Non Overlapping Intervals Minimum Removal in DSA Rust - Interactive Practice

Practice - 5 Tasks

1fill in blank

easy

Complete the code to sort intervals by their end time.

DSA Rust

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

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

Astart

Bmid

Clength

Dend

Attempts:

3 left

💡 Hint

Common Mistakes

Sorting by start time instead of end time.

Sorting by interval length which does not guarantee minimal removals.

Sorting intervals by their end time helps to greedily select intervals that finish earliest, minimizing overlaps.

2fill in blank

medium

Complete the code to initialize the count of removals to zero.

DSA Rust

let mut removals = [1];

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A0

B1

C-1

Dintervals.len()

Attempts:

3 left

💡 Hint

Common Mistakes

Starting removals at 1 which incorrectly assumes one interval is removed initially.

Starting removals at intervals.len() which is too large.

We start with zero removals because initially we haven't removed any intervals.

3fill in blank

hard

Fix the error in the condition to check if current interval overlaps with previous.

DSA Rust

if intervals[i].[1] <= prev_end { removals += 1; } else { prev_end = intervals[i].end; }

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

Alength

Bend

Cstart

Dmid

Attempts:

3 left

💡 Hint

Common Mistakes

Comparing current end with previous end instead of current start.

Using length or mid which are not relevant for overlap check.

We check if the start of the current interval is less than or equal to the end of the previous to detect overlap.

4fill in blank

hard

Fill both blanks to update the previous end and increment removals correctly.

DSA Rust

if intervals[i].start <= prev_end { removals [1] 1; } else { prev_end [2] intervals[i].end; }

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A+=

B-=

C=

D*=

Attempts:

3 left

💡 Hint

Common Mistakes

Using '-=' or '*=' which incorrectly changes removals or prev_end.

Using '=' to increment removals instead of '+='.

We add 1 to removals when overlap is found and update prev_end to current interval's end when no overlap.

5fill in blank

hard

Fill all three blanks to complete the function that returns minimum removals.

DSA Rust

fn erase_overlap_intervals(mut intervals: Vec<Interval>) -> i32 {
    if intervals.is_empty() { return 0; }
    intervals.sort_by_key(|interval| interval.[1]);
    let mut removals = [2];
    let mut prev_end = intervals[0].end;
    for i in 1..intervals.len() {
        if intervals[i].start <= prev_end {
            removals [3] 1;
        } else {
            prev_end = intervals[i].end;
        }
    }
    removals
}

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

Astart

B0

C+=

Dend

Attempts:

3 left

💡 Hint

Common Mistakes

Sorting by start instead of end in this implementation.

Initializing removals incorrectly.

Using '=' instead of '+=' to increment removals.

Sort by end, initialize removals to 0, and increment removals with '+=' when overlap is found.

Practice

1. What is the main goal of the Non Overlapping Intervals Minimum Removal problem?

easy

A. To find the maximum number of overlapping intervals

B. To merge all intervals into one big interval

C. To remove the minimum number of intervals so that the remaining intervals do not overlap

D. To sort intervals by their start time

Solution

1. Step 1: Understand the problem goal

   The problem asks to remove as few intervals as possible so that no two intervals overlap.

2. Step 2: Clarify what non-overlapping means

   Non-overlapping means intervals do not share any common points except possibly touching ends.

3. Final Answer:

   To remove the minimum number of intervals so that the remaining intervals do not overlap -> Option C

4. Quick Check:

   Minimum removal = non-overlapping intervals [OK]

Hint: Focus on removing intervals, not merging or counting overlaps [OK]

Common Mistakes:

• Thinking the goal is to merge intervals
• Confusing maximum overlap with minimum removal
• Sorting intervals without removing any

2. Which Rust code snippet correctly declares a vector of intervals as tuples for this problem?

easy

A. let intervals = vec![[1, 3], [2, 4], [3, 5]];

B. let intervals = [(1, 3), (2, 4), (3, 5)];

C. let intervals = vec!{(1, 3), (2, 4), (3, 5)};

D. let intervals = vec![(1, 3), (2, 4), (3, 5)];

Solution

1. Step 1: Identify correct vector syntax in Rust

   Vectors use vec![] macro with parentheses for tuples.

