Recursion vs Iteration When Each Wins in DSA Javascript - Interactive Comparison Practice

Practice - 5 Tasks

1fill in blank

easy

Complete the code to calculate factorial using recursion.

DSA Javascript

function factorial(n) {
  if (n === 0) return 1;
  return n * factorial([1]);
}

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

An

Bn + 1

Cn - 1

D1

Attempts:

3 left

💡 Hint

Common Mistakes

Using n + 1 causes infinite recursion.

Using n directly causes infinite recursion.

The factorial function calls itself with n - 1 to reduce the problem size until it reaches 0.

2fill in blank

medium

Complete the code to calculate factorial using iteration.

DSA Javascript

function factorialIter(n) {
  let result = 1;
  for (let i = 1; i [1]= n; i++) {
    result *= i;
  }
  return result;
}

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

A<

B>

C>=

D<=

Attempts:

3 left

💡 Hint

Common Mistakes

Using < excludes n, giving wrong factorial.

Using > or >= causes the loop to not run correctly.

The loop runs from 1 up to and including n, multiplying each number to get factorial.

3fill in blank

hard

Fix the error in the recursive Fibonacci function to avoid infinite recursion.

DSA Javascript

function fib(n) {
  if (n === 0) return 0;
  if (n === 1) return 1;
  return fib([1]) + fib(n - 2);
}

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

An - 1

Bn + 1

Cn

D1

Attempts:

3 left

💡 Hint

Common Mistakes

Using n + 1 causes infinite recursion.

Using n or 1 does not correctly compute Fibonacci.

The function calls fib(n - 1) and fib(n - 2) to get the previous two Fibonacci numbers.

4fill in blank

hard

Fill both blanks to create a recursive function that sums numbers from n down to 1.

DSA Javascript

function sumTo(n) {
  if (n === 0) return 0;
  return n [1] sumTo([2]);
}

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

A+

Bn - 1

C*

D/

Attempts:

3 left

💡 Hint

Common Mistakes

Using multiplication instead of addition.

Not reducing n in the recursive call.

The function adds n to the sum of numbers from n-1 down to 1.

5fill in blank

hard

Fill all three blanks to iteratively create an object that maps numbers to their squares for numbers greater than 2.

DSA Javascript

const squares = {};
for (let [3] = 1; [3] <= 5; [3]++) {
  if ([3] > 2) {
    squares[[1]] = [2] * [2];
  }
};

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

Ai

Dj

Attempts:

3 left

💡 Hint

Common Mistakes

Using different variables for key and value.

Using a variable not declared in the loop.

The variable i is used as key and value base. The loop maps i to i*i for i > 2.

Practice

1. Which scenario is best suited for using recursion instead of iteration?

easy

A. When the problem requires simple repeated loops

B. When the problem can be divided into smaller similar subproblems

C. When you want to avoid function call overhead

D. When memory usage must be minimal

Solution

1. Step 1: Understand recursion use case

   Recursion works well when a problem can be broken down into smaller versions of itself, like tree traversals or factorial calculation.

2. Step 2: Compare with iteration scenarios

   Iteration is better for simple repeated tasks without nested subproblems, so recursion is preferred for divide-and-conquer problems.

3. Final Answer:

   When the problem can be divided into smaller similar subproblems -> Option B

4. Quick Check:

   Recursion fits divide-and-conquer = C [OK]

Hint: Recursion fits problems breaking into smaller similar parts [OK]

Common Mistakes:

• Thinking iteration is always faster
• Ignoring function call overhead in recursion
• Confusing simple loops with recursive problems

2. Which of the following is the correct syntax for a recursive function in JavaScript?

easy

A. function recurse(n) { while(n > 0) { n--; } return n; }

B. function recurse(n) { for(let i=0; i

C. function recurse(n) { if (n <= 0) return 0; return n + recurse(n - 1); }

D. function recurse(n) { if(n > 0) { return n * recurse(n + 1); } else { return 1; } }

Solution

1. Step 1: Identify recursive call and base case

   function recurse(n) { if (n <= 0) return 0; return n + recurse(n - 1); } has a base case (n <= 0) and calls itself with a smaller argument (n - 1), which is correct recursion.

2. Step 2: Check other options for recursion errors

   function recurse(n) { while(n > 0) { n--; } return n; } uses a loop, not recursion. function recurse(n) { for(let i=0; i 0) { return n * recurse(n + 1); } else { return 1; } } calls recurse with n + 1, causing infinite recursion.

3. Final Answer:

   function recurse(n) { if (n <= 0) return 0; return n + recurse(n - 1); } -> Option C

4. Quick Check:

   Base case + smaller recursive call = B [OK]

Hint: Look for base case and smaller recursive call [OK]

Common Mistakes:

• Missing base case causing infinite recursion
• Using loops instead of recursive calls
• Increasing argument in recursive call causing stack overflow

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

function factorial(n) {
  if (n === 0) return 1;
  return n * factorial(n - 1);
}
console.log(factorial(4));

medium

A. 24

B. 10

C. 16

D. Error

Solution

1. Step 1: Trace recursive calls for factorial(4)

   factorial(4) calls factorial(3), factorial(3) calls factorial(2), factorial(2) calls factorial(1), factorial(1) calls factorial(0) which returns 1.

2. Step 2: Calculate return values step-by-step

   factorial(1) = 1 * 1 = 1, factorial(2) = 2 * 1 = 2, factorial(3) = 3 * 2 = 6, factorial(4) = 4 * 6 = 24.

3. Final Answer:

   24 -> Option A

4. Quick Check:

   4! = 24 [OK]

Hint: Factorial n! = n * (n-1)! [OK]

Common Mistakes:

• Confusing factorial with sum
• Forgetting base case returns 1
• Miscomputing multiplication order

4. Identify the error in this recursive function and choose the fix:

function sumTo(n) {
  if (n === 1) return 1;
  return n + sumTo(n);
}

medium

A. Change return to console.log

B. Add a base case for n === 0

C. Replace recursion with a for loop

D. Change sumTo(n) to sumTo(n - 1) in the return statement

Solution

1. Step 1: Analyze recursive call argument

   The function calls sumTo(n) again without changing n, causing infinite recursion and stack overflow.

2. Step 2: Fix recursive call to reduce problem size

   Changing sumTo(n) to sumTo(n - 1) reduces n each call, eventually reaching base case n === 1.

3. Final Answer:

   Change sumTo(n) to sumTo(n - 1) in the return statement -> Option D

4. Quick Check:

   Recursive call must reduce argument = A [OK]

Hint: Recursive calls must progress toward base case [OK]

Common Mistakes:

• Calling function with same argument causing infinite loop
• Missing or wrong base case
• Confusing recursion with iteration fixes

5. You want to process a large list of numbers to find their sum. Which approach is better and why?

hard

A. Iteration is better because it uses less memory and avoids stack overflow

B. Recursion is better because it is easier to write for large lists

C. Recursion with memoization is best for summing lists

D. Recursive function without base case is best for performance

Solution

1. Step 1: Consider recursion limits for large lists

   Recursion uses call stack for each call, which can cause stack overflow for large lists.

2. Step 2: Compare iteration benefits

   Iteration uses constant memory and avoids stack overflow, making it safer and more efficient for large data.

3. Final Answer:

   Iteration is better because it uses less memory and avoids stack overflow -> Option A

4. Quick Check:

   Iteration safer for large data = A [OK]

Hint: Use iteration for large data to avoid stack overflow [OK]

Common Mistakes:

• Assuming recursion always simpler for large data
• Ignoring stack overflow risk
• Misusing memoization for simple sums