Non Overlapping Intervals Minimum Removal in DSA Go - Build from Scratch

Non Overlapping Intervals Minimum Removal

📖 Scenario: You are managing a schedule of meetings represented as time intervals. Some meetings overlap, causing conflicts. You want to remove the minimum number of meetings so that the remaining meetings do not overlap.

🎯 Goal: Build a Go program that finds the minimum number of intervals to remove to make the rest non-overlapping.

📋 What You'll Learn

Create a slice of intervals with exact values

Add a variable to count removals

Implement the logic to find minimum removals using sorting and iteration

Print the final count of intervals to remove

💡 Why This Matters

🌍 Real World

Scheduling meetings, booking rooms, or managing time slots without conflicts.

💼 Career

Common interview problem testing understanding of greedy algorithms and interval scheduling.

1

Create the intervals slice

Create a slice of intervals called intervals with these exact values: {1, 3}, {2, 4}, {3, 5}, {6, 7}, {8, 10}, {9, 11}. Use a struct type Interval with start and end fields of type int.

DSA Go

package main

import "fmt"

type Interval struct {
    start int
    end   int
}

func main() {
    // Create the intervals slice with given values
    // Your code here
}

Hint

Define a struct Interval with start and end fields. Then create a slice intervals with the exact intervals listed.

2

Add a removal counter variable

Add an integer variable called removeCount and set it to 0 to count how many intervals need to be removed.

DSA Go

package main

import "fmt"

type Interval struct {
    start int
    end   int
}

func main() {
    intervals := []Interval{
        {1, 3},
        {2, 4},
        {3, 5},
        {6, 7},
        {8, 10},
        {9, 11},
    }
    // Add a variable removeCount and set it to 0
    // Your code here
}

Hint

Just create a variable removeCount and assign it zero.

3

Implement the minimum removal logic

Sort the intervals slice by the end field in ascending order. Then use a for loop with index i starting from 1 to iterate through intervals. Keep track of the end of the last added interval in a variable end initialized to intervals[0].end. For each interval, if its start is less than end, increment removeCount. Otherwise, update end to the current interval's end.

DSA Go

package main

import (
    "fmt"
    "sort"
)

type Interval struct {
    start int
    end   int
}

func main() {
    intervals := []Interval{
        {1, 3},
        {2, 4},
        {3, 5},
        {6, 7},
        {8, 10},
        {9, 11},
    }
    removeCount := 0

    // Sort intervals by end time
    // Use a for loop from i=1 to len(intervals)-1
    // Track end of last added interval
    // Increment removeCount if overlap found
    // Your code here
}

Hint

Sort intervals by their end time. Then loop from the second interval, compare start with last end, count removals if overlapping.

4

Print the minimum number of removals

Print the value of removeCount using fmt.Println(removeCount).

DSA Go

package main

import (
    "fmt"
    "sort"
)

type Interval struct {
    start int
    end   int
}

func main() {
    intervals := []Interval{
        {1, 3},
        {2, 4},
        {3, 5},
        {6, 7},
        {8, 10},
        {9, 11},
    }
    removeCount := 0

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

    end := intervals[0].end
    for i := 1; i < len(intervals); i++ {
        if intervals[i].start < end {
            removeCount++
        } else {
            end = intervals[i].end
        }
    }

    // Print the removeCount
    // Your code here
}

Hint

Use fmt.Println(removeCount) to show the final answer.

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