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

Non Overlapping Intervals Minimum Removal

📖 Scenario: You are organizing a schedule with several time intervals. Some intervals overlap, and you want to remove the minimum number of intervals so that no intervals overlap.Think of it like planning meetings in a room where no two meetings can happen at the same time.

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

📋 What You'll Learn

Create an array of intervals where each interval is a tuple of two numbers representing start and end times.

Add a variable to count the number of intervals removed.

Implement logic to sort intervals and find overlaps.

Print the minimum number of intervals to remove.

💡 Why This Matters

🌍 Real World

Scheduling meetings, booking rooms, or managing tasks where overlapping times cause conflicts.

💼 Career

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

1

Create the intervals array

Create a variable called intervals and assign it the array of intervals: [[1, 3], [2, 4], [3, 5], [6, 7]].

DSA Typescript

// Your code here

Hint

Use const intervals: [number, number][] to declare the array of intervals.

2

Add a variable to count removals

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

DSA Typescript

const intervals: [number, number][] = [[1, 3], [2, 4], [3, 5], [6, 7]];
// Your code here

Hint

Use let removeCount = 0; to start counting removals.

3

Sort intervals and find overlaps

Sort the intervals array by the end time of each interval using intervals.sort((a, b) => a[1] - b[1]). Then, create a variable prevEnd set to the end time of the first interval. Use a for loop starting from index 1 with variable i to check if the start time of intervals[i] is less than prevEnd. If yes, increment removeCount. Otherwise, update prevEnd to intervals[i][1].

DSA Typescript

const intervals: [number, number][] = [[1, 3], [2, 4], [3, 5], [6, 7]];
let removeCount = 0;
// Your code here

Hint

Sort by end time to keep intervals that finish earliest. Then check overlaps by comparing start time with previous end time.

4

Print the minimum number of intervals to remove

Print the value of removeCount using console.log(removeCount).

DSA Typescript

const intervals: [number, number][] = [[1, 3], [2, 4], [3, 5], [6, 7]];
let removeCount = 0;
intervals.sort((a, b) => a[1] - b[1]);
let prevEnd = intervals[0][1];
for (let i = 1; i < intervals.length; i++) {
    if (intervals[i][0] < prevEnd) {
        removeCount++;
    } else {
        prevEnd = intervals[i][1];
    }
}
// Your code here

Hint

Use console.log(removeCount) to show the result.

Practice

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

easy

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

B. To find the maximum number of overlapping intervals

C. To merge all intervals into one large interval

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 A

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 of the following TypeScript function signatures correctly represents a function to solve the Non Overlapping Intervals Minimum Removal problem?

easy

A. function eraseOverlapIntervals(intervals: string[][]): string

B. function eraseOverlapIntervals(intervals: number[]): number[]

C. function eraseOverlapIntervals(intervals: number[][]): number

D. function eraseOverlapIntervals(intervals: number[][]): void

Solution

1. Step 1: Identify input and output types

   The input is an array of intervals, each interval is an array of two numbers. The output is the count of intervals removed, a number.

2. Step 2: Match the correct signature

   function eraseOverlapIntervals(intervals: number[][]): number matches input as number[][] and output as number, which fits the problem.

3. Final Answer:

   function eraseOverlapIntervals(intervals: number[][]): number -> Option C

4. Quick Check:

   Input array of number arrays, output number [OK]

Hint: Input is array of number pairs, output is a number count [OK]

Common Mistakes:

• Using wrong input type like number[] instead of number[][]
• Returning array instead of number
• Using void return type

3. What is the output of the following TypeScript code?

function eraseOverlapIntervals(intervals: number[][]): number {
  intervals.sort((a, b) => a[1] - b[1]);
  let count = 0;
  let end = -Infinity;
  for (const interval of intervals) {
    if (interval[0] >= end) {
      end = interval[1];
    } else {
      count++;
    }
  }
  return count;
}

console.log(eraseOverlapIntervals([[1,2],[2,3],[3,4],[1,3]]));

medium

A. 1

B. 2

C. 3

D. 0

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 removals

   Start with end = -Infinity
   Check [1,2]: 1 >= -Infinity true, end=2
   [2,3]: 2 >= 2 true, end=3
   [1,3]: 1 >= 3 false, count=1
   [3,4]: 3 >= 3 true, end=4

3. Final Answer:

   1 -> Option A

4. Quick Check:

   One interval removed to avoid overlap [OK]

Hint: Sort by end, count intervals starting before current 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 kept

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

function eraseOverlapIntervals(intervals: number[][]): number {
  intervals.sort((a, b) => a[0] - b[0]);
  let count = 0;
  let end = intervals[0][1];
  for (let i = 1; i < intervals.length; i++) {
    if (intervals[i][0] < end) {
      count++;
    } else {
      end = intervals[i][1];
    }
  }
  return count;
}

medium

A. Initializing end with intervals[0][0] instead of intervals[0][1]

B. Sorting intervals by start time instead of end time causes incorrect removals

C. Loop should start from i = 0 instead of i = 1

D. The function should return the count of intervals kept, not removed

Solution

1. Step 1: Check sorting criteria

   The function sorts intervals by start time, but the greedy approach requires sorting by end time to minimize removals.

2. Step 2: Understand impact on removal count

   Sorting by start time can cause incorrect counting of overlapping intervals, leading to wrong removal count.

3. Final Answer:

   Sorting intervals by start time instead of end time causes incorrect removals -> Option B

4. Quick Check:

   Sort by end time for correct minimum removal [OK]

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

Common Mistakes:

• Sorting by start time instead of end time
• Wrong initialization of end variable
• Incorrect loop start index

5. Given intervals [[0,2],[1,3],[2,4],[3,5],[4,6]], what is the minimum number of intervals to remove to make the rest non-overlapping?

hard

A. 0

B. 3

C. 1

D. 2

Solution

1. Step 1: Sort intervals by end time

   Sorted intervals: [0,2], [1,3], [2,4], [3,5], [4,6]

2. Step 2: Select intervals greedily and count removals

   Start with end = -Infinity, count = 0
   [0,2]: 0 >= -Infinity true, end=2
   [1,3]: 1 < 2 overlap, count=1
   [2,4]: 2 >= 2 true, end=4
   [3,5]: 3 < 4 overlap, count=2
   [4,6]: 4 >= 4 true, end=6

3. Final Answer:

   2 -> Option D

4. Quick Check:

   Two intervals removed to avoid overlap [OK]

Hint: Sort by end, remove intervals starting before current end [OK]

Common Mistakes:

• Not sorting intervals before processing
• Counting intervals kept instead of removed
• Misunderstanding overlap conditions