Why Greedy Works and When It Fails in DSA C++ - Challenge Your Understanding

Challenge - 5 Problems

❓ Predict Output

intermediate

Output of Greedy Coin Change Algorithm

What is the output of the following C++ code that uses a greedy approach to make change with coin denominations 1, 5, and 10 for amount 18?

DSA C++

#include <iostream>
#include <vector>
using namespace std;

int main() {
    vector<int> coins = {10, 5, 1};
    int amount = 18;
    vector<int> result;
    for (int coin : coins) {
        while (amount >= coin) {
            amount -= coin;
            result.push_back(coin);
        }
    }
    for (int c : result) cout << c << " ";
    return 0;
}

A 10 5 1 1 1

B 10 5 5

C 5 5 5 1 1 1

D 10 1 1 1 1 1 1 1 1

💡 Hint

Think about how the greedy algorithm picks the largest coin first until it cannot anymore.

The greedy algorithm picks 10 first (amount becomes 8), then 5 (amount becomes 3), then three 1s to make the total 18.

🧠 Conceptual

intermediate

Why Greedy Fails for Some Coin Sets

Why does the greedy algorithm fail to find the minimum number of coins for the amount 6 when coin denominations are 1, 3, and 4?

A Because greedy picks the largest coin first, it chooses 4 + 1 + 1 = 3 coins instead of 3 + 3 = 2 coins.

B Because greedy always picks the smallest coin first, it ends up with more coins than needed.

C Because greedy uses dynamic programming internally, it sometimes overcounts coins.

D Because greedy sorts coins in ascending order, it misses the optimal combination.

💡 Hint

Consider how greedy picks coins and compare with the optimal solution.

Greedy picks the largest coin (4), then two 1s, totaling 3 coins. The optimal is two 3s, totaling 2 coins.

🔧 Debug

advanced

Identify the Bug in Greedy Interval Scheduling

What error does the following C++ code produce when trying to select the maximum number of non-overlapping intervals using a greedy approach?

DSA C++

#include <iostream>
#include <vector>
#include <algorithm>
using namespace std;

int main() {
    vector<pair<int,int>> intervals = {{1,4}, {3,5}, {0,6}, {5,7}, {8,9}};
    sort(intervals.begin(), intervals.end());
    int count = 0;
    int end_time = 0;
    for (auto interval : intervals) {
        if (interval.first >= end_time) {
            count++;
            end_time = interval.second;
        }
    }
    cout << count << endl;
    return 0;
}

A The code outputs 4, which is the correct maximum number of intervals.

B The code outputs 2, which is less than the maximum possible.

C The code causes a runtime error due to invalid iterator usage.

D The code outputs 3, which is incorrect for maximum intervals.

💡 Hint

Check how intervals are sorted and what property greedy needs to work correctly.

The code sorts intervals by start time, but greedy interval scheduling requires sorting by finish time to maximize intervals selected.

🚀 Application

advanced

Choosing Tasks with Deadlines and Profits

Given tasks with deadlines and profits, which greedy strategy correctly maximizes total profit?

A Schedule tasks randomly and pick the highest profit tasks after scheduling.

B Sort tasks by earliest deadline and schedule them in that order regardless of profit.

C Sort tasks by decreasing profit and schedule each task at the latest available slot before its deadline.

D Sort tasks by increasing profit and schedule each at the earliest available slot.

💡 Hint

Think about how to maximize profit while respecting deadlines.

Sorting by profit descending and placing tasks as late as possible before their deadline ensures maximum profit without conflicts.

🧠 Conceptual

expert

When Does Greedy Guarantee Optimality?

Which condition ensures that a greedy algorithm always produces an optimal solution?

A The problem can be solved by brute force in polynomial time.

B The problem has overlapping subproblems and uses memoization.

C The problem requires backtracking to explore all solutions.

D The problem exhibits the greedy-choice property and optimal substructure.

💡 Hint

Recall the two main properties that justify greedy algorithms.

Greedy algorithms work optimally when the problem has the greedy-choice property (local choices lead to global optimum) and optimal substructure (optimal solution contains optimal subsolutions).

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