Non Overlapping Intervals Minimum Removal in DSA C# - Time & Space Complexity

Time Complexity: Non Overlapping Intervals Minimum Removal

O(n log n)

Understanding Time Complexity

We want to find out how long it takes to decide the minimum number of intervals to remove so that no intervals overlap.

How does the time needed grow as we have more intervals?

Scenario Under Consideration

Analyze the time complexity of the following code snippet.

public int EraseOverlapIntervals(int[][] intervals) {
    Array.Sort(intervals, (a, b) => a[1].CompareTo(b[1]));
    int count = 0, end = int.MinValue;
    foreach (var interval in intervals) {
        if (interval[0] >= end) {
            end = interval[1];
        } else {
            count++;
        }
    }
    return count;
}
    

This code sorts intervals by their end time, then counts how many intervals must be removed to avoid overlaps.

Identify Repeating Operations

Identify the loops, recursion, array traversals that repeat.

• Primary operation: Sorting the intervals array.
• How many times: Sorting takes multiple comparisons, roughly proportional to n log n for n intervals.
• Secondary operation: One loop through all intervals to count removals, runs n times.

How Execution Grows With Input

Sorting grows faster than the loop as input size increases.

Input Size (n)	Approx. Operations
10	About 10 * log(10) ≈ 33 operations for sorting + 10 for loop
100	About 100 * log(100) ≈ 664 operations for sorting + 100 for loop
1000	About 1000 * log(1000) ≈ 9966 operations for sorting + 1000 for loop

Pattern observation: The sorting step dominates and grows faster than the simple loop.

Final Time Complexity

Time Complexity: O(n log n)

This means as the number of intervals grows, the time needed grows a bit faster than just the number of intervals, mainly due to sorting.

Common Mistake

[X] Wrong: "The loop that counts removals is the slowest part, so time is O(n)."

[OK] Correct: Sorting takes more time than the loop, so it controls the overall speed.

Interview Connect

Understanding this time complexity helps you explain how sorting affects performance and why greedy choices after sorting are efficient.

Self-Check

"What if the intervals were already sorted by end time? How would the time complexity change?"

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