2. Step 2: Check tuple and vector syntax

   let intervals = vec![(1, 3), (2, 4), (3, 5)]; uses vec![(1, 3), (2, 4), (3, 5)] which is correct for vector of tuples.

3. Final Answer:

   let intervals = vec![(1, 3), (2, 4), (3, 5)]; -> Option D

4. Quick Check:

   vec! macro + tuples = correct syntax [OK]

Hint: Use vec![] with parentheses for tuples, not brackets or braces [OK]

Common Mistakes:

• Using square brackets for tuples
• Using braces {} instead of brackets [] for vec!
• Declaring array without vec! macro

3. What is the output of the following Rust code that finds the minimum number of intervals to remove?

fn erase_overlap_intervals(mut intervals: Vec<(i32, i32)>) -> i32 {
    intervals.sort_by_key(|k| k.1);
    let mut count = 0;
    let mut end = std::i32::MIN;
    for interval in intervals {
        if interval.0 >= end {
            end = interval.1;
        } else {
            count += 1;
        }
    }
    count
}

fn main() {
    let intervals = vec![(1, 2), (2, 3), (3, 4), (1, 3)];
    println!("{}", erase_overlap_intervals(intervals));
}

medium

A. 0

B. 1

C. 3

D. 2

Solution

1. Step 1: Sort intervals by end time

   Sorted intervals: (1,2), (2,3), (1,3), (3,4)

2. Step 2: Iterate and count overlaps

   Check intervals: - (1,2): end = 2, count=0 - (2,3): start 2 >= end 2, end=3, count=0 - (1,3): start 1 < end 3, count=1 (remove this) - (3,4): start 3 >= end 3, end=4, count=1

3. Final Answer:

   1 -> Option B

4. Quick Check:

   One interval removed to avoid overlap [OK]

Hint: Sort by end time, count intervals starting before last end [OK]

Common Mistakes:

• Not sorting intervals by end time
• Counting intervals incorrectly when start equals end
• Confusing count of removals with count of intervals

4. Identify the bug in this Rust function for minimum interval removal:

fn erase_overlap_intervals(mut intervals: Vec<(i32, i32)>) -> i32 {
    intervals.sort_by_key(|k| k.0);
    let mut count = 0;
    let mut end = std::i32::MIN;
    for interval in intervals {
        if interval.0 >= end {
            end = interval.1;
        } else {
            count += 1;
        }
    }
    count
}

medium

A. Sorting intervals by start time instead of end time causes incorrect removal count

B. Using std::i32::MIN for end causes overflow

C. The count variable should be initialized to 1

D. The function should return a boolean instead of i32

Solution

1. Step 1: Check sorting criteria

   The function sorts intervals by start time, but correct approach is sorting by end time to minimize removals.

2. Step 2: Understand impact of sorting

   Sorting by start time can cause wrong intervals to be kept, increasing removals incorrectly.

3. Final Answer:

   Sorting intervals by start time instead of end time causes incorrect removal count -> Option A

4. Quick Check:

   Sort by end time, not start time [OK]

Hint: Always sort intervals by their end time for this problem [OK]

Common Mistakes:

• Sorting by start time instead of end time
• Misinitializing end variable
• Returning wrong data type

5. Given intervals [(1, 5), (2, 3), (4, 6), (7, 8), (5, 7)], what is the minimum number of intervals to remove to make the rest non-overlapping?

hard

A. 2

B. 1

C. 3

D. 0

Solution

1. Step 1: Sort intervals by end time

   Sorted intervals: (2,3), (1,5), (4,6), (5,7), (7,8)

2. Step 2: Greedily select non-overlapping intervals and count removals

   Initialize end = MIN, count = 0
   • (2,3): 2 >= MIN, keep, end=3
   • (1,5): 1 < 3, remove, count=1
   • (4,6): 4 >= 3, keep, end=6
   • (5,7): 5 < 6, remove, count=2
   • (7,8): 7 >= 6, keep, end=8
   Total removals: 2. This greedy approach by earliest end time yields the minimum.

3. Final Answer:

   2 -> Option A

4. Quick Check:

   Minimum removals = 2 [OK]

Hint: Sort by end, greedily keep intervals, count removals [OK]

Common Mistakes:

• Assuming zero removals without checking overlaps
• Not sorting intervals before processing
• Counting removals incorrectly when intervals touch