Non Overlapping Intervals Minimum Removal in DSA Go - Interactive Practice

Practice - 5 Tasks

1fill in blank

easy

Complete the code to sort intervals by their end time.

DSA Go

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

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

A<

B==

C>

D!=

Attempts:

3 left

💡 Hint

Common Mistakes

Using '>' instead of '<' causes sorting in descending order, which breaks the algorithm.

Using '==' or '!=' does not sort properly.

We sort intervals by their end time in ascending order to efficiently find non-overlapping intervals.

2fill in blank

medium

Complete the code to skip overlapping intervals by checking start time.

DSA Go

if intervals[i].Start [1] end {
    count++
} else {
    end = intervals[i].End
}

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

A<=

B>

C<

D>=

Attempts:

3 left

💡 Hint

Common Mistakes

Using '<=' causes wrong overlap detection.

Using '<' or '>' reverses the logic.

If the current interval starts before the last selected interval ends, it overlaps and should be removed.

3fill in blank

hard

Fix the error in updating the end time after selecting a non-overlapping interval.

DSA Go

end = intervals[[1]].End

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

Ai - 1

Bi

C0

Dlen(intervals) - 1

Attempts:

3 left

💡 Hint

Common Mistakes

Using i - 1 causes index errors or wrong interval end time.

Using 0 or last index does not update correctly.

We update end to the end time of the current interval at index i after confirming no overlap.

4fill in blank

hard

Fill both blanks to initialize variables for counting removals and tracking end time.

DSA Go

count, end := 0, intervals[[1]].End
for i := [2]; i < len(intervals); i++ {
    // loop body
}

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

A0

B1

Clen(intervals) - 1

D2

Attempts:

3 left

💡 Hint

Common Mistakes

Starting loop at 0 causes comparing the first interval with itself.

Setting end to intervals[1].End skips the first interval.

We start count at 0 and set end to the end time of the first interval at index 0. The loop starts from index 1.

5fill in blank

hard

Fill all three blanks to complete the function that returns minimum removals for non-overlapping intervals.

DSA Go

func eraseOverlapIntervals(intervals []Interval) int {
    if len(intervals) == 0 {
        return 0
    }
    sort.Slice(intervals, func(i, j int) bool {
        return intervals[i].[1] < intervals[j].[2]
    })
    count, end := 0, intervals[0].End
    for i := 1; i < len(intervals); i++ {
        if intervals[i].Start < end {
            count++
        } else {
            end = intervals[i].[3]
        }
    }
    return count
}

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

AEnd

BStart

Attempts:

3 left

💡 Hint

Common Mistakes

Sorting by Start time breaks the greedy approach.

Updating end with Start time causes wrong overlap checks.

We sort intervals by their End time, so blanks 1 and 2 are 'End'. When updating end, we use the current interval's End time.

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