Non Overlapping Intervals Minimum Removal in DSA C# - Interactive Practice

Practice - 5 Tasks

1fill in blank

easy

Complete the code to sort intervals by their end time.

DSA C#

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

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

A1

B0

C2

Da

Attempts:

3 left

💡 Hint

Common Mistakes

Using the wrong index for sorting (start instead of end).

Confusing the order of a and b in CompareTo.

We sort intervals by their end time, which is at index 1. The code compares a[{{BLANK_1}}] with b[{{BLANK_1}}], so {{BLANK_1}} should be 1.

2fill in blank

medium

Complete the code to initialize the count of removals to zero.

DSA C#

int removals = [1];

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A-1

B1

C0

Dintervals.Length

Attempts:

3 left

💡 Hint

Common Mistakes

Starting removals at 1 instead of 0.

Initializing removals with the length of intervals.

We start with zero removals because initially we haven't removed any intervals.

3fill in blank

hard

Fix the error in the if condition to detect overlapping intervals.

DSA C#

if (intervals[i][[1]] < intervals[prev][1])

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A1

B0

Ci

Dprev

Attempts:

3 left

💡 Hint

Common Mistakes

Comparing end time with end time.

Using wrong indices for start or end.

To check if the current interval starts before the previous interval ends, compare the start time of current (index 0) with end time of previous (index 1).

4fill in blank

hard

Fill both blanks to update the previous interval index and increment removals when overlap is found.

DSA C#

if (intervals[i][1] < intervals[prev][1]) prev = [1];
removals[2];

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Ai

B++

C+=

Dprev

Attempts:

3 left

💡 Hint

Common Mistakes

Using += without a number.

Updating prev incorrectly.

When current interval ends earlier, update prev to i. Then increment removals by 1 using ++.

5fill in blank

hard

Fill all three blanks to complete the loop that checks intervals and returns the removal count.

DSA C#

for (int [1] = 1; [2] < intervals.Length; [3]++) {
    if (intervals[i][0] < intervals[prev][1]) {
        if (intervals[i][1] < intervals[prev][1]) prev = i;
        removals++;
    } else {
        prev = i;
    }
}
return removals;

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

Ai

Dj

Attempts:

3 left

💡 Hint

Common Mistakes

Using different variable names in loop parts.

Starting loop from 0 instead of 1.

The loop variable is i, starting from 1, running while i < intervals.Length, incrementing i each time.

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