.push([new_start, new_end]); while i < n { result.push(intervals[i]); i += 1; } result }

A The code causes a runtime panic due to index out of bounds.

B The code produces incorrect merging because it uses '<=' and '<' instead of '<' and '<=' in the while conditions.

C The code produces a syntax error because of missing semicolons.

D The code runs correctly and produces [[1, 5], [6, 9]].

💡 Hint

Check the comparison operators in the while loops carefully.

The conditions in the while loops are swapped. Using '<=' and '<' causes incorrect merging behavior.

🧠 Conceptual

advanced

Number of intervals after inserting a fully overlapping interval

Given a sorted list of intervals [[1, 2], [3, 5], [6, 7], [8, 10], [12, 16]], what is the number of intervals after inserting [4, 9]?

A 3

B 5

C 4

D 6

💡 Hint

The new interval overlaps with multiple existing intervals and merges them.

The new interval [4, 9] overlaps with [3, 5], [6, 7], and [8, 10], merging them into one interval. The final list has 3 intervals.

🚀 Application

expert

Final intervals after multiple insertions

Starting with an empty list, insert intervals in this order: [5, 7], [1, 3], [2, 6], [10, 12]. What is the final list of intervals?

A [[1, 7], [2, 6], [10, 12]]

B [[1, 3], [2, 6], [5, 7], [10, 12]]

C [[1, 6], [5, 7], [10, 12]]

D [[1, 7], [10, 12]]

💡 Hint

Merge intervals as you insert if they overlap.

Inserting [5,7] then [1,3] gives [[1,3],[5,7]]. Inserting [2,6] merges [1,3],[2,6],[5,7] into [1,7]. Finally, [10,12] is separate.

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