DP vs Recursion vs Greedy Choosing the Right Tool in DSA C++ - Compare & Choose

Challenge - 5 Problems

❓ Predict Output

intermediate

Output of Recursive Fibonacci Function

What is the output of this C++ code that calculates Fibonacci numbers using recursion?

DSA C++

#include <iostream>

int fib(int n) {
    if (n <= 1) return n;
    return fib(n-1) + fib(n-2);
}

int main() {
    int result = fib(5);
    std::cout << result << std::endl;
    return 0;
}

A 5

B 8

C 13

D 3

💡 Hint

Trace the recursive calls for fib(5).

The Fibonacci sequence starts 0,1,1,2,3,5,8,... fib(5) = fib(4)+fib(3) = 3+2 = 5.

🧠 Conceptual

intermediate

Choosing Between Greedy and Dynamic Programming

Which problem characteristic suggests that a greedy algorithm is NOT suitable and dynamic programming should be used instead?

A The problem can be solved by making locally optimal choices.

B The problem has no overlapping subproblems.

C The problem has optimal substructure and overlapping subproblems.

D The problem requires sorting input before processing.

💡 Hint

Think about when greedy fails and DP helps.

Dynamic programming is needed when subproblems overlap and optimal substructure exists, which greedy alone cannot handle.

🔧 Debug

advanced

Why Does This Greedy Coin Change Fail?

What is the output of this code that tries to find minimum coins for amount 6 using greedy with coins {1, 3, 4}?

DSA C++

#include <iostream>

int minCoins(int amount) {
    int coins[] = {4, 3, 1};
    int count = 0;
    for (int i = 0; i < 3; i++) {
        while (amount >= coins[i]) {
            amount -= coins[i];
            count++;
        }
    }
    return count;
}

int main() {
    std::cout << minCoins(6) << std::endl;
    return 0;
}

A 3

B 2

C 4

D 1

💡 Hint

Check how greedy picks coins for 6.

Greedy picks 4 once, then 1 twice, total 3 coins. But optimal is 2 coins (3+3).

❓ Predict Output

advanced

Output of Memoized Fibonacci (DP) Implementation

What is the output of this C++ code using memoization to compute fib(6)?

DSA C++

#include <iostream>
#include <vector>

int fib(int n, std::vector<int>& memo) {
    if (n <= 1) return n;
    if (memo[n] != -1) return memo[n];
    memo[n] = fib(n-1, memo) + fib(n-2, memo);
    return memo[n];
}

int main() {
    int n = 6;
    std::vector<int> memo(n+1, -1);
    std::cout << fib(n, memo) << std::endl;
    return 0;
}

A 5

B 8

C 21

D 13

💡 Hint

Memoization avoids repeated calls.

fib(6) = fib(5)+fib(4) = 5+3 = 8.

🧠 Conceptual

expert

Identifying When Greedy Algorithm Guarantees Optimal Solution

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

A The problem requires exploring all possible solutions.

B The problem has overlapping subproblems and no optimal substructure.

C The problem has no subproblems.

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

💡 Hint

Greedy works when local choices lead to global optimum.

Greedy algorithms work optimally when the greedy-choice property and optimal substructure hold.

Practice

1. Which approach is best suited when a problem has overlapping subproblems and optimal substructure?

easy

A. Dynamic Programming

B. Greedy Algorithm

C. Simple Recursion without memoization

D. Brute Force

Solution

1. Step 1: Understand problem characteristics

   Overlapping subproblems means the same smaller problems repeat. Optimal substructure means the best solution can be built from best solutions of smaller problems.

2. Step 2: Match approach to characteristics

   Dynamic Programming stores results of subproblems to avoid recomputation, ideal for overlapping subproblems and optimal substructure.

