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

Recall & Review

beginner

What is the main 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 instead of start time in this problem?

Sorting by end time helps to keep intervals that finish earliest, maximizing the number of non-overlapping intervals and minimizing removals.

beginner

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

'count' represents the number of intervals removed to achieve a set of non-overlapping intervals.

intermediate

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

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

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 all intervals that overlap

D Count the total number of intervals

Sorting intervals by their end time helps to pick intervals that finish earliest, allowing more intervals to fit without overlapping.

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

A The interval with the earlier end time

B The interval with the earlier start time

C The interval with the later end time

D The interval with the longer duration

Removing the interval with the later end time allows keeping the interval that finishes earlier, maximizing non-overlapping intervals.

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

A Number of intervals kept

B Number of intervals removed

C Total number of intervals

D Sum of interval lengths

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

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

A Array of arrays or tuples

B Linked list

C Stack

D Hash map

Intervals are usually stored as arrays or tuples with start and end values.

What is the main reason the greedy approach works for this problem?

A Sorting intervals by length solves the problem

B Choosing intervals with latest start time removes overlaps

C Randomly removing intervals reduces complexity

D Choosing intervals with earliest end time leaves room for more intervals

Selecting intervals that finish earliest allows fitting the maximum number of intervals without overlap.

Explain step-by-step how to find the minimum number of intervals to remove so that no intervals overlap.

Think about sorting and comparing intervals one by one.

Describe why sorting intervals by their end time helps in minimizing the number of removals.

Consider how choosing intervals that end early affects the rest.

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 sort intervals by their start time

D. To merge all overlapping intervals into one

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 intervals mean no two intervals share any common points except possibly at the boundary.

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:

• Confusing removal with merging intervals
• Thinking the goal is to count overlaps
• Assuming intervals must be sorted only

2. Which of the following JavaScript code snippets correctly sorts intervals by their end time for the minimum removal problem?

easy

A. intervals.sort((a, b) => b[0] - a[0]);

B. intervals.sort((a, b) => a[0] - b[0]);

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

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

Solution

1. Step 1: Identify sorting by end time

   Sorting by end time means comparing the second element of each interval, which is a[1] and b[1].

2. Step 2: Check the sorting order

   We want ascending order, so subtract a[1] - b[1] to sort from smallest to largest end time.

3. Final Answer:

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

4. Quick Check:

   Sort by end time ascending = intervals.sort((a, b) => a[1] - b[1]); [OK]

Hint: Sort intervals by end time ascending for greedy approach [OK]

Common Mistakes:

• Sorting by start time instead of end time
• Sorting in descending order by mistake
• Using wrong array indices for sorting

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

const intervals = [[1,3],[2,4],[3,5]];
intervals.sort((a,b) => a[1] - b[1]);
let count = 0;
let end = -Infinity;
for (const [start, finish] of intervals) {
  if (start >= end) {
    end = finish;
  } else {
    count++;
  }
}
console.log(count);

medium

A. 3

B. 1

C. 0

D. 2

Solution

1. Step 1: Sort intervals by end time

   Sorted intervals: [[1,3],[2,4],[3,5]] (already sorted by end time 3,4,5).

2. Step 2: Iterate and count removals

   Start with end = -Infinity.
   Interval [1,3]: start 1 >= -Infinity, update end=3, count=0.
   Interval [2,4]: start 2 < end 3, overlap, count=1.
   Interval [3,5]: start 3 >= end 3, update end=5, count=1.

3. Final Answer:

   1 -> Option B

4. Quick Check:

   One interval removed to avoid overlap [OK]

Hint: Count overlaps after sorting by end time [OK]

Common Mistakes:

• Not updating end after choosing interval
• Counting non-overlapping intervals as removals
• Incorrectly sorting intervals

4. Find the bug in this JavaScript code for minimum interval removal:

function eraseOverlapIntervals(intervals) {
  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. Sorting by start time instead of end time causes incorrect removal count

B. The loop should start from i = 0, not 1

C. The variable 'count' should be initialized to 1

D. The 'end' variable should be updated inside the if condition

Solution

1. Step 1: Check sorting criteria

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

2. Step 2: Understand impact of sorting

   Sorting by start time can cause counting more removals than necessary because it doesn't prioritize earliest finishing intervals.

3. Final Answer:

   Sorting by start time instead of end time causes incorrect removal count -> Option A

4. Quick Check:

   Sort by end time for correct removal count [OK]

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

Common Mistakes:

• Sorting by start time instead of end time
• Not updating 'end' correctly after removal
• Starting loop from wrong index

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

hard

A. 0

B. 2

C. 1

D. 3

Solution

1. Step 1: Sort intervals by end time

   Sorted intervals: [[1,2],[2,3],[1,3],[3,4]] by end times 2,3,3,4.

2. Step 2: Select intervals greedily and count removals

   Choose [1,2], end=2.
   [2,3] start 2 >= 2, choose, end=3.
   [1,3] start 1 < 3, overlap, remove count=1.
   [3,4] start 3 >= 3, choose, end=4.

3. Final Answer:

   1 -> Option C

4. Quick Check:

   One interval removed to avoid overlap [OK]

Hint: Sort by end time, pick intervals greedily, count overlaps [OK]

Common Mistakes:

• Not sorting intervals before processing
• Counting non-overlapping intervals as removals
• Ignoring intervals with same end time