DP vs Recursion vs Greedy Choosing the Right Tool in DSA C - Interactive Comparison Practice

Practice - 5 Tasks

1fill in blank

easy

Complete the code to calculate Fibonacci numbers using simple recursion.

DSA C

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

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

A+

B-

C*

D/

Attempts:

3 left

💡 Hint

Common Mistakes

Using subtraction or multiplication instead of addition.

Forgetting to return the sum.

The Fibonacci number is the sum of the two previous Fibonacci numbers, so we add the results of fib(n-1) and fib(n-2).

2fill in blank

medium

Complete the code to implement memoization in Fibonacci calculation.

DSA C

int fibMemo(int n, int memo[]) {
    if (n <= 1) {
        return n;
    }
    if (memo[n] != -1) {
        return memo[n];
    }
    memo[n] = fibMemo(n - 1, memo) [1] fibMemo(n - 2, memo);
    return memo[n];
}

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

A*

B-

C+

D/

Attempts:

3 left

💡 Hint

Common Mistakes

Not storing the sum in memo array.

Using wrong operators causing incorrect results.

Memoization stores the sum of fib(n-1) and fib(n-2) to avoid repeated calculations.

3fill in blank

hard

Fix the error in the greedy coin change function to return minimum coins needed.

DSA C

int minCoins(int coins[], int n, int amount) {
    int count = 0;
    for (int i = n - 1; i >= 0; i--) {
        while (amount >= coins[i]) {
            amount [1] coins[i];
            count++;
        }
    }
    return amount == 0 ? count : -1;
}

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

A-=

B/=

C*=

D+=

Attempts:

3 left

💡 Hint

Common Mistakes

Using '+=' which increases amount causing infinite loop.

Using multiplication or division operators incorrectly.

To reduce the amount by the coin value, we subtract coins[i] from amount using '-=' operator.

4fill in blank

hard

Fill both blanks to implement bottom-up DP for Fibonacci numbers.

DSA C

int fibDP(int n) {
    int dp[n + 1];
    dp[0] = 0;
    dp[1] = 1;
    for (int i = 2; i <= n; i++) {
        dp[i] = dp[i [1] 1] [2] dp[i - 2];
    }
    return dp[n];
}

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

A-

B+

C*

D/

Attempts:

3 left

💡 Hint

Common Mistakes

Using wrong operators causing index errors.

Mixing multiplication or division instead of addition.

We calculate dp[i] by adding dp[i-1] and dp[i-2]. The first blank is '-' to access previous index, second blank is '+' to add values.

5fill in blank

hard

Fill all three blanks to create a greedy algorithm for activity selection problem.

DSA C

void selectActivities(int start[], int finish[], int n) {
    int i = 0;
    printf("Activity %d selected\n", i);
    for (int j = 1; j < n; j++) {
        if (start[j] [1] finish[[3]]) {
            printf("Activity %d selected\n", j);
            i = [2];
        }
    }
}

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

A>=

Bj

C<=

Di

Attempts:

3 left

💡 Hint

Common Mistakes

Using wrong comparison operators causing wrong selections.

Not updating the last selected activity index correctly.

We select activity j if its start time is greater or equal to finish time of last selected activity i. Then update i to j.

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 mean 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 repeated work, 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 memoization causing slow code
• Confusing optimal substructure with greedy choice property

2. Which of the following is the correct syntax to implement memoization in C for a recursive Fibonacci function?

easy

A. Use a global array to store results and check before recursive calls

B. Use a local variable inside the function to store results

C. Use a static variable without initialization inside the function

D. Use a pointer without allocating memory for memoization

Solution

1. Step 1: Understand memoization storage

   Memoization requires storing results outside the recursive calls to reuse them. A global or static array is suitable.

2. Step 2: Check options for correct usage

   Only a global array checked before recursion ensures stored results are reused properly. Local variables reset each call, static without init may cause errors, pointers need allocation.

3. Final Answer:

   Use a global array to store results and check before recursive calls -> Option A

4. Quick Check:

   Memoization needs persistent storage = global array [OK]

Hint: Memoize with global/static array checked before recursion [OK]

Common Mistakes:

• Using local variables that reset each call
• Not initializing static variables properly
• Using pointers without memory allocation

3. What will be the output of the following C code snippet implementing a greedy approach for coin change?

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++;
    }
}
printf("%d\n", count);

medium

A. 3

B. 2

C. 4

D. 5

Solution

1. Step 1: Trace the greedy coin selection

   Start with 30 amount. Use 25 coin once (30-25=5), count=1. Then 10 coin can't be used (5<10). Use 5 coin once (5-5=0), count=2.

2. Step 2: Count total coins used

   Coins used: 25 (1), 5 (1), total count = 2 coins. The code increments count inside while loop, so count=2.

3. Step 3: Check for remaining coins

   Amount is now 0, so no more coins used.

4. Final Answer:

   2 -> Option B

5. Quick Check:

   Greedy coin count for 30 = 2 coins [OK]

Hint: Subtract largest coin repeatedly until amount zero [OK]

Common Mistakes:

• Counting coins incorrectly by missing loops
• Assuming greedy always gives minimum coins
• Confusing coin values and counts

4. Identify the error in this recursive function for factorial and suggest the fix:

int factorial(int n) {
    if (n == 0) return 0;
    return n * factorial(n - 1);
}

medium

A. Recursive call should be factorial(n + 1)

B. Function should return void

C. Base case should return 1 instead of 0

D. No error, code is correct

Solution

1. Step 1: Analyze base case correctness

   Factorial of 0 is 1, but code returns 0, which will cause all results to be zero.

2. Step 2: Suggest correct base case

   Change base case to return 1 when n == 0 to correctly end recursion and compute factorial.

3. Final Answer:

   Base case should return 1 instead of 0 -> Option C

4. Quick Check:

   Factorial(0) = 1 base case [OK]

Hint: Factorial base case returns 1, not 0 [OK]

Common Mistakes:

• Returning 0 in base case causing zero product
• Changing recursion direction incorrectly
• Ignoring base case leading to infinite recursion

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. Brute force, to try all combinations without optimization

B. Recursion without memoization, to explore all possibilities

C. Greedy, because always picking largest coin works

D. Dynamic Programming, to store and reuse subproblem results

Solution

1. Step 1: Understand problem requirements

   Minimum coins for amount with coins {1,3,4} needs checking all combinations to find optimal solution.

2. Step 2: Evaluate approaches

   Greedy fails here because picking largest coin first may not yield minimum coins. Recursion without memoization is slow. Dynamic Programming stores results of smaller amounts to build optimal solution efficiently.

3. Final Answer:

   Dynamic Programming, to store and reuse subproblem results -> Option D

4. Quick Check:

   Minimum coins with overlapping subproblems = 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
• Ignoring overlapping subproblems in coin change