Non Overlapping Intervals Minimum Removal in DSA Rust - Execution Trace

Concept Flow - Non Overlapping Intervals Minimum Removal

Sort intervals by end time

↓

Initialize count = 0, prev_end = -∞

↓

For each interval

↓

Check if current start >= prev_end?

No→Remove interval, count += 1

↓

Skip removal

↓

Update prev_end = current end

↓

Repeat

↓

Return count

Sort intervals by their end time, then iterate to count how many intervals must be removed to avoid overlaps.

Execution Sample

DSA Rust

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

This Rust function sorts intervals by end time and counts minimum removals to avoid overlaps.

Execution Table

Step	Operation	Current Interval	prev_end	Removal Count	Action	Intervals Visual State
1	Sort intervals	[(1,3), (2,4), (3,5), (6,7)]	-∞	0	Sorted by end time	(1,3) -> (2,4) -> (3,5) -> (6,7) -> null
2	Check interval	(1,3)	-∞	0	1 >= -∞, keep	1,3 -> 2,4 -> 3,5 -> 6,7 -> null
3	Update prev_end	(1,3)	3	0	prev_end = 3	1,3 -> 2,4 -> 3,5 -> 6,7 -> null
4	Check interval	(2,4)	3	0	2 < 3, remove	1,3 -> 2,4 (removed) -> 3,5 -> 6,7 -> null
5	Increment removal	(2,4)	3	1	count = 1	1,3 -> 3,5 -> 6,7 -> null
6	Check interval	(3,5)	3	1	3 >= 3, keep	1,3 -> 3,5 -> 6,7 -> null
7	Update prev_end	(3,5)	5	1	prev_end = 5	1,3 -> 3,5 -> 6,7 -> null
8	Check interval	(6,7)	5	1	6 >= 5, keep	1,3 -> 3,5 -> 6,7 -> null
9	Update prev_end	(6,7)	7	1	prev_end = 7	1,3 -> 3,5 -> 6,7 -> null
10	End	N/A	7	1	All intervals processed	1,3 -> 3,5 -> 6,7 -> null

💡 All intervals checked; minimum removal count is 1.

Variable Tracker

Variable	Start	After Step 2	After Step 4	After Step 6	After Step 8	Final
prev_end	-∞	3	3	5	7	7
count	0	0	1	1	1	1

Key Moments - 3 Insights

Why do we sort intervals by their end time instead of start time?

Sorting by end time ensures we always pick the interval that finishes earliest, leaving more room for following intervals. This is shown in execution_table step 1 where intervals are sorted by end time.

Why do we remove an interval when current start < prev_end?

If current interval starts before the previous interval ends, they overlap. We remove the current one to keep the earliest finishing interval, as seen in step 4 where removal count increases.

Why update prev_end only when we keep an interval?

prev_end tracks the end of the last kept interval to compare with next intervals. Updating it only when we keep an interval ensures correct overlap checks, as in steps 3, 7, and 9.

Visual Quiz - 3 Questions

Look at the execution_table at step 4, what is the removal count after processing interval (2,4)?

A 0

B 1

C 2

D 3

💡 Hint

Check the 'Removal Count' column at step 4 in execution_table.

At which step does prev_end get updated to 5?

A Step 7

B Step 6

C Step 3

D Step 9

💡 Hint

Look at the 'prev_end' column and 'Action' at steps 6 and 7.

If intervals were not sorted by end time, what would likely happen to the removal count?

A It would decrease

B It would stay the same

C It could increase

D It would be zero

💡 Hint

Refer to concept_flow where sorting by end time is the first step to minimize removals.

Concept Snapshot

Non Overlapping Intervals Minimum Removal:
- Sort intervals by end time.
- Initialize removal count = 0 and prev_end = -∞.
- For each interval, if start >= prev_end, update prev_end.
- Else, increment removal count.
- Return removal count as minimum intervals to remove.

Full Transcript

This concept solves the problem of removing the minimum number of intervals to avoid overlaps. We first sort intervals by their end times to prioritize intervals that finish earlier. We keep track of the end time of the last kept interval (prev_end). For each interval, if its start time is at least prev_end, we keep it and update prev_end. Otherwise, we remove it and increment the removal count. This approach ensures the minimum removals needed. The execution table shows step-by-step how intervals are checked, when removals happen, and how prev_end and count change. Key moments clarify why sorting by end time is crucial, why removals happen on overlaps, and why prev_end updates only on kept intervals. The visual quiz tests understanding of removal counts, prev_end updates, and the importance of sorting.

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