Why Intervals Are a Common Problem Pattern in DSA C# - Test Your Knowledge

Practice - 5 Tasks

1fill in blank

easy

Complete the code to create an interval class with start and end properties.

DSA C#

public class Interval {
    public int Start { get; set; }
    public int End { get; set; }

    public Interval(int [1], int end) {
        Start = start;
        End = end;
    }
}

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

Abegin

Bfrom

CintervalStart

Dstart

Attempts:

3 left

💡 Hint

Common Mistakes

Using a parameter name that does not match the property, causing confusion.

Forgetting to assign the parameter to the property.

The constructor parameter should be named 'start' to match the assignment to the Start property.

2fill in blank

medium

Complete the code to check if two intervals overlap.

DSA C#

public bool Overlaps(Interval other) {
    return this.Start <= other.End && other.Start [1] this.End;
}

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

A>

B>=

C<=

D<

Attempts:

3 left

💡 Hint

Common Mistakes

Using strict inequality which misses touching intervals.

Reversing the comparison operators.

Two intervals overlap if this.Start <= other.End and other.Start <= this.End.

3fill in blank

hard

Fix the error in the method that merges two overlapping intervals.

DSA C#

public Interval Merge(Interval other) {
    if (!this.Overlaps(other)) {
        throw new ArgumentException("Intervals do not overlap");
    }
    int mergedStart = Math.Min(this.Start, other.Start);
    int mergedEnd = Math.Max(this.End, [1]);
    return new Interval(mergedStart, mergedEnd);
}

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

Aother.Start

Bother.End

Cthis.Start

Dthis.End

Attempts:

3 left

💡 Hint

Common Mistakes

Using the start values instead of end values for merging.

Using the minimum instead of maximum for the end.

To merge intervals, take the maximum end between this.End and other.End.

4fill in blank

hard

Fill both blanks to create a method that checks if one interval is completely inside another.

DSA C#

public bool IsInside(Interval other) {
    return this.Start [1] other.Start && this.End [2] other.End;
}

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

A>=

B<=

C>

D<

Attempts:

3 left

💡 Hint

Common Mistakes

Using strict inequalities which exclude equal boundaries.

Mixing up the comparison directions.

An interval is inside another if its start is greater or equal and its end is less or equal to the other's.

5fill in blank

hard

Fill both blanks to create a method that returns a dictionary of intervals with their lengths, only for intervals longer than 5.

DSA C#

public Dictionary<Interval, int> GetLongIntervals(List<Interval> intervals) {
    return intervals.Where(i => (i.End - i.Start) [1] 5)
                    .ToDictionary(i => i, i => i.End [2] i.Start);
}

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

A>

B-

C -

D<

Attempts:

3 left

💡 Hint

Common Mistakes

Using '<' instead of '>' in the filter condition.

Using incorrect subtraction or spacing causing syntax errors.

Using the interval object incorrectly as dictionary key.

Filter intervals longer than 5 using '>' and create dictionary with interval as key and length (end - start) as value.

Practice

1. Why are interval problems common in data structures and algorithms?

easy

A. Because intervals do not overlap in any problem

B. Because intervals are easy to sort alphabetically

C. Because intervals always have fixed lengths

D. Because many real-world problems involve ranges or periods of time

Solution

1. Step 1: Understand what intervals represent

   Intervals represent ranges such as time periods or numeric ranges, common in scheduling and resource allocation.

2. Step 2: Recognize real-world applications

   Many problems like meeting rooms, calendar events, or booking systems use intervals to represent availability or usage.

3. Final Answer:

   Because many real-world problems involve ranges or periods of time -> Option D

4. Quick Check:

   Intervals = ranges in real life [OK]

Hint: Think about real-life ranges like meetings or bookings [OK]

Common Mistakes:

• Assuming intervals always have fixed lengths
• Thinking intervals never overlap
• Believing intervals are sorted alphabetically

2. Which of the following is the correct C# syntax to declare an interval as a tuple with start and end integers?

easy

A. var interval = (start: 5, end: 10);

B. var interval = [5, 10];

C. var interval = {5, 10};

D. var interval = (5; 10);

Solution

1. Step 1: Recall tuple syntax in C#

   Tuples use parentheses with comma-separated values, optionally named like (start: 5, end: 10).

