Non Overlapping Intervals Minimum Removal in DSA Typescript - 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

Which greedy strategy is commonly used to solve the Non Overlapping Intervals Minimum Removal problem?

Sort intervals by their end time and iteratively select intervals that start after the last selected interval ends.

intermediate

Why do we sort intervals by their end time in this problem?

Sorting by end time helps to maximize the number of non-overlapping intervals by always choosing the interval that finishes earliest.

beginner

What does the minimum removal count equal in terms of total intervals and selected intervals?

Minimum removal = Total intervals - Maximum number of non-overlapping intervals selected.

beginner

In TypeScript, what is the type signature of the function solving this problem if intervals are given as an array of [start, end] pairs?

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

What is the first step in the greedy approach to solve the Non Overlapping Intervals Minimum Removal problem?

A Sort intervals by their start time

B Sort intervals by their end time

C Remove intervals with the longest duration first

D Select intervals randomly

Sorting intervals by their end time allows us to pick intervals that finish earliest, maximizing non-overlapping intervals.

If intervals are [[1,3],[2,4],[3,5]], what is the minimum number of intervals to remove to avoid overlap?

A 1

B 2

C 0

D 3

Removing [2,4] leaves [1,3] and [3,5] which do not overlap.

What does the variable 'count' usually represent in the solution code?

A Number of intervals removed

B Sum of interval lengths

C Total number of intervals

D Number of intervals selected without overlap

'count' tracks how many intervals are selected that do not overlap.

Which data structure is primarily used to store intervals in this problem?

A Hash map

B Linked list

C Array of arrays

D Stack

Intervals are stored as arrays of two numbers representing start and end times.

What is the time complexity of the greedy solution after sorting?

A O(n log n)

B O(n^2)

C O(n)

D O(log n)

Sorting takes O(n log n), and the single pass to select intervals is O(n), so overall O(n log n).

Explain the greedy approach to solve the Non Overlapping Intervals Minimum Removal problem step-by-step.

Think about how choosing intervals that finish earliest helps.

Describe how to implement the solution in TypeScript including sorting and counting non-overlapping intervals.

Focus on array methods and simple loop logic.

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