Why Intervals Are a Common Problem Pattern in DSA Rust - Test Your Knowledge

Practice - 5 Tasks

1fill in blank

easy

Complete the code to create a vector of intervals represented as tuples.

DSA Rust

let intervals = vec![(1, 3), (6, 9), (2, [1])];

Drag options to blanks, or click blank then click option'

A4

B5

C7

D8

Attempts:

3 left

💡 Hint

Common Mistakes

Choosing an end value less than or equal to the start value.

Selecting a number that does not overlap with existing intervals.

The interval (2, 4) fits the common example of overlapping intervals.

2fill in blank

medium

Complete the code to sort intervals by their start time.

DSA Rust

intervals.sort_by_key(|interval| interval.[1]);

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A1

B0

Cstart

Dend

Attempts:

3 left

💡 Hint

Common Mistakes

Using 1 instead of 0, which refers to the end time.

Using string keys which are invalid for tuple indexing.

Tuples in Rust are indexed starting at 0, so interval.0 is the start time.

3fill in blank

hard

Fix the error in the code to merge overlapping intervals.

DSA Rust

if intervals[i].1 >= intervals[[1]].0 {
    merged.push((intervals[[1]].0, intervals[i].1));
}

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Ai - 1

Bi + 1

Ci

D0

Attempts:

3 left

💡 Hint

Common Mistakes

Using i + 1 which may cause out-of-bound errors.

Using i which compares the same interval.

We compare the current interval's end with the previous interval's start, so index i-1 is needed.

4fill in blank

hard

Fill both blanks to filter intervals that start after 5 and end before 10.

DSA Rust

let filtered: Vec<_> = intervals.iter().filter(|&&([1], [2])| [1] > 5 && [2] < 10).collect();

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Astart

Bend

Cs

De

Attempts:

3 left

💡 Hint

Common Mistakes

Using invalid variable names that don't match the tuple structure.

Not destructuring the tuple properly.

The closure destructures the tuple into variables s and e representing start and end.

5fill in blank

hard

Fill all three blanks to create a map from interval start to interval length for intervals longer than 2.

DSA Rust

let interval_map: std::collections::HashMap<_, _> = intervals.iter()
    .filter(|&&(start, end)| end - start [1] 2)
    .map(|&(start, end)| (start, end [2] start))
    .collect::<std::collections::HashMap<_, _>>();

Drag options to blanks, or click blank then click option'

A>

B>=

C-

D+

Attempts:

3 left

💡 Hint

Common Mistakes

Using >= instead of > changes the filter condition.

Using + instead of - for length calculation.

We filter intervals longer than 2 (end - start > 2) and map start to length (end - start).

Practice

1. Why are interval problems common in data structures and algorithms?

easy

A. Because intervals are easy to sort alphabetically

B. Because many real-world problems involve ranges or periods of time

C. Because intervals always have fixed lengths

D. Because intervals never overlap in practice

Solution

1. Step 1: Understand what intervals represent

   Intervals represent ranges, such as time periods or numeric ranges, which appear often in scheduling, booking, or resource allocation problems.

2. Step 2: Recognize real-world relevance

   Many practical problems require handling overlapping or merging intervals, making them a common pattern in algorithms.

3. Final Answer:

   Because many real-world problems involve ranges or periods of time -> Option B

4. Quick Check:

   Intervals = ranges in real life [OK]

Hint: Think about real-life ranges like meetings or bookings [OK]

Common Mistakes:

• Assuming intervals always have fixed lengths
• Thinking intervals never overlap
• Confusing intervals with single points

2. Which Rust syntax correctly defines a struct to represent an interval with start and end as integers?

easy

A. struct Interval { start: i32, end: i32 }

B. struct Interval(start: i32, end: i32);

C. struct Interval { start i32; end i32 }

D. struct Interval = { start: i32, end: i32 }

Solution

1. Step 1: Recall Rust struct syntax

   Rust structs use curly braces with field names and types separated by colons and commas.

2. Step 2: Check each option

   struct Interval { start: i32, end: i32 } matches correct syntax: fields with colons and commas inside braces. Options A, C, and D have syntax errors.