3. Final Answer:

   Dynamic Programming -> Option A

4. Quick Check:

   Overlapping subproblems + optimal substructure = Dynamic Programming [OK]

Hint: Use DP when subproblems repeat and build optimal solutions [OK]

Common Mistakes:

• Choosing greedy when subproblems overlap
• Using simple recursion without memo causing slow code
• Confusing optimal substructure with greedy choice property

2. Which of the following is the correct syntax to implement memoization in C++ recursion?

easy

A. Use a while loop instead of recursion

B. Use a for loop inside the recursive function

C. Call the recursive function twice without storing results

D. Use a global array or vector to store results and check before recursion

Solution

1. Step 1: Understand memoization concept

   Memoization stores results of recursive calls to avoid repeated work.

2. Step 2: Identify correct syntax usage

   Using a global or passed array/vector to save results and checking it before recursion is the correct way.

3. Final Answer:

   Use a global array or vector to store results and check before recursion -> Option D

4. Quick Check:

   Memoization = store results before recursion [OK]

Hint: Memoize by storing results in array before recursive calls [OK]

Common Mistakes:

• Not storing results causing repeated calculations
• Using loops inside recursion incorrectly
• Calling recursion multiple times without memo

3. What is the output of the following C++ code snippet?

int fib(int n) {
  if (n <= 1) return n;
  return fib(n-1) + fib(n-2);
}

int main() {
  std::cout << fib(4);
  return 0;
}

medium

A. 4

B. 5

C. 3

D. 7

Solution

1. Step 1: Trace fib(4) calls

   fib(4) = fib(3) + fib(2) - fib(3) = fib(2) + fib(1) = (fib(1)+fib(0)) + 1 = (1+0)+1=2 - fib(2) = fib(1) + fib(0) = 1 + 0 = 1 So fib(4) = 2 + 1 = 3

2. Step 2: Confirm output

   The program prints 3 as the result of fib(4).

3. Final Answer:

   3 -> Option C

4. Quick Check:

   fib(4) = 3 [OK]

Hint: Trace small inputs stepwise to find Fibonacci output [OK]

Common Mistakes:

• Confusing fib(4) with fib(5)
• Adding wrong Fibonacci numbers
• Assuming fib(4) equals 4

4. Identify the error in the following code snippet implementing a greedy approach for coin change:

int coinChange(int amount, vector<int>& coins) {
  int count = 0;
  for (int i = 0; i < coins.size(); i++) {
    while (amount >= coins[i]) {
      amount -= coins[i];
      count++;
    }
  }
  return count;
}

medium

A. Coins vector is not sorted in descending order

B. Using while loop inside for loop is incorrect

C. Function should return amount, not count

D. No error, code is correct

Solution

1. Step 1: Analyze greedy coin change logic

   Greedy coin change requires coins sorted descending to pick largest coin first.

2. Step 2: Identify error in code

   The code loops coins in given order; if coins are not sorted descending, greedy fails to minimize count.

3. Final Answer:

   Coins vector is not sorted in descending order -> Option A

4. Quick Check:

   Greedy coin change needs descending sorted coins [OK]

Hint: Sort coins descending before greedy coin change [OK]

Common Mistakes:

• Not sorting coins descending
• Returning wrong variable
• Thinking while loop inside for is wrong

5. You want to find the minimum number of coins to make a given amount using coins of denominations [1, 3, 4]. Which approach is best and why?

hard

A. Greedy approach because picking largest coin first always works

B. Dynamic Programming to store minimum coins needed for sub-amounts

C. Simple recursion without memoization to try all combinations

D. Brute force by checking all subsets of coins

Solution

1. Step 1: Understand problem requirements

   We want minimum coins for amount using coins [1,3,4]. Greedy fails here because picking largest coin first may not give minimum coins.

2. Step 2: Choose best approach

   Dynamic Programming stores minimum coins for all smaller amounts, building up to the target efficiently and correctly.

3. Final Answer:

   Dynamic Programming to store minimum coins needed for sub-amounts -> Option B

4. Quick Check:

   Minimum coins with tricky denominations = DP [OK]

Hint: Use DP when greedy fails to find minimum coins [OK]

Common Mistakes:

• Assuming greedy always works for coin change
• Using recursion without memo causing slow code
• Trying brute force which is inefficient