Fibonacci Using DP in DSA Typescript - Interactive Practice

Practice - 5 Tasks

1fill in blank

easy

Complete the code to initialize the memo array with -1.

DSA Typescript

const memo: number[] = new Array(n + 1).fill([1]);

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

Anull

B1

C0

D-1

Attempts:

3 left

💡 Hint

Common Mistakes

Using 0 causes incorrect returns since fib(0) = 0 and memo[2] would wrongly return 0.

Using 1 causes confusion with fib(1) = 1.

Using null causes type errors in TypeScript.

We fill the memo array with -1 to indicate that no Fibonacci values are computed yet.

2fill in blank

medium

Complete the base case check to return n if it is 0 or 1.

DSA Typescript

if (n === 0 || n === [1]) return n;

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

A1

B2

C3

Dn

Attempts:

3 left

💡 Hint

Common Mistakes

Checking for n === 2 instead of 1.

Returning a fixed number instead of n.

The Fibonacci of 0 is 0 and of 1 is 1, so we return n when n is 0 or 1.

3fill in blank

hard

Fix the error in the memoization check to return the stored value if already computed.

DSA Typescript

if (memo[n] !== [1]) return memo[n];

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

A0

B-1

Cnull

Dundefined

Attempts:

3 left

💡 Hint

Common Mistakes

Checking for undefined, 0, or null instead of -1.

Using == instead of !== causing wrong logic.

We check if memo[n] !== -1 to know if the value is already computed.

4fill in blank

hard

Fill both blanks to compute and store the Fibonacci value in memo.

DSA Typescript

memo[n] = fibonacci[1](n - 1) + fibonacci[2](n - 2);

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

A(

B[

C]

D)

Attempts:

3 left

💡 Hint

Common Mistakes

Using square brackets [] which are for arrays, not function calls.

Missing parentheses causing syntax errors.

We call the fibonacci function recursively with parentheses around the arguments.

5fill in blank

hard

Fill all three blanks to complete the memoized Fibonacci function.

DSA Typescript

function fibonacci[1](n: number): number {
  if (n === 0 || n === 1) return n;
  if (memo[n] !== -1) return memo[n];
  memo[n] = fibonacci[2](n - 1) + fibonacci[3](n - 2);
  return memo[n];
}

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

A(

B)

C[

D]

Attempts:

3 left

💡 Hint

Common Mistakes

Using square brackets instead of parentheses.

Omitting parentheses causing syntax errors.

Function calls and definitions use parentheses () around parameters and arguments.

Practice

1. What is the main advantage of using Dynamic Programming (DP) to calculate Fibonacci numbers compared to a simple recursive approach?

easy

A. It uses less memory by not storing any results.

B. It avoids repeated calculations by storing intermediate results.

C. It calculates Fibonacci numbers randomly.

D. It only works for small Fibonacci numbers.

Solution

1. Step 1: Understand recursive Fibonacci calculation

   Simple recursion recalculates the same Fibonacci numbers many times, causing inefficiency.

2. Step 2: Understand DP optimization

   DP stores previously calculated Fibonacci numbers in a table (memoization or tabulation) to avoid repeated work.

3. Final Answer:

   It avoids repeated calculations by storing intermediate results. -> Option B

4. Quick Check:

   DP = stores results to save time [OK]

Hint: DP saves time by remembering past answers [OK]

Common Mistakes:

• Thinking DP uses less memory than recursion
• Believing DP calculates randomly
• Assuming DP only works for small inputs

2. Which of the following TypeScript code snippets correctly initializes the DP array for Fibonacci calculation up to n?

easy

A. const dp: number[] = new Array(n + 1).fill(0); dp[1] = 1;

B. const dp: number[] = new Array(n).fill(1);

C. const dp: number[] = []; dp[0] = 1; dp[1] = 1;

D. const dp: number[] = new Array(n + 1); dp[1] = 1;

Solution

1. Step 1: Check array size and initialization

   We need an array of size n+1 to store Fibonacci numbers from 0 to n. Filling with 0 and setting dp[1] = 1 is correct.

2. Step 2: Validate each option

   const dp: number[] = new Array(n + 1).fill(0); dp[1] = 1; correctly creates dp with size n+1, fills with 0, and sets dp[1] = 1. Others have wrong sizes or initial values.

3. Final Answer:

   const dp: number[] = new Array(n + 1).fill(0); dp[1] = 1; -> Option A

4. Quick Check:

   Array size = n+1, dp[1] = 1 [OK]

Hint: DP array size must be n+1 with dp[1] = 1 [OK]

Common Mistakes:

• Using array size n instead of n+1
• Initializing dp[0] or dp[1] incorrectly
• Leaving dp array uninitialized

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

function fib(n: number): number {
  const dp: number[] = new Array(n + 1).fill(0);
  dp[1] = 1;
  for (let i = 2; i <= n; i++) {
    dp[i] = dp[i - 1] + dp[i - 2];
  }
  return dp[n];
}
console.log(fib(5));

medium

A. 13

B. 8

C. 3

D. 5

Solution

1. Step 1: Trace DP array values for n=5

   dp[0]=0, dp[1]=1; then dp[2]=1, dp[3]=2, dp[4]=3, dp[5]=5.

2. Step 2: Return dp[5]

   The function returns dp[5] which is 5.

3. Final Answer:

   5 -> Option D

4. Quick Check:

   Fibonacci(5) = 5 [OK]

Hint: Remember Fibonacci sequence: 0,1,1,2,3,5... [OK]

Common Mistakes:

• Confusing index starting point
• Returning dp[n-1] instead of dp[n]
• Miscomputing dp values in loop

4. Identify the error in this TypeScript DP Fibonacci code and select the fix:

function fib(n: number): number {
  const dp: number[] = new Array(n).fill(0);
  dp[0] = 0;
  dp[1] = 1;
  for (let i = 2; i <= n; i++) {
    dp[i] = dp[i - 1] + dp[i - 2];
  }
  return dp[n];
}

medium

A. Change array size to n+1.

B. Initialize dp[2] to 1 before the loop.

C. Start loop from i = 1 instead of 2.

D. Remove dp[0] initialization.

Solution

1. Step 1: Analyze array size and access

   Array dp has size n, but loop accesses dp[n] beyond the initial length.

2. Step 2: Fix array size

   Increase dp size to n+1 to safely access dp[n].

3. Final Answer:

   Change array size to n+1 to avoid out-of-bounds error. -> Option A

4. Quick Check:

   Array size must cover dp[n] [OK]

Hint: Array size must be n+1 for dp[n] access [OK]

Common Mistakes:

• Ignoring array size mismatch
• Starting loop incorrectly
• Unnecessary dp initialization changes

5. You want to optimize the space used by the DP Fibonacci function. