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-to-weight ratio in the greedy algorithm for the fractional knapsack problem.

DSA C++

int selectNextItem(vector<Item>& items) {
    int bestIndex = -1;
    double bestRatio = 0.0;
    for (int i = 0; i < items.size(); i++) {
        if (!items[i].taken) {
            double ratio = items[i].value [1] items[i].weight;
            if (ratio > bestRatio) {
                bestRatio = ratio;
                bestIndex = i;
            }
        }
    }
    return bestIndex;
}

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

A/

B*

C+

D-

Attempts:

3 left

💡 Hint

Common Mistakes

Using multiplication instead of division for ratio calculation.

Using addition or subtraction which does not represent ratio.

The greedy choice is based on the value-to-weight ratio, which is calculated by dividing value by weight.

2fill in blank

medium

Complete the code to check if the greedy algorithm can pick the entire item or only a fraction of it based on remaining capacity.

DSA C++

if (items[i].weight [1] capacity) {
    totalValue += items[i].value;
    capacity -= items[i].weight;
} else {
    totalValue += items[i].value * ((double)capacity / items[i].weight);
    capacity = 0;
}

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

A>

B<

C<=

D>=

Attempts:

3 left

💡 Hint

Common Mistakes

Using greater than or greater than or equal which reverses the logic.

Using less than which excludes the exact fit case.

We check if the item's weight is less than or equal to the remaining capacity to take it fully.

3fill in blank

hard

Fix the error in the greedy coin change algorithm to pick the largest coin not exceeding the remaining amount.

DSA C++

for (int i = 0; i < coins.size(); i++) {
    while (coins[i] [1] amount) {
        amount -= coins[i];
        count++;
    }
}

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

A<=

B>

C<

D>=

Attempts:

3 left

💡 Hint

Common Mistakes

Using greater than which never allows picking coins.

Using less than which excludes coins equal to amount.

We pick coins while the coin value is less than or equal to the remaining amount.

4fill in blank

hard

Fill both blanks to create a dictionary comprehension that maps words to their lengths only if the length is greater than 3.

DSA C++

{word: [1] for word in words if len(word) [2] 3}

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

Alen(word)

B>

C<

Dword

Attempts:

3 left

💡 Hint

Common Mistakes

Using the word itself as value instead of length.

Using less than instead of greater than in condition.

We map each word to its length and filter words with length greater than 3.

5fill in blank

hard

Fill all three blanks to create a dictionary comprehension that maps uppercase keys to values only if the value is positive.

DSA C++

result = { [1]: [2] for k, v in data.items() if v [3] 0 }

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

Ak.upper()

Bv

C>

Dk.lower()

Attempts:

3 left

💡 Hint

Common Mistakes

Using lowercase keys instead of uppercase.

Using less than or equal instead of greater than zero.

Keys are converted to uppercase, values are kept as is, and only positive values are included.

Practice

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

easy

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

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

C. Because it uses recursion to explore all paths

D. Because it sorts the input data before processing

Solution

1. Step 1: Understand greedy choice property

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

2. Step 2: Connect greedy choice to global optimum

   This local best choice leads to a global optimum only if the problem has the greedy choice property and optimal substructure.

3. Final Answer:

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

4. Quick Check:

   Greedy = local best choice [OK]

Hint: Greedy picks best now, not all possibilities [OK]

Common Mistakes:

• Thinking greedy tries all solutions
• Confusing greedy with backtracking
• Assuming sorting alone makes greedy work

2. Which of the following is the correct syntax to select the next item in a greedy algorithm loop in C++?

easy

A. for (auto item : items) { choose item; }

B. for (int i = 0 i < n; i++) { choose item[i]; }

C. while (i <= n) { choose item[i]; i++; }

D. do { choose item[i]; i--; } while (i > 0);

Solution

1. Step 1: Identify correct C++ loop syntax for iteration

   The range-based for loop for (auto item : items) is modern and correct for iterating over a container.

2. Step 2: Check other options for correctness

   for (int i = 0 i < n; i++) { choose item[i]; } is missing semicolons and proper syntax. while (i <= n) { choose item[i]; i++; } uses while but lacks initialization shown. do { choose item[i]; i--; } while (i > 0); decrements i incorrectly.

