Why Dynamic Programming and When Recursion Alone Fails in DSA Javascript - Test Your Knowledge

Practice - 5 Tasks

1fill in blank

easy

Complete the code to calculate Fibonacci numbers using simple recursion.

DSA Javascript

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

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

A*

B-

C+

D/

Attempts:

3 left

💡 Hint

Common Mistakes

Using subtraction or multiplication instead of addition.

Forgetting the base case for n <= 1.

The Fibonacci sequence is calculated by adding the two previous numbers, so the operator must be '+'.

2fill in blank

medium

Complete the code to store computed Fibonacci values in a cache object.

DSA Javascript

function fibMemo(n, cache = {}) {
  if (n <= 1) return n;
  if (cache[n] !== undefined) return cache[n];
  cache[n] = fibMemo(n - 1, cache) [1] fibMemo(n - 2, cache);
  return cache[n];
}

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

A+

B-

C*

D/

Attempts:

3 left

💡 Hint

Common Mistakes

Not storing the result in cache.

Using wrong operators causing incorrect results.

We add the two previous Fibonacci numbers and store the result in cache[n].

3fill in blank

hard

Fix the error in the recursive Fibonacci function to avoid repeated calculations.

DSA Javascript

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

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

Acache[n] == null

Bcache[n] !== undefined

Ccache[n] === null

Dcache[n] == undefined

Attempts:

3 left

💡 Hint

Common Mistakes

Using loose equality which can cause false negatives.

Checking for null instead of undefined.

We check if cache[n] is not undefined to know if the value was computed before.

4fill in blank

hard

Fill both blanks to create a bottom-up dynamic programming solution for Fibonacci.

DSA Javascript

function fibDP(n) {
  const dp = [0, 1];
  for (let i = 2; i [1] n; i++) {
    dp[i] = dp[i - 1] [2] dp[i - 2];
  }
  return dp[n];
}

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

A<=

B+

C-

D<

Attempts:

3 left

💡 Hint

Common Mistakes

Using < causes an off-by-one error.

Using subtraction instead of addition.

The loop runs while i <= n, and each dp[i] is the sum of dp[i-1] and dp[i-2].

5fill in blank

hard

Fill all three blanks to optimize Fibonacci calculation using constant space.

DSA Javascript

function fibOptimized(n) {
  if (n <= 1) return n;
  let a = 0, b = 1, temp;
  for (let i = 2; i [1] n; i++) {
    temp = a;
    a = b;
    b = b [2] temp;
  }
  return [3];
}

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

A<=

B+

Cb

D<

Attempts:

3 left

💡 Hint

Common Mistakes

Returning a instead of b.

Using < causing insufficient iterations.

Using wrong operator instead of +.

Loop runs while i <= n, b is updated by adding temp (old a), and we return b as the nth Fibonacci number.

Practice

1. Why does recursion alone sometimes fail or become inefficient for certain problems like Fibonacci sequence calculation?

easy

A. Because recursion can repeat the same calculations many times, causing slow performance

B. Because recursion always uses loops which are slow

C. Because recursion cannot handle base cases

D. Because recursion does not use any memory

Solution

1. Step 1: Understand recursion behavior in repeated calculations

   Recursion calls itself with smaller inputs but may recompute the same values multiple times, like Fibonacci(5) calling Fibonacci(4) and Fibonacci(3), and Fibonacci(4) again calling Fibonacci(3).

2. Step 2: Identify inefficiency due to repeated work

   This repeated work causes exponential time complexity, making recursion alone slow for large inputs.

3. Final Answer:

   Because recursion can repeat the same calculations many times, causing slow performance -> Option A

4. Quick Check:

   Repeated calculations cause inefficiency = C [OK]

Hint: Look for repeated calls with same inputs causing slow recursion [OK]

Common Mistakes:

• Thinking recursion always uses loops
• Believing recursion can't handle base cases
• Assuming recursion never uses memory

2. Which of the following is the correct syntax to implement memoization in a recursive JavaScript function?

easy

A. Use a cache object to store results and check before recursive calls

B. Use a for loop inside the recursive function

C. Call the recursive function without any parameters

D. Return undefined for base cases

Solution

1. Step 1: Understand memoization concept

   Memoization stores results of expensive function calls in a cache object to avoid repeated calculations.

