Recursion Base Case and Recursive Case in DSA C++ - Cheat Sheet & Quick Revision

Recall & Review

beginner

What is the base case in recursion?

The base case is the condition that stops the recursion. It prevents infinite calls by providing a simple answer without further recursion.

beginner

What is the recursive case in recursion?

The recursive case is the part where the function calls itself with a smaller or simpler input, moving towards the base case.

intermediate

Why do we need both base case and recursive case in a recursive function?

The base case stops the recursion to avoid infinite loops, and the recursive case breaks the problem into smaller parts to solve it step by step.

beginner

Consider this code snippet:

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

Identify the base case and recursive case.

Base case: if (n == 0) return 1;
Recursive case: return n * factorial(n - 1);

beginner

What happens if a recursive function does not have a base case?

The function will keep calling itself forever, causing a stack overflow error because it never stops.

What is the purpose of the base case in recursion?

A To print the result

B To call the function again

C To increase the input size

D To stop the recursion

The base case stops the recursion to prevent infinite calls.

In the recursive case, the function usually calls itself with:

A A larger input

B The same input

C A smaller or simpler input

D No input

The recursive case calls the function with a smaller or simpler input to approach the base case.

What will happen if a recursive function lacks a base case?

A It will run once and stop

B It will cause infinite recursion and crash

C It will return zero

D It will print an error message

Without a base case, recursion never stops, causing a crash due to stack overflow.

In the factorial function, what is the base case?

A n == 0

B n == 1

C n > 0

D n < 0

The base case for factorial is when n equals 0, returning 1.

Which part of recursion breaks the problem into smaller parts?

A Recursive case

B Return statement

C Base case

D Input validation

The recursive case breaks the problem into smaller parts by calling itself.

Explain the roles of the base case and recursive case in a recursive function.

Think about how recursion stops and how it progresses.

Describe what happens if a recursive function does not have a base case.

Consider what stops recursion normally.

Practice

1. What is the purpose of the base case in a recursive function?

easy

A. To print the result of the recursion

B. To call the function again with a smaller input

C. To increase the input size for the next call

D. To stop the recursion and prevent infinite calls

Solution

1. Step 1: Understand recursion basics

   Recursion works by calling itself repeatedly until a stopping point is reached.

2. Step 2: Identify the role of base case

   The base case stops further recursive calls to avoid infinite loops.

3. Final Answer:

   To stop the recursion and prevent infinite calls -> Option D

4. Quick Check:

   Base case = stop recursion [OK]

Hint: Base case stops recursion to avoid infinite loops [OK]

Common Mistakes:

• Confusing base case with recursive case
• Thinking base case calls recursion again
• Ignoring the need for a stopping condition

2. Which of the following is the correct syntax for a base case in this recursive function to calculate factorial?

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

easy

A. n > 0

B. n == 0

C. n == 1

D. n < 0

Solution

1. Step 1: Identify factorial base case

   Factorial of 0 is defined as 1, so base case is when n == 0.

2. Step 2: Check options for correct base case

   n == 0 matches n == 0, which returns 1 stopping recursion correctly.

3. Final Answer:

   n == 0 -> Option B

4. Quick Check:

   Factorial base case = n == 0 [OK]

Hint: Factorial base case is n equals zero [OK]

Common Mistakes:

• Using n > 0 as base case causes infinite recursion
• Using n < 0 which is invalid for factorial
• Confusing base case with recursive case

3. What will be the output of the following code?

int sum(int n) {
  if (n == 0) return 0;
  return n + sum(n - 1);
}

int main() {
  std::cout << sum(3);
  return 0;
}

medium

A. 0

B. 3

C. 6

D. Error

Solution

1. Step 1: Trace recursive calls for sum(3)

   sum(3) = 3 + sum(2), sum(2) = 2 + sum(1), sum(1) = 1 + sum(0), sum(0) = 0.

2. Step 2: Calculate total sum

   sum(3) = 3 + 2 + 1 + 0 = 6.

3. Final Answer:

   6 -> Option C

4. Quick Check:

   Sum 1 to 3 = 6 [OK]

Hint: Sum from n to 0 adds all numbers [OK]

Common Mistakes:

• Returning n instead of n + sum(n-1)
• Missing base case causing infinite recursion
• Confusing output with just n

4. Identify the error in this recursive function to find factorial:

int factorial(int n) {
  if (n == 1) return 1;
  else return n * factorial(n + 1);
}

medium

A. Recursive call increases n, causing infinite recursion

B. Function should return void, not int

C. Missing return statement in base case

D. Base case should be n == 0, not n == 1

Solution

1. Step 1: Analyze recursive call

   The function calls factorial(n + 1), which increases n each time, never reaching base case.

2. Step 2: Identify infinite recursion cause

   Because n grows, base case n == 1 is never met, causing infinite recursion.

3. Final Answer:

   Recursive call increases n, causing infinite recursion -> Option A

4. Quick Check:

   Recursive call must reduce n [OK]

Hint: Recursive calls must reduce input to reach base case [OK]

Common Mistakes:

• Increasing n instead of decreasing in recursion
• Wrong base case value
• Ignoring infinite recursion risk

5. You want to write a recursive function to compute the nth Fibonacci number where fib(0) = 0 and fib(1) = 1. Which of the following correctly defines the base cases and recursive case?

hard

A. if (n == 0) return 0; if (n == 1) return 1; else return fib(n - 1) + fib(n - 2);

B. if (n == 1) return 0; if (n == 2) return 1; else return fib(n - 1) + fib(n - 2);

C. if (n <= 1) return n; else return fib(n + 1) + fib(n - 2);

D. if (n == 0) return 1; if (n == 1) return 1; else return fib(n - 1) + fib(n - 2);

Solution

1. Step 1: Identify correct base cases for Fibonacci

   By definition, fib(0) = 0 and fib(1) = 1 are base cases.

2. Step 2: Check recursive case correctness

   Recursive case sums fib(n-1) and fib(n-2), matching Fibonacci definition.

3. Final Answer:

   if (n == 0) return 0; if (n == 1) return 1; else return fib(n - 1) + fib(n - 2); -> Option A

4. Quick Check:

   Fibonacci base and recursive cases match definition [OK]

Hint: Base cases fib(0)=0, fib(1)=1; recursive sums previous two [OK]

Common Mistakes:

• Wrong base case values
• Increasing n in recursive calls
• Returning wrong values for base cases