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

Practice - 5 Tasks

1fill in blank

easy

Complete the code to calculate Fibonacci numbers using recursion.

DSA Go

func fib(n int) int {
    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.

The Fibonacci sequence adds the two previous numbers to get the next one, so use the '+' operator.

2fill in blank

medium

Complete the code to implement a greedy approach for coin change problem.

DSA Go

func coinChange(coins []int, amount int) int {
    count := 0
    for i := len(coins) - 1; i >= 0; i-- {
        count += amount [1] coins[i]
        amount = amount % coins[i]
    }
    return count
}

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

A+

B/

C*

D-

Attempts:

3 left

💡 Hint

Common Mistakes

Using multiplication or addition instead of division.

We divide the amount by the coin value to find how many coins of that type we can use.

3fill in blank

hard

Fix the error in the recursive factorial function by completing the return statement.

DSA Go

func factorial(n int) int {
    if n == 0 {
        return 1
    }
    return n [1] factorial(n-1)
}

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

A*

B-

C/

D+

Attempts:

3 left

💡 Hint

Common Mistakes

Using addition or subtraction instead of multiplication.

Factorial multiplies n by factorial of (n-1), so use '*'.

4fill in blank

hard

Fill both blanks to create a dynamic programming solution for the climbing stairs problem.

DSA Go

func climbStairs(n int) int {
    dp := make([]int, n+1)
    dp[0], dp[1] = 1, 1
    for i := 2; i <= n; i++ {
        dp[i] = dp[i-[1]] + dp[i-[2]]
    }
    return dp[n]
}

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

A1

B2

C3

D0

Attempts:

3 left

💡 Hint

Common Mistakes

Using wrong indices like 0 or 3.

Each step can be reached from one step before or two steps before, so use 1 and 2.

5fill in blank

hard

Fill all three blanks to create a dictionary comprehension that filters items with value greater than zero.

DSA Go

result := map[string]int{}
for k, v := range data {
    if v [1] 0 {
        result[[2]] = [3]
    }
}

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

A>

Bk

Cv

D<

Attempts:

3 left

💡 Hint

Common Mistakes

Using '<' instead of '>' or swapping key and value.

We check if value is greater than zero, then add key and value to result map.

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 parts.

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
• Confusing brute force with DP

2. Which of the following Go function signatures correctly represents a recursive function to compute factorial?

easy

A. func factorial(n int) { if n == 0 { return 1 } return n * factorial(n-1) }

B. func factorial(n int) int { if n == 0 { return 1 } return n * factorial(n-1) }

C. func factorial(n int) int { if n == 0 { return 0 } return n * factorial(n-1) }

D. func factorial(n int) int { if n == 1 { return 0 } return n * factorial(n-1) }

Solution

1. Step 1: Check function return type and base case

   Factorial returns int, so function must return int. Base case for factorial is n == 0 returning 1.

2. Step 2: Verify recursive call and return statement

   Recursive call multiplies n by factorial(n-1). func factorial(n int) int { if n == 0 { return 1 } return n * factorial(n-1) } matches correct signature and logic.

3. Final Answer:

   func factorial(n int) int { if n == 0 { return 1 } return n * factorial(n-1) } -> Option B

4. Quick Check:

   Correct return type and base case = func factorial(n int) int { if n == 0 { return 1 } return n * factorial(n-1) } [OK]

Hint: Check base case and return type carefully in recursion [OK]

Common Mistakes:

• Missing return type in function signature
• Incorrect base case value
• Returning wrong type or no return

3. What is the output of this Go code using recursion to compute Fibonacci numbers?

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

func main() {
  fmt.Println(fib(5))
}

medium

A. 3

B. 8

C. 5

D. 13

Solution

1. Step 1: Trace fib(5) calls

   fib(5) = fib(4) + fib(3); fib(4) = fib(3) + fib(2); fib(3) = fib(2) + fib(1); fib(2) = fib(1) + fib(0)

2. Step 2: Calculate values bottom-up

   fib(0)=0, fib(1)=1; fib(2)=1; fib(3)=2; fib(4)=3; fib(5)=5

3. Final Answer:

   5 -> Option C

4. Quick Check:

   Fibonacci(5) = 5 [OK]

Hint: Remember Fibonacci sequence: 0,1,1,2,3,5,8... [OK]

Common Mistakes:

• Confusing index starting point
• Miscomputing recursive calls
• Assuming fib(1) = 0

4. Identify the error in this greedy algorithm code snippet for coin change in Go:

func coinChange(coins []int, amount int) int {
  count := 0
  for i := len(coins) - 1; i >= 0; i-- {
    for amount >= coins[i] {
      amount -= coins[i]
      count++
    }
  }
  if amount != 0 {
    return -1
  }
  return count
}

medium

A. Greedy may fail if coins are not canonical denominations

B. The inner loop should increment amount instead of decrement

C. Coins array must be sorted in ascending order

D. Return value should be amount instead of count

Solution

1. Step 1: Understand greedy coin change logic

   The code tries largest coins first, subtracting from amount until no longer possible.

2. Step 2: Identify limitation of greedy approach

   Greedy works only if coin denominations are canonical (like US coins). Otherwise, it may not find minimum coins or any solution.

3. Final Answer:

   Greedy may fail if coins are not canonical denominations -> Option A

4. Quick Check:

   Greedy correctness depends on coin set [OK]

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

Common Mistakes:

• Assuming greedy always finds minimum coins
• Ignoring coin order importance
• Misunderstanding return values

5. You want to find the longest increasing subsequence (LIS) in an array. Which approach is best and why? A) Simple recursion tries all subsequences but is slow. B) Greedy picks local best but may miss global LIS. C) Dynamic Programming stores LIS lengths to avoid recomputation. D) Brute force enumerates all subsequences without optimization.

hard

A. Simple recursion is best for LIS

B. Greedy approach always finds LIS

C. Brute force is efficient for LIS

D. Dynamic Programming is best for LIS

Solution

1. Step 1: Analyze LIS problem nature

   LIS requires checking subsequences and their lengths. Simple recursion tries all but is exponential time.

2. Step 2: Compare approaches

   Greedy fails because local choices don't guarantee global LIS. DP stores LIS lengths for subproblems, improving efficiency.

3. Final Answer:

   Dynamic Programming is best for LIS -> Option D

4. Quick Check:

   LIS problem = DP best approach [OK]

Hint: Use DP for LIS to avoid exponential recursion [OK]

Common Mistakes:

• Choosing greedy for LIS
• Ignoring exponential time of recursion
• Thinking brute force is efficient