Non Overlapping Intervals Minimum Removal in DSA Rust - Cheat Sheet & Quick Revision

Recall & Review

beginner

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

To find the minimum number of intervals to remove from a list so that the remaining intervals do not overlap.

intermediate

Why do we sort intervals by their end time in the Non Overlapping Intervals Minimum Removal problem?

Sorting by end time helps us greedily select intervals that finish earliest, maximizing the number of non-overlapping intervals.

beginner

In the Non Overlapping Intervals Minimum Removal problem, what does it mean if the start of the current interval is less than the end of the last selected interval?

It means the current interval overlaps with the last selected interval and must be removed or skipped.

intermediate

What is the time complexity of the common greedy solution for the Non Overlapping Intervals Minimum Removal problem?

O(n log n) due to sorting the intervals, where n is the number of intervals.

beginner

Show the Rust function signature for solving the Non Overlapping Intervals Minimum Removal problem given intervals as Vec<(i32, i32)>.

fn erase_overlap_intervals(intervals: Vec<(i32, i32)>) -> i32

What is the first step in solving the Non Overlapping Intervals Minimum Removal problem?

A Sort intervals by their end time

B Sort intervals by their start time

C Remove intervals randomly

D Count all intervals

Sorting intervals by their end time allows a greedy approach to select the maximum number of non-overlapping intervals.

If two intervals overlap, which one should be removed to minimize removals?

A The one with the earlier start time

B The one with the later end time

C The one with the earlier end time

D The one with the longer duration

Removing the interval with the later end time keeps more room for future intervals, minimizing removals.

What does the output represent in the Non Overlapping Intervals Minimum Removal problem?

A Number of intervals remaining

B Maximum overlap count

C Sum of interval lengths

D Number of intervals to remove

The problem asks for the minimum number of intervals to remove to avoid overlaps.

Which data structure is commonly used to store intervals for this problem?

A Hash map

B Stack

C Array or vector of tuples

D Queue

Intervals are typically stored as arrays or vectors of (start, end) tuples.

What is the main algorithmic technique used to solve this problem efficiently?

A Greedy algorithm

B Dynamic programming

C Backtracking

D Divide and conquer

A greedy approach selecting intervals by earliest end time is the efficient solution.

Explain step-by-step how to find the minimum number of intervals to remove to avoid overlaps.

Think about choosing intervals that finish earliest to keep more intervals.

Describe why sorting intervals by their end time is better than sorting by start time for this problem.

Consider how choosing intervals affects future choices.

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