Recursion Concept and Call Stack Visualization in DSA Go - Practice Problems & Challenges

Challenge - 5 Problems

❓ Predict Output

intermediate

Output of Recursive Factorial Function

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

DSA Go

package main
import "fmt"

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

func main() {
    fmt.Println(factorial(4))
}

A 24

B 10

C 120

D 1

💡 Hint

Recall factorial of 4 is 4*3*2*1.

The function calls itself multiplying n by factorial of n-1 until n is 0, returning 1. So factorial(4) = 4*3*2*1 = 24.

❓ Predict Output

intermediate

Output of Recursive Fibonacci Function

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

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 5

B 8

C 3

D 13

💡 Hint

Fibonacci sequence starts 0,1,1,2,3,5...

fib(5) = fib(4) + fib(3) = 3 + 2 = 5.

🔧 Debug

advanced

Identify the Error in Recursive Sum Function

What error does this Go code produce when trying to sum numbers recursively?

DSA Go

package main
import "fmt"

func sum(n int) int {
    if n == 0 {
        return 0
    }
    return n + sum(n)
}

func main() {
    fmt.Println(sum(3))
}

A Syntax error: missing return type

B Stack overflow (infinite recursion)

C Compile error: undefined function sum

D Output: 6

💡 Hint

Check the recursive call argument.

The function calls sum(n) instead of sum(n-1), causing infinite recursion and stack overflow.

🧠 Conceptual

advanced

Call Stack Depth in Recursive Function

If a recursive function calls itself with n-1 until n is 0, what is the maximum call stack depth when called with n=7?

A 7

B 6

C 8

D 1

💡 Hint

Count the initial call plus each recursive call until base case.

The call stack depth is n+1 because it includes the initial call plus calls down to 0.

🚀 Application

expert

Output of Recursive String Reverse Function

What is the output of this Go code that reverses a string recursively?

DSA Go

package main
import (
    "fmt"
)

func reverse(s string) string {
    if len(s) == 0 {
        return ""
    }
    return reverse(s[1:]) + string(s[0])
}

func main() {
    fmt.Println(reverse("go"))
}

A go

B og\n

C error: invalid string indexing

D og

💡 Hint

The function moves first character to the end recursively.

The function takes the first character and appends it after reversing the rest, so "go" becomes "og".

Practice

1. What happens when a recursive function calls itself without a stopping condition?

easy

A. It converts into a normal loop automatically

B. It immediately returns without doing anything

C. It runs only once and stops

D. It causes an infinite loop and eventually a stack overflow error

Solution

1. Step 1: Understand recursion without base case

   Without a stopping condition, the function keeps calling itself endlessly.

2. Step 2: Effect on call stack

   Each call adds a new frame to the call stack, which grows until memory runs out causing stack overflow.

3. Final Answer:

   It causes an infinite loop and eventually a stack overflow error -> Option D

4. Quick Check:

   Missing base case = infinite recursion [OK]

Hint: Always check for a base case to stop recursion [OK]

Common Mistakes:

• Thinking recursion stops automatically
• Confusing recursion with loops
• Ignoring stack overflow possibility

2. Which of the following is the correct syntax to define a recursive function in Go that calculates factorial of n?

easy

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

B. func factorial(n 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 signature and return type

   Factorial returns an int, so function must return int type.

2. Step 2: Verify base case and recursive call

   Base case is n == 0 returning 1; recursive call multiplies n by factorial(n-1).

3. Final Answer:

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

4. Quick Check:

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

Hint: Base case returns 1 for factorial(0) [OK]

Common Mistakes:

• Missing return type in function signature
• Returning 0 instead of 1 in base case
• Not returning value from function

3. What is the output of the following Go code?

func printCount(n int) {
    if n == 0 {
        return
    }
    fmt.Print(n, " ")
    printCount(n-1)
}

func main() {
    printCount(3)
}

medium

A. 3 2 1

B. 1 2 3

C. 3 2 1 0

D. 0 1 2 3

Solution

1. Step 1: Trace recursive calls and prints

   printCount(3) prints 3, then calls printCount(2), which prints 2, then calls printCount(1), which prints 1, then calls printCount(0) which returns.

2. Step 2: Confirm output sequence

   Output is printed before recursive call, so order is 3 2 1 with spaces.

3. Final Answer:

   3 2 1 -> Option A

4. Quick Check:

   Print before recursion = descending order [OK]

Hint: Print before recursion prints descending order [OK]

Common Mistakes:

• Assuming print happens after recursion
• Including 0 in output
• Confusing order of prints

4. Identify the error in this recursive Go function that calculates sum of numbers from 1 to n:

func sum(n int) int {
    if n == 1 {
        return 1
    }
    return n + sum(n)
}

medium

A. Function does not return any value

B. Missing base case for n == 0

C. Recursive call uses sum(n) instead of sum(n-1), causing infinite recursion

D. Base case returns wrong value

Solution

1. Step 1: Analyze recursive call parameter

   The function calls sum(n) again without decreasing n, causing infinite recursion.

2. Step 2: Correct recursive call

   It should call sum(n-1) to reduce n and reach base case.

3. Final Answer:

   Recursive call uses sum(n) instead of sum(n-1), causing infinite recursion -> Option C

4. Quick Check:

   Recursive call must reduce n [OK]

Hint: Recursive call parameter must change to approach base case [OK]

Common Mistakes:

• Calling sum(n) again without decrement
• Wrong base case value
• Not returning result from recursion

5. Given the recursive function below, what will be the output when calling mystery(4)? Explain the call stack behavior.

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

func main() {
    fmt.Println(mystery(4))
}

hard

A. 5

B. 3

C. 7

D. 6

Solution

1. Step 1: Recognize the function as Fibonacci calculation

   The function returns n if n <= 1, else sum of mystery(n-1) and mystery(n-2), matching Fibonacci logic.

2. Step 2: Calculate mystery(4) step-by-step

   mystery(4) = mystery(3) + mystery(2) mystery(3) = mystery(2) + mystery(1) = (mystery(1)+mystery(0)) + 1 = (1+0)+1=2 mystery(2) = mystery(1) + mystery(0) = 1 + 0 = 1 Sum: 2 + 1 = 3 Wait, re-check carefully: mystery(0)=0, mystery(1)=1 mystery(2)=1+0=1 mystery(3)=1+1=2 mystery(4)=2+1=3 But options show 3 and 5, so re-check Fibonacci sequence: 0,1,1,2,3,5... Actually mystery(4) = 3

3. Final Answer:

   3 -> Option B

4. Quick Check:

   Fibonacci(4) = 3 [OK]

Hint: Recognize Fibonacci pattern in recursion [OK]

Common Mistakes:

• Miscounting Fibonacci numbers
• Confusing base cases
• Ignoring call stack repeated calls