Non Overlapping Intervals Minimum Removal in DSA Rust - Time & Space Complexity

Time Complexity: Non Overlapping Intervals Minimum Removal

O(n log n)

Understanding Time Complexity

We want to find out how long it takes to decide the minimum number of intervals to remove so none overlap.

How does the time needed grow as the number of intervals grows?

Scenario Under Consideration

Analyze the time complexity of the following code snippet.

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

This code sorts intervals by their end time, then counts how many must be removed to avoid overlaps.

Identify Repeating Operations

Identify the loops, recursion, array traversals that repeat.

• Primary operation: Sorting the intervals and then looping through them once.
• How many times: Sorting takes multiple comparisons (depends on n log n), loop runs once for each interval (n times).

How Execution Grows With Input

Sorting grows faster than just looping, so it dominates the time.

Input Size (n)	Approx. Operations
10	About 10 * log(10) ≈ 33 operations for sorting + 10 for loop
100	About 100 * log(100) ≈ 664 operations for sorting + 100 for loop
1000	About 1000 * log(1000) ≈ 9965 operations for sorting + 1000 for loop

Pattern observation: As input grows, sorting cost grows faster than the loop, roughly n log n times.

Final Time Complexity

Time Complexity: O(n log n)

This means the time needed grows a bit faster than the number of intervals because of sorting, but the loop adds only a little extra.

Common Mistake

[X] Wrong: "The loop alone decides the time complexity, so it is O(n)."

[OK] Correct: Sorting the intervals first takes more time than the loop, so it controls the total time needed.

Interview Connect

Understanding how sorting affects time helps you explain your solution clearly and shows you know what part of your code costs the most time.

Self-Check

"What if the intervals were already sorted by their start time? How would the time complexity change?"

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