Why Greedy Works and When It Fails in DSA C# - Test Your Knowledge

Practice - 5 Tasks

1fill in blank

easy

Complete the code to select the next item with the highest value per weight ratio.

DSA C#

items.Sort((a, b) => b.Value / b.Weight [1] a.Value / a.Weight);

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

A>

B-

C<

D+

Attempts:

3 left

💡 Hint

Common Mistakes

Using '+' or '-' instead of comparison operators.

Sorting in ascending order by mistake.

We sort items so that those with higher value-to-weight ratio come first. Using > ensures descending order.

2fill in blank

medium

Complete the code to pick items greedily until the capacity is full.

DSA C#

while (capacity > 0 && i < items.Count) { var take = Math.Min(capacity, items[i].Weight); result += take * items[i].Value / items[i].Weight; capacity [1]= take; i++; }

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

A+

B/

C*

D-

Attempts:

3 left

💡 Hint

Common Mistakes

Adding to capacity instead of subtracting.

Multiplying or dividing capacity incorrectly.

We reduce capacity by the weight taken, so we subtract 'take' from capacity.

3fill in blank

hard

Fix the error in the greedy coin change algorithm to pick the largest coin first.

DSA C#

coins.Sort(); coins.Reverse(); foreach (var coin in coins) { while (amount [1] coin) { amount -= coin; count++; } }

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

A>

B<=

C>=

D<

Attempts:

3 left

💡 Hint

Common Mistakes

Using < or <= which stops too early.

Using > which excludes equal amounts.

We pick coins while the amount is at least the coin value, so use >=.

4fill in blank

hard

Fill both blanks to create a dictionary comprehension that filters items with value greater than 10.

DSA C#

var filtered = items.Where(item => item.Value [1] 10).ToDictionary(item => item.Key, item => item.Value [2] 0);

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

A>

B<

C+

D-

Attempts:

3 left

💡 Hint

Common Mistakes

Using < to filter incorrectly.

Subtracting value instead of adding.

We filter items with value > 10 and add 0 to the value when creating the dictionary (no change).

5fill in blank

hard

Fill all three blanks to implement a greedy interval scheduling algorithm selecting intervals with earliest finish time.

DSA C#

intervals.Sort((a, b) => a.End [1] b.End); var count = 0; int lastEnd = int.MinValue; foreach (var interval in intervals) { if (interval.Start [2] lastEnd) { count[3]1; lastEnd = interval.End; } }

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

A<

B>

C++

D+=

Attempts:

3 left

💡 Hint

Common Mistakes

Sorting descending instead of ascending.

Using < instead of > in condition.

Using ++ incorrectly on count.

Sort by earliest end time using <. Select intervals starting after last selected ends using >. Increment count by 1 using +=.

Practice

1. What is the main reason a greedy algorithm works correctly for some problems?

easy

A. Because it sorts the input data before processing

B. Because it tries all possible solutions and picks the best

C. Because it makes the best choice at each step hoping for a global optimum

D. Because it uses recursion to explore all paths

Solution

1. Step 1: Understand greedy choice property

   A greedy algorithm picks the best local option at each step without reconsidering previous choices.

2. Step 2: Connect local choices to global optimum

   For some problems, these local best choices lead to the overall best solution, which is why greedy works.

3. Final Answer:

   Because it makes the best choice at each step hoping for a global optimum -> Option C

4. Quick Check:

   Greedy = best local choice each step [OK]

Hint: Greedy picks best now, hoping for best overall [OK]

Common Mistakes:

• Thinking greedy tries all solutions
• Confusing greedy with backtracking
• Assuming sorting always means greedy

2. Which of the following is the correct syntax to start a greedy algorithm loop in C# to process a sorted array named arr?

easy

A. foreach (int i in arr.Length)

B. for (int i = 0; i < arr.Length; i++)

C. while (i <= arr.Length)

D. for (int i = 1; i <= arr.Length; i++)

Solution

1. Step 1: Identify correct for loop syntax in C#

   The standard for loop iterates from 0 to less than array length using i < arr.Length.

2. Step 2: Check each option for syntax correctness

   for (int i = 0; i < arr.Length; i++) uses correct syntax; B uses foreach incorrectly; C uses while with undefined i; D uses 1-based index which is off by one.