3. Final Answer:

   struct Interval { start: i32, end: i32 } -> Option A

4. Quick Check:

   Rust struct = braces + colons + commas [OK]

Hint: Remember Rust structs use braces and colons for fields [OK]

Common Mistakes:

• Using parentheses instead of braces
• Missing colons between field names and types
• Using equal sign instead of braces

3. What is the output of this Rust code that merges overlapping intervals?

fn merge_intervals(mut intervals: Vec<(i32, i32)>) -> Vec<(i32, i32)> {
    intervals.sort_by_key(|k| k.0);
    let mut merged = Vec::new();
    for interval in intervals {
        if let Some(last) = merged.last_mut() {
            if interval.0 <= last.1 {
                last.1 = last.1.max(interval.1);
            } else {
                merged.push(interval);
            }
        } else {
            merged.push(interval);
        }
    }
    merged
}

fn main() {
    let intervals = vec![(1, 3), (2, 6), (8, 10), (15, 18)];
    let result = merge_intervals(intervals);
    println!("{:?}", result);
}

medium

A. [(1, 6), (2, 6), (8, 10), (15, 18)]

B. [(1, 3), (2, 6), (8, 10), (15, 18)]

C. [(1, 3), (8, 10), (15, 18)]

D. [(1, 6), (8, 10), (15, 18)]

Solution

1. Step 1: Sort intervals by start

   Sorted intervals: (1,3), (2,6), (8,10), (15,18)

2. Step 2: Merge overlapping intervals

   First two intervals overlap (2 ≤ 3), merge to (1,6). Others do not overlap, so final merged list is [(1,6), (8,10), (15,18)].

3. Final Answer:

   [(1, 6), (8, 10), (15, 18)] -> Option D

4. Quick Check:

   Merged intervals = [(1,6), (8,10), (15,18)] [OK]

Hint: Sort intervals by start, then merge overlapping ones [OK]

Common Mistakes:

• Not sorting intervals before merging
• Pushing intervals without merging overlaps
• Merging intervals incorrectly

4. Identify the bug in this Rust function that attempts to merge intervals:

fn merge(mut intervals: Vec<(i32, i32)>) -> Vec<(i32, i32)> {
    intervals.sort_by_key(|k| k.1);
    let mut merged = Vec::new();
    for interval in intervals {
        if let Some(last) = merged.last_mut() {
            if interval.0 <= last.1 {
                last.1 = last.1.max(interval.1);
            } else {
                merged.push(interval);
            }
        } else {
            merged.push(interval);
        }
    }
    merged
}

medium

A. Missing return statement at the end

B. Using max instead of min when merging intervals

C. Sorting intervals by end instead of start causes incorrect merging

D. Not initializing merged vector before loop

Solution

1. Step 1: Analyze sorting key

   The function sorts intervals by their end value, but merging requires sorting by start to process intervals in order.

2. Step 2: Understand merging logic impact

   Sorting by end breaks the assumption that intervals are processed from earliest start, causing incorrect merges.

3. Final Answer:

   Sorting intervals by end instead of start causes incorrect merging -> Option C

4. Quick Check:

   Sort by start, not end, for correct merging [OK]

Hint: Sort intervals by start, not end, before merging [OK]

Common Mistakes:

• Sorting by wrong interval boundary
• Confusing max and min when merging
• Forgetting to push first interval

5. Given a list of intervals representing booked meeting times, how would you find the minimum number of meeting rooms required to hold all meetings without overlap? Choose the best approach.

hard

A. Sort intervals by start time, use a min-heap to track earliest end time, allocate rooms accordingly

B. Sort intervals by end time, merge all overlapping intervals into one

C. Count total intervals and subtract overlapping pairs

D. Sort intervals by length and assign rooms based on shortest first

Solution

1. Step 1: Understand the problem

   We need to find how many rooms are needed so no meetings overlap in the same room.

2. Step 2: Use sorting and min-heap

   Sort intervals by start time. Use a min-heap to track the earliest finishing meeting. If a meeting starts after the earliest ends, reuse that room; else allocate a new room.

3. Final Answer:

   Sort intervals by start time, use a min-heap to track earliest end time, allocate rooms accordingly -> Option A

4. Quick Check:

   Min rooms = track earliest end with heap [OK]

Hint: Use min-heap on end times after sorting by start [OK]

Common Mistakes:

• Merging intervals instead of counting overlaps
• Sorting by length instead of start time
• Ignoring the need to track multiple rooms