2. Step 2: Check each option

   var interval = (start: 5, end: 10); uses correct tuple syntax with named elements. Options A and C use invalid collection syntax. var interval = (5; 10); uses a semicolon which is invalid.

3. Final Answer:

   var interval = (start: 5, end: 10); -> Option A

4. Quick Check:

   Tuple syntax = parentheses with commas [OK]

Hint: Tuples use parentheses and commas, not brackets or braces [OK]

Common Mistakes:

• Using square brackets instead of parentheses
• Using braces which are for collections
• Separating tuple elements with semicolons

3. What is the output of the following C# code that merges overlapping intervals?

var intervals = new List<(int start, int end)> { (1, 3), (2, 6), (8, 10), (15, 18) };
var merged = new List<(int start, int end)>();
intervals.Sort((a, b) => a.start.CompareTo(b.start));
foreach (var interval in intervals) {
  if (merged.Count == 0 || merged[^1].end < interval.start) {
    merged.Add(interval);
  } else {
    merged[^1] = (merged[^1].start, Math.Max(merged[^1].end, interval.end));
  }
}
foreach (var m in merged) {
  Console.Write($"({m.start},{m.end}) ");
}

medium

A. (1,3) (2,6) (8,10) (15,18)

B. (1,6) (8,10) (15,18)

C. (1,6) (2,6) (8,10) (15,18)

D. (1,3) (8,10) (15,18)

Solution

1. Step 1: Sort intervals by start

   Intervals sorted: (1,3), (2,6), (8,10), (15,18).

2. Step 2: Merge overlapping intervals

   First interval (1,3) added. Next (2,6) overlaps with (1,3), merge to (1,6). Then (8,10) no overlap, add. Then (15,18) no overlap, add.

3. Final Answer:

   (1,6) (8,10) (15,18) -> Option B

4. Quick Check:

   Merged intervals = (1,6), (8,10), (15,18) [OK]

Hint: Sort intervals first, then merge overlapping by comparing ends [OK]

Common Mistakes:

• Not sorting intervals before merging
• Overwriting intervals incorrectly
• Printing original intervals without merging

4. Identify the bug in this C# code snippet that tries to find if two intervals overlap:

bool Overlaps((int start, int end) a, (int start, int end) b) {
  return a.end < b.start && b.end > a.start;
}

medium

A. The condition logic is reversed; it should check if intervals do not overlap

B. The condition should use >= and <= for inclusive overlap

C. The condition should use || instead of &&

D. The function should return void, not bool

Solution

1. Step 1: Understand overlap condition

   Two intervals overlap if they share any points. The given condition checks for non-overlap incorrectly.

2. Step 2: Correct the logic

   Correct overlap check is !(a.end < b.start || b.end < a.start), meaning intervals do overlap if they are not completely apart.

3. Final Answer:

   The condition logic is reversed; it should check if intervals do not overlap -> Option A

4. Quick Check:

   Overlap means intervals share points, not separated [OK]

Hint: Overlap means intervals are not completely apart [OK]

Common Mistakes:

• Using && instead of || in condition
• Ignoring inclusive boundaries
• Returning wrong data type

5. Given a list of intervals representing booked meeting times, how would you find the minimum number of meeting rooms required to hold all meetings without overlap?

hard

A. Merge all intervals and count the total length

B. Sort intervals by end time, count intervals with same start time

C. Sort intervals by start time, use a min-heap to track end times, count max concurrent meetings

D. Use a stack to push intervals and pop when they overlap

Solution

1. Step 1: Sort intervals by start time

   Sorting helps process meetings in chronological order.

2. Step 2: Use a min-heap to track earliest ending meeting

   For each meeting, if it starts after the earliest ending meeting, reuse room; else allocate new room. The heap size tracks rooms needed.

3. Final Answer:

   Sort intervals by start time, use a min-heap to track end times, count max concurrent meetings -> Option C

4. Quick Check:

   Min rooms = max concurrent meetings [OK]

Hint: Track earliest end times with min-heap to count rooms [OK]

Common Mistakes:

• Merging intervals instead of counting overlaps
• Using stack which doesn't track end times properly
• Counting intervals with same start time only