Non Overlapping Intervals Minimum Removal in DSA C# - Build from Scratch

Non Overlapping Intervals Minimum Removal

📖 Scenario: Imagine you have a schedule with several meetings, each represented as a time interval. Some meetings overlap, and you want to remove the fewest number of meetings so that the remaining meetings do not overlap.

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

📋 What You'll Learn

Create a list of intervals with exact start and end times

Add a variable to count removals

Sort intervals by their end time

Use a loop to find overlapping intervals and count removals

Print the minimum number of intervals to remove

💡 Why This Matters

🌍 Real World

Scheduling meetings, booking rooms, or planning events without conflicts.

💼 Career

Helps in understanding greedy algorithms and interval scheduling problems common in software engineering.

1

Create the list of intervals

Create a list of integer arrays called intervals with these exact intervals: [1, 3], [2, 4], [3, 5], [6, 7], [8, 10], [9, 11].

DSA C#

// Your code here

Hint

Use List<int[]> and add each interval as new int[] {start, end}.

2

Add a variable to count removals

Add an integer variable called removals and set it to 0 to count how many intervals we remove.

DSA C#

using System;
using System.Collections.Generic;

class Program
{
    static void Main()
    {
        List intervals = new List
        {
            new int[] {1, 3},
            new int[] {2, 4},
            new int[] {3, 5},
            new int[] {6, 7},
            new int[] {8, 10},
            new int[] {9, 11}
        };
        // Your code here
    }
}

Hint

Declare int removals = 0; after the intervals list.

3

Sort intervals and count removals

Sort the intervals list by the end time of each interval using intervals.Sort((a, b) => a[1].CompareTo(b[1])). Then create an integer variable prevEnd set to the end time of the first interval. Use a for loop starting from index 1 to the end of intervals. Inside the loop, if the start time of the current interval is less than prevEnd, increment removals. Otherwise, update prevEnd to the end time of the current interval.

DSA C#

using System;
using System.Collections.Generic;

class Program
{
    static void Main()
    {
        List intervals = new List
        {
            new int[] {1, 3},
            new int[] {2, 4},
            new int[] {3, 5},
            new int[] {6, 7},
            new int[] {8, 10},
            new int[] {9, 11}
        };
        int removals = 0;
        // Your code here
    }
}

Hint

Sort by end time, then loop from second interval checking if start is less than previous end.

4

Print the minimum removals

Print the value of removals using Console.WriteLine(removals);.

DSA C#

using System;
using System.Collections.Generic;

class Program
{
    static void Main()
    {
        List intervals = new List
        {
            new int[] {1, 3},
            new int[] {2, 4},
            new int[] {3, 5},
            new int[] {6, 7},
            new int[] {8, 10},
            new int[] {9, 11}
        };
        int removals = 0;
        intervals.Sort((a, b) => a[1].CompareTo(b[1]));
        int prevEnd = intervals[0][1];
        for (int i = 1; i < intervals.Count; i++)
        {
            if (intervals[i][0] < prevEnd)
            {
                removals++;
            }
            else
            {
                prevEnd = intervals[i][1];
            }
        }
        // Your code here
    }
}

Hint

Use Console.WriteLine(removals); to show the result.

Practice

1. What is the main goal of the Non Overlapping Intervals Minimum Removal problem?

easy

A. To sort intervals by their start time

B. To remove the minimum number of intervals so that the remaining intervals do not overlap

C. To merge all intervals into one big interval

D. To find the maximum number of overlapping intervals

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 for any two intervals, one ends before the other starts.

3. Final Answer:

   To remove the minimum number of intervals so that the remaining intervals do not overlap -> Option B

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 overlap with minimum removal
• Sorting intervals without removing any

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

easy

A. Array.Sort(intervals, (a, b) => b[1].CompareTo(a[1]));

B. Array.Sort(intervals, (a, b) => a[0].CompareTo(b[0]));

C. Array.Sort(intervals, (a, b) => b[0].CompareTo(a[0]));

D. Array.Sort(intervals, (a, b) => a[1].CompareTo(b[1]));

Solution

1. Step 1: Identify sorting criteria for minimum removal

   We sort intervals by their end time ascending to greedily keep intervals that finish earliest.

2. Step 2: Match code to sorting by end time ascending

   Array.Sort(intervals, (a, b) => a[1].CompareTo(b[1])); uses a[1].CompareTo(b[1]) which sorts by end time ascending correctly.

3. Final Answer:

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

4. Quick Check:

   Sort by end time ascending = Array.Sort(intervals, (a, b) => a[1].CompareTo(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 comparison

3. What is the output of the following C# code snippet?

int[][] intervals = new int[][] {
  new int[] {1, 3},
  new int[] {2, 4},
  new int[] {3, 5}
};

Array.Sort(intervals, (a, b) => a[1].CompareTo(b[1]));

int count = 0;
int end = intervals[0][1];
for (int i = 1; i < intervals.Length; i++) {
  if (intervals[i][0] < end) {
    count++;
  } else {
    end = intervals[i][1];
  }
}
Console.WriteLine(count);

medium

A. 1

B. 2

C. 0

D. 3

Solution

1. Step 1: Sort intervals by end time

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

2. Step 2: Count overlaps by comparing start with current end

   Check intervals: - i=1: start=2 < end=3 -> overlap, count=1 - i=2: start=3 < end=3? No, update end=5

3. Final Answer:

   1 -> Option A

4. Quick Check:

   One overlap counted = 1 [OK]

Hint: Count intervals starting before current end to find removals [OK]

Common Mistakes:

• Not updating end when no overlap
• Counting intervals that do not overlap
• Sorting incorrectly before loop

4. Identify the bug in the following C# code for minimum interval removal:

int[][] intervals = new int[][] {
  new int[] {1, 2},
  new int[] {1, 3},
  new int[] {2, 4}
};

Array.Sort(intervals, (a, b) => a[0].CompareTo(b[0]));

int count = 0;
int end = intervals[0][1];
for (int i = 1; i < intervals.Length; i++) {
  if (intervals[i][0] < end) {
    count++;
  } else {
    end = intervals[i][1];
  }
}
Console.WriteLine(count);

medium

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

B. Not initializing count to 1 causes wrong output

C. Loop should start from i = 0 instead of i = 1

D. Using intervals[i][0] instead of intervals[i][1] in comparison

Solution

1. Step 1: Check sorting criteria

   The code sorts intervals by start time, but greedy approach requires sorting by end time.

2. Step 2: Understand impact on counting removals

   Sorting by start time can cause wrong overlaps detection and incorrect count.

3. Final Answer:

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

4. Quick Check:

   Sort by end time needed, not start time [OK]

Hint: Always sort by end time ascending for correct greedy removal [OK]

Common Mistakes:

• Sorting by start time instead of end time
• Starting loop at wrong index
• Confusing interval start and end in comparisons

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 ascending

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

2. Step 2: Count overlaps and removals

   Start with end=2 (from [1,2]) - [1,3]: start=1 < end=2 overlap, count=1 - [2,3]: start=2 <= end=2 no overlap, update end=3 - [3,4]: start=3 <= end=3 no overlap, update end=4 Total removals = 1

3. Step 3: Re-check for minimal removals

   Check if removing [1,3] is enough. Yes, only 1 removal needed.

4. Final Answer:

   1 -> Option C

5. Quick Check:

   Remove overlapping [1,3] only = 1 [OK]

Hint: Sort by end time, greedily keep intervals with earliest end [OK]

Common Mistakes:

• Counting all overlapping intervals instead of minimal removals
• Not sorting intervals before processing
• Confusing intervals with same start or end times