3. Final Answer:

   for (int i = 0; i < arr.Length; i++) -> Option B

4. Quick Check:

   for loop with i from 0 to length-1 [OK]

Hint: Use for loop from 0 to less than length [OK]

Common Mistakes:

• Using foreach with arr.Length instead of arr
• Starting loop at 1 instead of 0
• Using while without initializing i

3. Consider the following C# code that tries to select the maximum number of non-overlapping intervals using a greedy approach:

var intervals = new List<(int start, int end)> { (1,3), (2,5), (4,7), (6,9) };
intervals.Sort((a,b) => a.end.CompareTo(b.end));
var count = 0;
int lastEnd = int.MinValue;
foreach (var interval in intervals) {
  if (interval.start >= lastEnd) {
    count++;
    lastEnd = interval.end;
  }
}
Console.WriteLine(count);

What is the output of this code?

medium

A. 2

B. 1

C. 4

D. 3

Solution

1. Step 1: Sort intervals by end time

   Sorted intervals: (1,3), (2,5), (4,7), (6,9)

2. Step 2: Select intervals greedily

   Pick (1,3) first (count=1, lastEnd=3). Next (2,5) starts before 3, skip. Next (4,7) starts after 3, pick (count=2, lastEnd=7). Next (6,9) starts after 7, pick (count=3, lastEnd=9).

3. Final Answer:

   3 -> Option D

4. Quick Check:

   Greedy interval selection count = 3 [OK]

Hint: Sort by end, pick intervals starting after last end [OK]

Common Mistakes:

• Counting overlapping intervals
• Not sorting by end time
• Updating lastEnd incorrectly

4. The following C# code tries to solve the coin change problem using a greedy approach but sometimes fails. What is the bug?

int[] coins = {1, 3, 4};
int amount = 6;
int count = 0;
for (int i = coins.Length - 1; i >= 0; i--) {
  while (amount >= coins[i]) {
    amount -= coins[i];
    count++;
  }
}
Console.WriteLine(count);

medium

A. Greedy fails because coin denominations are not standard, so it misses optimal solution

B. The loop should start from 0, not from coins.Length - 1

C. The while condition should be amount > coins[i], not >=

D. The count variable is not initialized to zero

Solution

1. Step 1: Understand greedy coin change approach

   Greedy picks largest coin first repeatedly until amount is less, then smaller coins.

2. Step 2: Check coin denominations and optimality

   With coins {1,3,4} and amount 6, greedy picks 4 + 1 + 1 = 3 coins, but optimal is 3 + 3 = 2 coins.

3. Final Answer:

   Greedy fails because coin denominations are not standard, so it misses optimal solution -> Option A

4. Quick Check:

   Non-standard coins break greedy optimality [OK]

Hint: Greedy fails if coin set is non-standard [OK]

Common Mistakes:

• Thinking loop direction causes failure
• Confusing while condition with off-by-one
• Ignoring coin denomination impact

5. You want to schedule the maximum number of tasks without overlap. Each task has a start and end time. Which approach guarantees the optimal solution and why?

Options:
A) Sort tasks by start time and pick earliest starting tasks
B) Sort tasks by duration and pick shortest tasks first
C) Sort tasks by end time and pick tasks finishing earliest first
D) Pick tasks randomly and check for overlaps

hard

A. Sort tasks by end time and pick tasks finishing earliest first

B. Sort tasks by duration and pick shortest tasks first

C. Sort tasks by start time and pick earliest starting tasks

D. Pick tasks randomly and check for overlaps

Solution

1. Step 1: Understand the interval scheduling problem

   The goal is to select maximum non-overlapping tasks.

2. Step 2: Identify greedy strategy that works

   Sorting by end time and picking earliest finishing tasks ensures maximum room for future tasks, proven optimal.

3. Step 3: Evaluate other options

   Sorting by start time or duration can lead to suboptimal selections; random picks are unreliable.

4. Final Answer:

   Sort tasks by end time and pick tasks finishing earliest first -> Option A

5. Quick Check:

   Earliest finish time greedy = optimal schedule [OK]

Hint: Pick tasks finishing earliest first for max schedule [OK]

Common Mistakes:

• Choosing earliest start time instead of end time
• Picking shortest duration tasks ignoring overlaps
• Assuming random picks can be optimal