Non Overlapping Intervals Minimum Removal in DSA Javascript - Execution Trace

Concept Flow - Non Overlapping Intervals Minimum Removal

Sort intervals by end time

↓

Initialize count = 0, end = -∞

↓

For each interval

↓

Check if interval start >= end?

No→Remove interval, count++

Yes↓

Update end = interval end

↓

Repeat for all intervals

↓

Return count of removals

Sort intervals by their end time, then iterate to count how many intervals must be removed to avoid overlaps.

Execution Sample

DSA Javascript

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

This code sorts intervals by end time and counts minimum removals to avoid overlaps.

Execution Table

Step	Operation	Current Interval	End Pointer	Count Removals	Visual State
1	Sort intervals	[1,2], [2,3], [1,3], [3,4]	-∞	0	[1,2] -> [2,3] -> [1,3] -> [3,4]
2	Check interval	[1,2]	-∞	0	[1,2] -> [2,3] -> [1,3] -> [3,4]
3	start >= end? 1 >= -∞	Yes	Update end = 2	0	[1,2] -> [2,3] -> [1,3] -> [3,4]
4	Check interval	[2,3]	2	0	[1,2] -> [2,3] -> [1,3] -> [3,4]
5	start >= end? 2 >= 2	Yes	Update end = 3	0	[1,2] -> [2,3] -> [1,3] -> [3,4]
6	Check interval	[1,3]	3	0	[1,2] -> [2,3] -> [1,3] -> [3,4]
7	start >= end? 1 >= 3	No	Count removal++	1	[1,2] -> [2,3] -> [1,3] -> [3,4]
8	Check interval	[3,4]	3	1	[1,2] -> [2,3] -> [1,3] -> [3,4]
9	start >= end? 3 >= 3	Yes	Update end = 4	1	[1,2] -> [2,3] -> [1,3] -> [3,4]
10	End of intervals	-	4	1	[1,2] -> [2,3] -> [1,3] -> [3,4]

💡 All intervals processed; minimum removals counted as 1.

Variable Tracker

Variable	Start	After Step 3	After Step 5	After Step 7	After Step 9	Final
end	-∞	2	3	3	4	4
count	0	0	0	1	1	1

Key Moments - 3 Insights

Why do we sort intervals by their end time instead of start time?

Sorting by end time ensures we always pick the interval that finishes earliest, leaving more room for following intervals. This is shown in execution_table step 1 where intervals are sorted by their end values.

Why do we update 'end' only when the current interval does not overlap?

We update 'end' to the current interval's end only if it doesn't overlap, so we keep track of the last interval's end that we accepted. See execution_table steps 3, 5, and 9 where 'end' updates only when start >= end.

What happens when an interval overlaps?

When an interval overlaps (start < end), we increment removal count but do not update 'end' to keep the earliest finishing interval. This is shown in step 7 of execution_table.

Visual Quiz - 3 Questions

Look at the execution_table at step 5, what is the value of 'end'?

A 2

B 4

C 3

D -∞

💡 Hint

Check the 'End Pointer' column at step 5 in execution_table.

At which step does the removal count increase?

A Step 7

B Step 9

C Step 3

D Step 5

💡 Hint

Look at the 'Count Removals' column in execution_table to find when it changes from 0 to 1.

If the intervals were not sorted by end time, how would the algorithm's behavior change?

A It would still find the minimum removals correctly.

B It might remove more intervals than necessary.

C It would remove fewer intervals than necessary.

D It would not terminate.

💡 Hint

Refer to the concept_flow and key_moments about why sorting by end time is important.

Concept Snapshot

Non Overlapping Intervals Minimum Removal:
- Sort intervals by end time ascending.
- Initialize count=0, end=-∞.
- For each interval:
  - If start >= end, update end.
  - Else, increment count (remove interval).
- Return count as minimum removals.

Full Transcript

This concept solves the problem of removing the minimum number of intervals to avoid overlaps. The key step is sorting intervals by their end time so that we always pick the interval that finishes earliest. We keep a pointer 'end' to track the end of the last accepted interval. For each interval, if its start is greater or equal to 'end', we accept it and update 'end'. Otherwise, we remove it and increment the removal count. This process ensures the minimum removals. The execution table shows step-by-step how intervals are checked and how 'end' and 'count' change. Key moments clarify why sorting by end time is crucial and how overlaps are handled.

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