Non Overlapping Intervals Minimum Removal in DSA Go - Execution Trace

Concept Flow - Non Overlapping Intervals Minimum Removal

Sort intervals by end time

↓

Initialize count=0, prev_end = -∞

↓

For each interval

↓

Check if current start >= prev_end?

No→Remove interval, count++

Yes↓

Update prev_end = current 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 Go

import "math"

intervals := [][]int{{1,2},{2,3},{1,3}}
count := 0
prevEnd := math.MinInt32
for _, iv := range intervals {
  if iv[0] < prevEnd {
    count++
  } else {
    prevEnd = iv[1]
  }
}
return count

This code counts minimum intervals to remove so remaining intervals do not overlap.

Execution Table

Step	Operation	Current Interval	prevEnd	Count Removed	Action	Intervals Visual
1	Sort intervals	[1,2]	-∞	0	Intervals sorted by end time	[1,2] -> [2,3] -> [1,3]
2	Check interval	[1,2]	-∞	0	1 >= -∞, keep interval	[1,2]
3	Update prevEnd	[1,2]	2	0	prevEnd updated to 2	[1,2]
4	Check interval	[2,3]	2	0	2 >= 2, keep interval	[1,2] -> [2,3]
5	Update prevEnd	[2,3]	3	0	prevEnd updated to 3	[1,2] -> [2,3]
6	Check interval	[1,3]	3	0	1 < 3, remove interval	[1,2] -> [2,3]
7	Increment count	[1,3]	3	1	Count incremented to 1	[1,2] -> [2,3]
8	End	-	3	1	All intervals processed	[1,2] -> [2,3]

💡 All intervals checked; minimum removal count is 1 because [1,3] overlaps with [2,3]

Variable Tracker

Variable	Start	After Step 2	After Step 3	After Step 4	After Step 5	After Step 6	After Step 7	Final
prevEnd	-∞	-∞	2	2	3	3	3	3
count	0	0	0	0	0	0	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 keep the interval that finishes earliest, leaving room for more intervals to fit without overlap. See execution_table step 1 where sorting arranges intervals by end time.

Why do we compare current interval's start with prevEnd?

We check if the current interval starts before the previous interval ends to detect overlap. If start < prevEnd, it overlaps and must be removed. See execution_table step 6 where 1 < 3 triggers removal.

Why do we update prevEnd only when we keep an interval?

prevEnd tracks the end of the last kept interval to compare with next intervals. If we remove an interval, prevEnd stays the same to keep the last valid end. See steps 3, 5 where prevEnd updates only after keeping intervals.

Visual Quiz - 3 Questions

Look at the execution_table at step 6, what is the reason for removing the interval [1,3]?

A It is the first interval in the list

B Its end time is less than prevEnd

C Its start time is less than prevEnd, causing overlap

D It has the earliest end time

💡 Hint

Check the 'Action' and 'prevEnd' columns at step 6 in execution_table

At which step does the count of removed intervals increase?

A Step 7

B Step 4

C Step 3

D Step 5

💡 Hint

Look at the 'Count Removed' and 'Action' columns in execution_table

If the intervals were sorted by start time instead of end time, what would likely happen?

A The algorithm would still find the minimum removals correctly

B The count of removals might increase due to suboptimal choices

C The count of removals would always be zero

D The intervals would not overlap

💡 Hint

Refer to key_moments about why sorting by end time is important

Concept Snapshot

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

Full Transcript

To find the minimum number of intervals to remove so that no intervals overlap, first sort all intervals by their end time. This ensures we always pick the interval that finishes earliest, maximizing space for others. Initialize a count of removals to zero and a variable prevEnd to negative infinity. Iterate through each interval: if the current interval's start time is less than prevEnd, it overlaps with the last kept interval, so increment the removal count. Otherwise, update prevEnd to the current interval's end time. After processing all intervals, the count variable holds the minimum removals needed to avoid overlaps.

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 big 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 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 = no overlaps [OK]

Hint: Focus on removing intervals, not merging or counting overlaps [OK]

Common Mistakes:

• Thinking the goal is to merge intervals
• Confusing maximum overlaps with minimum removals
• Sorting intervals without removing any

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

easy

A. sort.Slice(intervals, func(i, j int) bool { return intervals[i][0] > intervals[j][0] })

B. sort.Slice(intervals, func(i, j int) bool { return intervals[i][0] < intervals[j][0] })

C. sort.Slice(intervals, func(i, j int) bool { return intervals[i][1] < intervals[j][1] })

D. sort.Slice(intervals, func(i, j int) bool { return intervals[i][1] > intervals[j][1] })

Solution

1. Step 1: Identify sorting by end time

   We need to sort intervals by their end time ascending to apply greedy approach.

