Non Overlapping Intervals Minimum Removal in DSA Rust - Build from Scratch

Non Overlapping Intervals Minimum Removal

📖 Scenario: You are organizing a schedule with several time intervals. Some intervals overlap, causing conflicts. You want to remove the fewest intervals possible so that the remaining intervals do not overlap.

🎯 Goal: Build a Rust program that finds the minimum number of intervals to remove to make the rest non-overlapping.

📋 What You'll Learn

Create a vector of tuples called intervals with exact intervals

Add a variable count to track removals

Implement logic to find minimum removals for non-overlapping intervals

Print the final count of intervals to remove

💡 Why This Matters

🌍 Real World

Scheduling meetings, booking rooms, or planning events without conflicts requires removing overlapping time slots.

💼 Career

This problem is common in software engineering roles involving scheduling systems, calendar apps, and resource allocation.

1

Create the intervals vector

Create a vector called intervals with these exact tuples: (1, 3), (2, 4), (3, 5), (6, 7), (8, 10), (9, 11).

DSA Rust

# Create the intervals vector with given tuples
// Your code here

Hint

Use let intervals = vec![(1, 3), (2, 4), (3, 5), (6, 7), (8, 10), (9, 11)]; to create the vector.

2

Add a removal counter

Add a mutable variable called count and set it to 0 to count how many intervals we remove.

DSA Rust

fn main() {
    let intervals = vec![(1, 3), (2, 4), (3, 5), (6, 7), (8, 10), (9, 11)];
    // Add a mutable count variable set to 0
    // Your code here
}

Hint

Use let mut count = 0; to create a mutable counter.

3

Implement minimum removal logic

Sort intervals by their end time using sort_by_key. Then create a variable prev_end set to i32::MIN. Use a for loop with variable interval to iterate over intervals. Inside the loop, if interval.0 is less than prev_end, increment count. Otherwise, update prev_end to interval.1.

DSA Rust

fn main() {
    let mut intervals = vec![(1, 3), (2, 4), (3, 5), (6, 7), (8, 10), (9, 11)];
    let mut count = 0;
    // Sort intervals by end time
    // Create prev_end variable
    // Loop through intervals and update count and prev_end
    // Your code here
}

Hint

Sort by end time, then check if current interval starts before previous end. If yes, increment count. Else update prev_end.

4

Print the minimum removal count

Print the value of count using println!.

DSA Rust

fn main() {
    let mut intervals = vec![(1, 3), (2, 4), (3, 5), (6, 7), (8, 10), (9, 11)];
    intervals.sort_by_key(|k| k.1);
    let mut count = 0;
    let mut prev_end = i32::MIN;
    for interval in intervals {
        if interval.0 < prev_end {
            count += 1;
        } else {
            prev_end = interval.1;
        }
    }
    // Print the count
    // Your code here
}

Hint

Use println!("{}", count); to print the number of intervals to remove.

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