2. Step 2: Apply memoization in recursion

   Before making a recursive call, check if the result exists in cache; if yes, return it; else compute and store it.

3. Final Answer:

   Use a cache object to store results and check before recursive calls -> Option A

4. Quick Check:

   Memoization uses cache to avoid repeats = A [OK]

Hint: Use a cache object to save and reuse results [OK]

Common Mistakes:

• Using loops inside recursion unnecessarily
• Not passing parameters to recursive calls
• Returning undefined for base cases

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

function fib(n, memo = {}) {
  if (n <= 1) return n;
  if (memo[n]) return memo[n];
  memo[n] = fib(n - 1, memo) + fib(n - 2, memo);
  return memo[n];
}
console.log(fib(5));

medium

A. Undefined

B. 8

C. 5

D. 3

Solution

1. Step 1: Trace fib(5) with memoization

   fib(5) calls fib(4) and fib(3), storing results in memo. fib(4) = 3, fib(3) = 2, so fib(5) = 3 + 2 = 5.

2. Step 2: Confirm memo usage avoids repeated calls

   Memo stores intermediate results, so no repeated calculations, final output is 5.

3. Final Answer:

   5 -> Option C

4. Quick Check:

   fib(5) = 5 with memoization [OK]

Hint: Memoization returns correct Fibonacci quickly [OK]

Common Mistakes:

• Confusing Fibonacci(5) with Fibonacci(6)
• Ignoring memo cache effect
• Assuming output is sum of all calls

4. Identify the error in this recursive function for factorial and fix it:

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

medium

A. Recursive call should be factorial(n + 1)

B. No error, output is correct

C. Function should return n + factorial(n - 1)

D. Base case should return 1 instead of 0

Solution

1. Step 1: Check base case correctness

   Factorial of 0 is 1, but function returns 0, causing all results to be zero.

2. Step 2: Fix base case return value

   Change base case to return 1 so multiplication works correctly.

3. Final Answer:

   Base case should return 1 instead of 0 -> Option D

4. Quick Check:

   Factorial base case = 1 [OK]

Hint: Factorial(0) must return 1, not 0 [OK]

Common Mistakes:

• Returning 0 for base case
• Using wrong recursive call direction
• Adding instead of multiplying

5. You want to find the minimum number of coins needed to make a certain amount using coin denominations [1, 3, 4]. Why is dynamic programming better than plain recursion here, and how would you apply it?

hard

A. Dynamic programming uses loops only; recursion cannot solve coin problems

B. Dynamic programming stores results of subproblems to avoid repeated work; apply bottom-up or memoized recursion

C. Recursion is faster because it tries all combinations without storing results

D. Dynamic programming ignores smaller amounts and solves only the final amount

Solution

1. Step 1: Understand problem complexity with recursion

   Recursion tries all combinations repeatedly, causing exponential time and slow performance.

2. Step 2: Use dynamic programming to store subproblem results

   DP saves minimum coins needed for smaller amounts, reusing them to build solution for larger amounts efficiently.

3. Step 3: Apply bottom-up or memoized recursion approach

   Either build from smallest amount up or use recursion with memo cache to avoid repeated calculations.

4. Final Answer:

   Dynamic programming stores results of subproblems to avoid repeated work; apply bottom-up or memoized recursion -> Option B

5. Quick Check:

   DP avoids repeats with caching = A [OK]

Hint: Store subproblem results to avoid repeated calculations [OK]

Common Mistakes:

• Thinking recursion alone is faster
• Believing DP ignores smaller problems
• Confusing loops with DP necessity