Non Overlapping Intervals Minimum Removal in DSA Typescript - Interactive Practice

Practice - 5 Tasks

1fill in blank

easy

Complete the code to sort intervals by their end time.

DSA Typescript

intervals.sort((a, b) => a[[1]] - b[[1]]);

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A1

B0

C2

Dlength

Attempts:

3 left

💡 Hint

Common Mistakes

Sorting by start time instead of end time.

Using incorrect index for end time.

We sort intervals by their end time, which is at index 1 in each interval array.

2fill in blank

medium

Complete the code to initialize the variable that tracks the end of the last added interval.

DSA Typescript

let end = intervals[0][[1]];

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A1

Bstart

Clength

D0

Attempts:

3 left

💡 Hint

Common Mistakes

Using start time instead of end time.

Using an invalid index.

We initialize 'end' with the end time of the first interval, which is at index 1.

3fill in blank

hard

Fix the error in the condition that checks if the current interval overlaps with the last added interval.

DSA Typescript

if (intervals[i][0] < [1]) { count++; } else { end = intervals[i][1]; }

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Acount

Bend

Cintervals[i][0]

Dintervals[i][1]

Attempts:

3 left

💡 Hint

Common Mistakes

Comparing start with start or end with start incorrectly.

Using wrong variable in condition.

We compare the start of the current interval with the end of the last added interval to detect overlap.

4fill in blank

hard

Fill both blanks to complete the loop that counts intervals to remove.

DSA Typescript

for (let [1] = 1; [1] < intervals.[2]; i++) { /* loop body */ }

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Ai

Blength

Cj

Dsize

Attempts:

3 left

💡 Hint

Common Mistakes

Using wrong loop variable name.

Using incorrect property for array length.

We use 'i' as the loop variable starting from 1, and 'length' to get the number of intervals.

5fill in blank

hard

Fill all three blanks to complete the function that returns the minimum number of intervals to remove.

DSA Typescript

function eraseOverlapIntervals(intervals: number[][]): number {
  if (intervals.length === 0) return 0;
  intervals.sort((a, b) => a[[1]] - b[[2]]);
  let count = 0;
  let end = intervals[0][[3]];
  for (let i = 1; i < intervals.length; i++) {
    if (intervals[i][0] < end) {
      count++;
    } else {
      end = intervals[i][1];
    }
  }
  return count;
}

Drag options to blanks, or click blank then click option'

A1

B0

D2

Attempts:

3 left

💡 Hint

Common Mistakes

Mixing up start and end indices in sort.

Initializing 'end' with start time instead of end time.

We sort intervals by their end time, which is at index 1 in each interval array for both a and b. Initialize 'end' with the end time of the first interval, which is at index 1.

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