Recursion Base Case and Recursive Case in DSA Typescript - Interactive Practice

Practice - 5 Tasks

1fill in blank

easy

Complete the code to return 1 when n is 0, the base case of recursion.

DSA Typescript

function factorial(n: number): number {
  if (n === [1]) {
    return 1;
  }
  return n * factorial(n - 1);
}

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

An

B0

C-1

D1

Attempts:

3 left

💡 Hint

Common Mistakes

Using 1 instead of 0 as the base case condition.

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

2fill in blank

medium

Complete the code to call factorial recursively with n-1.

DSA Typescript

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

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

An - 1

Bn + 1

C1

Dn

Attempts:

3 left

💡 Hint

Common Mistakes

Using n + 1 causes infinite recursion.

The recursive call reduces n by 1 each time to reach the base case.

3fill in blank

hard

Fix the error in the base case condition to stop infinite recursion.

DSA Typescript

function sumToN(n: number): number {
  if (n [1] 0) {
    return 0;
  }
  return n + sumToN(n - 1);
}

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

A==

B<=

C>

D<

Attempts:

3 left

💡 Hint

Common Mistakes

Using < or > causes wrong base case detection.

The base case should check if n equals 0 to stop recursion.

4fill in blank

hard

Fill both blanks to create a recursive function that counts down to zero.

DSA Typescript

function countdown(n: number): void {
  if (n [1] 0) {
    console.log('Done');
    return;
  }
  console.log(n);
  countdown(n [2] 1);
}

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

A===

B>

C-

D<

Attempts:

3 left

💡 Hint

Common Mistakes

Using < or > incorrectly in base case or recursive call.

The base case checks if n is exactly 0, and the recursive call subtracts 1.

5fill in blank

hard

Fill all three blanks to create a recursive function that sums numbers from 1 to n.

DSA Typescript

function sumNumbers(n: number): number {
  if (n [1] 1) {
    return 1;
  }
  return n [2] sumNumbers(n [3] 1);
}

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

A===

B+

C-

D>

Attempts:

3 left

💡 Hint

Common Mistakes

Using > instead of === in base case.

Using subtraction instead of addition in return.

The base case checks if n equals 1, then returns 1. The recursive call adds n to sumNumbers(n - 1).

Practice

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

easy

A. To call the function again with a smaller input

B. To stop the recursion and prevent infinite calls

C. To increase the input size for the next call

D. To print the result of the recursion

Solution

1. Step 1: Understand what recursion does

   Recursion calls the same function repeatedly with smaller inputs.

2. Step 2: Identify the role of the base case

   The base case stops the recursion by providing a condition where the function returns without calling itself again.

3. Final Answer:

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

4. Quick Check:

   Base case = stop recursion [OK]

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

Common Mistakes:

• Thinking base case calls recursion again
• Confusing base case with recursive case
• Ignoring base case leads to infinite recursion

2. Which of the following is the correct syntax for a recursive function in TypeScript that calculates factorial?

easy

A. function factorial(n: number): number { if (n === 0) return 1; else return n * factorial(n - 1); }

B. function factorial(n: number): number { while (n > 0) return n * factorial(n - 1); }

C. function factorial(n: number): number { if (n === 1) return 0; else return n + factorial(n - 1); }

D. function factorial(n: number): number { return n * factorial(n + 1); }

Solution

1. Step 1: Check base case correctness

   function factorial(n: number): number { if (n === 0) return 1; else return n * factorial(n - 1); } uses if (n === 0) return 1; which is correct for factorial base case.

2. Step 2: Check recursive call correctness

   function factorial(n: number): number { if (n === 0) return 1; else return n * factorial(n - 1); } calls factorial(n - 1) and multiplies by n, which matches factorial definition.

3. Final Answer:

   function factorial(n: number): number { if (n === 0) return 1; else return n * factorial(n - 1); } -> Option A

4. Quick Check:

   Correct base and recursive case syntax = function factorial(n: number): number { if (n === 0) return 1; else return n * factorial(n - 1); } [OK]

Hint: Base case returns 1 for factorial, recursive calls with n-1 [OK]

Common Mistakes:

• Using while loop instead of recursion
• Wrong base case value
• Recursive call increasing n causing infinite recursion

3. What will be the output of this TypeScript code?

function sumTo(n: number): number {
  if (n === 0) return 0;
  return n + sumTo(n - 1);
}
console.log(sumTo(3));

medium

A. 6

B. 3

C. 0

D. Error

Solution

1. Step 1: Trace recursive calls

   sumTo(3) calls 3 + sumTo(2), sumTo(2) calls 2 + sumTo(1), sumTo(1) calls 1 + sumTo(0), sumTo(0) returns 0.

2. Step 2: Calculate return values

   sumTo(1) = 1 + 0 = 1, sumTo(2) = 2 + 1 = 3, sumTo(3) = 3 + 3 = 6.

3. Final Answer:

   6 -> Option A

4. Quick Check:

   Sum 1+2+3 = 6 [OK]

Hint: Add n plus sumTo(n-1) until base case 0 [OK]

Common Mistakes:

• Confusing base case return value
• Adding instead of multiplying
• Missing base case causing infinite recursion

4. Identify the error in this recursive function and select the fix.

function countdown(n: number): void {
  if (n === 0) return;
  console.log(n);
  countdown(n + 1);
}
countdown(3);

medium

A. Change console.log(n) to console.log(n - 1)

B. Add a base case for n === 1

C. Remove the base case to allow infinite recursion

D. Change countdown(n + 1) to countdown(n - 1)

Solution

1. Step 1: Analyze recursive call direction

   The function calls countdown(n + 1), which increases n, causing infinite recursion.

2. Step 2: Fix recursive call to decrease n

   Changing to countdown(n - 1) moves towards base case n === 0, stopping recursion.

3. Final Answer:

   Change countdown(n + 1) to countdown(n - 1) -> Option D

4. Quick Check:

   Recursive call must approach base case [OK]

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

Common Mistakes:

• Increasing input causing infinite recursion
• Missing or wrong base case
• Changing print statement instead of recursion

5. You want to write a recursive function flatten that takes a nested array of numbers (e.g., [1, [2, [3]], 4]) and returns a flat array [1, 2, 3, 4]. Which approach correctly uses base and recursive cases?

hard

A. Base case: always return input; Recursive case: never call flatten recursively

B. Base case: if input is number, return number; Recursive case: call flatten on input

C. Base case: if input is empty array, return empty array; Recursive case: for each element, if array, call flatten, else add element

D. Base case: if input is array, return input; Recursive case: if input is number, call flatten

Solution

1. Step 1: Define base case for empty array

   If the input array is empty, return an empty array to stop recursion.

2. Step 2: Define recursive case for nested arrays

   For each element, if it is an array, recursively call flatten; if not, add element to result.

3. Final Answer:

   Base case: if input is empty array, return empty array; Recursive case: for each element, if array, call flatten, else add element -> Option C

4. Quick Check:

   Empty array stops recursion, nested arrays flatten recursively [OK]

Hint: Empty array stops recursion; recurse on nested arrays [OK]

Common Mistakes:

• Wrong base case returning number instead of array
• Not checking element type before recursion
• Infinite recursion by missing base case