Insert Interval into Sorted List in DSA Rust - Interactive Practice

Practice - 5 Tasks

1fill in blank

easy

Complete the code to create a new interval struct.

DSA Rust

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

let new_interval = Interval { start: [1], end: 5 };

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

A0

Bstart

Cend

D3

Attempts:

3 left

💡 Hint

Common Mistakes

Using variable names instead of a number for start.

The new interval starts at 3 and ends at 5, so start must be 3.

2fill in blank

medium

Complete the code to push the new interval into the vector.

DSA Rust

let mut intervals = vec![Interval { start: 1, end: 2 }, Interval { start: 6, end: 9 }];
intervals.[1](new_interval);

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Aappend

Binsert

Cpush

Dextend

Attempts:

3 left

💡 Hint

Common Mistakes

Using insert without index.

Using append which requires another vector.

To add a single element at the end of a vector, use push.

3fill in blank

hard

Fix the error in the loop that merges overlapping intervals.

DSA Rust

for i in 1..intervals.len() {
    if intervals[i].start [1] intervals[i - 1].end {
        intervals[i - 1].end = intervals[i].end;
        intervals.remove(i);
    }
}

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

A>

B<=

C>=

D<

Attempts:

3 left

💡 Hint

Common Mistakes

Using > or < which misses overlapping intervals.

We merge intervals if the current start is less than or equal to the previous end.

4fill in blank

hard

Fill both blanks to correctly insert the new interval in sorted order.

DSA Rust

let mut i = 0;
while i < intervals.len() && intervals[i].start [1] new_interval.start {
    i [2] i + 1;
}
intervals.insert(i, new_interval);

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A<

B<=

C=

D+=

Attempts:

3 left

💡 Hint

Common Mistakes

Using <= causes wrong insertion point.

Using = instead of += for increment.

We move forward while interval start is less than new start, and increment i by 1.

5fill in blank

hard

Fill all three blanks to create a merged list of intervals after insertion.

DSA Rust

let mut merged = Vec::new();
for interval in intervals {
    if merged.is_empty() || merged.last().unwrap().end [1] interval.start {
        merged.[2](interval);
    } else {
        merged.last_mut().unwrap().end = merged.last().unwrap().end.[3] interval.end;
    }
}

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

A<=

Bpush

Cmax

D<

Attempts:

3 left

💡 Hint

Common Mistakes

Using <= causes wrong merge logic.

Using min instead of max for merging.

We add interval if no overlap (last end < current start), else merge by max end.

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