2. Step 2: Check the Go sort.Slice syntax

   sort.Slice(intervals, func(i, j int) bool { return intervals[i][1] < intervals[j][1] }) sorts by intervals[i][1] < intervals[j][1], which is correct for end time ascending.

3. Final Answer:

   sort.Slice(intervals, func(i, j int) bool { return intervals[i][1] < intervals[j][1] }) -> Option C

4. Quick Check:

   Sort by end time ascending = sort.Slice(intervals, func(i, j int) bool { return intervals[i][1] < intervals[j][1] }) [OK]

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

Common Mistakes:

• Sorting by start time instead of end time
• Sorting in descending order by mistake
• Using wrong slice indices

3. What is the output of the following Go code snippet that finds the minimum number of intervals to remove?

package main
import (
  "fmt"
  "sort"
)
func eraseOverlapIntervals(intervals [][]int) int {
  if len(intervals) == 0 {
    return 0
  }
  sort.Slice(intervals, func(i, j int) bool { return intervals[i][1] < intervals[j][1] })
  count := 0
  end := intervals[0][1]
  for i := 1; i < len(intervals); i++ {
    if intervals[i][0] < end {
      count++
    } else {
      end = intervals[i][1]
    }
  }
  return count
}
func main() {
  intervals := [][]int{{1,3},{2,4},{3,5}}
  fmt.Println(eraseOverlapIntervals(intervals))
}

medium

A. 2

B. 1

C. 0

D. 3

Solution

1. Step 1: Sort intervals by end time

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

2. Step 2: Count overlaps

   Start with end=3. Interval [2,4] starts at 2 < 3, so count=1. Interval [3,5] starts at 3 ≥ 3, update end=5.

3. Final Answer:

   1 -> Option B

4. Quick Check:

   One overlap removed = 1 [OK]

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

Common Mistakes:

• Not sorting intervals before counting
• Counting intervals that do not overlap
• Incorrectly updating end pointer

4. The following Go code is intended to solve the minimum removal problem but has a bug. What is the bug?

func eraseOverlapIntervals(intervals [][]int) int {
  if len(intervals) == 0 {
    return 0
  }
  sort.Slice(intervals, func(i, j int) bool { return intervals[i][1] < intervals[j][1] })
  count := 0
  end := intervals[0][1]
  for i := 1; i < len(intervals); i++ {
    if intervals[i][0] < end {
      count++
    } else {
      end = intervals[i][0]
    }
  }
  return count
}

medium

A. Updating end with intervals[i][0] instead of intervals[i][1] causes wrong comparisons

B. Missing check for empty intervals slice causes runtime error

C. Sorting by start time instead of end time causes incorrect removal count

D. Using count++ inside the loop is incorrect; should decrement count

Solution

1. Step 1: Check sorting and end update

   Sorting by end time is correct, but updating end incorrectly with intervals[i][0] is wrong.

2. Step 2: Explain why end update is wrong

   End should track the end time of last non-overlapping interval, so intervals[i][1] is correct.

3. Final Answer:

   Updating end with intervals[i][0] instead of intervals[i][1] causes wrong comparisons -> Option A

4. Quick Check:

   End must be interval end, not start [OK]

Hint: Update end with interval's end time, not start time [OK]

Common Mistakes:

• Confusing interval start and end when updating end
• Ignoring empty slice check
• Misunderstanding sorting impact

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. 2

B. 3

C. 0

D. 1

Solution

1. Step 1: Sort intervals by end time

   Sorted intervals: [1,2], [1,3], [2,3], [3,4]

2. Step 2: Count overlaps and removals

   Start end=2. Interval [1,3] starts at 1 < 2, count=1. Interval [2,3] starts at 2 ≥ 2, update end=3. Interval [3,4] starts at 3 ≥ 3, update end=4.

3. Final Answer:

   1 -> Option D

4. Quick Check:

   One removal needed = 1 [OK]

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

Common Mistakes:

• Not sorting intervals before counting
• Removing more intervals than needed
• Miscounting overlaps with boundary intervals