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

Challenge - 5 Problems

❓ Predict Output

intermediate

Output of Recursive Fibonacci Function

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

DSA Go

package main
import "fmt"
func fib(n int) int {
    if n <= 1 {
        return n
    }
    return fib(n-1) + fib(n-2)
}
func main() {
    fmt.Println(fib(5))
}

A 13

B 5

C 8

D 3

💡 Hint

Trace the recursive calls for fib(5).

The recursive Fibonacci function returns the sum of the two previous Fibonacci numbers. fib(5) = fib(4) + fib(3) = 3 + 2 = 5.

❓ Predict Output

intermediate

Output of Greedy Coin Change Algorithm

What is the output of this Go code that uses a greedy approach to find minimum coins for amount 27 with coins [10, 5, 1]?

DSA Go

package main
import "fmt"
func minCoins(coins []int, amount int) int {
    count := 0
    for _, coin := range coins {
        count += amount / coin
        amount %= coin
    }
    return count
}
func main() {
    coins := []int{10, 5, 1}
    fmt.Println(minCoins(coins, 27))
}

A 5

B 6

C 3

D 4

💡 Hint

Divide amount by largest coin first, then remainder by next coin.

27/10=2 coins, remainder 7; 7/5=1 coin, remainder 2; 2/1=2 coins; total 2+1+2=5 coins.

❓ Predict Output

advanced

Output of DP Fibonacci with Memoization

What is the output of this Go code that calculates Fibonacci numbers using dynamic programming with memoization for fib(7)?

DSA Go

package main
import "fmt"
func fibDP(n int, memo map[int]int) int {
    if n <= 1 {
        return n
    }
    if val, ok := memo[n]; ok {
        return val
    }
    memo[n] = fibDP(n-1, memo) + fibDP(n-2, memo)
    return memo[n]
}
func main() {
    memo := make(map[int]int)
    fmt.Println(fibDP(7, memo))
}

A 13

B 21

C 8

D 34

💡 Hint

Use memoization to avoid repeated calculations.

fib(7) = fib(6) + fib(5) = 8 + 5 = 13.

🧠 Conceptual

advanced

Choosing the Right Tool for the Job

Which approach is best suited for problems where the solution can be built from solutions of overlapping subproblems and optimal substructure?

A Greedy algorithm

B Brute force search

C Simple recursion without memoization

D Dynamic programming

💡 Hint

Think about reusing solutions to subproblems.

Dynamic programming is ideal when subproblems overlap and optimal substructure exists, allowing reuse of solutions.

🔧 Debug

expert

Identify the Error in Greedy Algorithm for Coin Change

This Go code tries to find minimum coins for amount 6 with coins [4, 3, 1] using a greedy approach. What error will it produce or what is the output?

DSA Go

package main
import "fmt"
func minCoinsGreedy(coins []int, amount int) int {
    count := 0
    for _, coin := range coins {
        count += amount / coin
        amount %= coin
    }
    if amount != 0 {
        return -1
    }
    return count
}
func main() {
    coins := []int{4, 3, 1}
    fmt.Println(minCoinsGreedy(coins, 6))
}

A 2

B -1

C 3

D 1

💡 Hint

Greedy may fail if coin denominations are not canonical.

The greedy approach picks 4 once (remainder 2), then 3 zero times, then 1 twice; total 3 coins. But minimum coins is 2 (3+3). The code returns 3, which is suboptimal as greedy fails for these denominations.

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