3. Final Answer:

   for (auto item : items) { choose item; } -> Option A

4. Quick Check:

   Range-based for loop is correct syntax [OK]

Hint: Use range-based for loops for clean iteration [OK]

Common Mistakes:

• Missing semicolons in for loop
• Using wrong loop conditions
• Decrementing index when should increment

3. Consider the following C++ code snippet for selecting intervals greedily by earliest finish time:

struct Interval {
  int start, end;
};

bool compare(Interval a, Interval b) {
  return a.end < b.end;
}

int main() {
  std::vector<Interval> intervals = {{1,4}, {3,5}, {0,6}, {5,7}, {8,9}};
  std::sort(intervals.begin(), intervals.end(), compare);
  int count = 0, last_end = -1;
  for (auto interval : intervals) {
    if (interval.start >= last_end) {
      count++;
      last_end = interval.end;
    }
  }
  std::cout << count << std::endl;
  return 0;
}

What is the output of this program?

medium

A. 5

B. 4

C. 3

D. 2

Solution

1. Step 1: Sort intervals by end time

   Sorted intervals by end: {1,4}, {3,5}, {0,6}, {5,7}, {8,9}

2. Step 2: Select intervals greedily

   Select {1,4} (count=1, last_end=4), skip {3,5} (start 3 < 4), skip {0,6} (start 0 < 4), select {5,7} (start 5 >= 4, count=2, last_end=7), select {8,9} (start 8 >= 7, count=3, last_end=9)

3. Step 3: Count total selected intervals

   Total selected intervals = 3

4. Final Answer:

   3 -> Option C

5. Quick Check:

   Greedy interval selection count = 3 [OK]

Hint: Sort by finish time, pick non-overlapping intervals [OK]

Common Mistakes:

• Counting overlapping intervals
• Not sorting intervals correctly
• Updating last_end incorrectly

4. The following code attempts to solve the coin change problem using a greedy approach:

int coins[] = {25, 10, 5, 1};
int amount = 30;
int count = 0;
for (int i = 0; i < 4; i++) {
  while (amount >= coins[i]) {
    amount -= coins[i];
    count++;
  }
}
std::cout << count << std::endl;

What is the main issue with this greedy approach?

medium

A. It always gives the minimum number of coins

B. It may fail for some coin sets where greedy is not optimal

C. It causes a syntax error due to missing semicolons

D. It runs infinitely because amount never decreases

Solution

1. Step 1: Understand greedy coin change approach

   The code picks largest coin repeatedly until amount is reduced to zero.

2. Step 2: Identify greedy failure cases

   Greedy works for US coins but fails for some coin sets where smaller coins combined give fewer coins than greedy picks.

3. Final Answer:

   It may fail for some coin sets where greedy is not optimal -> Option B

4. Quick Check:

   Greedy coin change can fail [OK]

Hint: Greedy coin change fails on non-canonical coin sets [OK]

Common Mistakes:

• Assuming greedy always gives minimum coins
• Confusing syntax errors with logic errors
• Ignoring infinite loop possibility

5. You want to select activities to attend from a list with start and end times. The greedy algorithm picks the earliest finishing activity first. However, if activities have the same end time, it picks the one with the earliest start time. Why might this tie-break rule cause the greedy algorithm to fail in some cases?

hard

A. Because tie-break rules do not affect greedy correctness

B. Because earliest start time always leads to the best solution

C. Because sorting by end time is not needed in greedy algorithms

D. Because picking earliest start time among ties may block more future activities

Solution

1. Step 1: Understand tie-breaking in activity selection

   Greedy sorts by earliest finish time. Ties in end time use earliest start as tie-breaker.

2. Step 2: Analyze effect on greedy selection

   Picking the earliest start time among tied end times may block more future activities, reducing the total number of activities selected.

3. Step 3: Confirm impact on optimality

   This tie-break rule can cause the greedy algorithm to fail to find the maximum set of non-overlapping activities.

4. Final Answer:

   Because picking earliest start time among ties may block more future activities -> Option D

5. Quick Check:

   Tie-break can affect future selections [OK]

Hint: Tie-breaks can affect greedy correctness when ends tie [OK]

Common Mistakes:

• Thinking tie-breaks do not affect optimality
• Assuming all tie-breaks are harmless
• Ignoring impact of start times on future choices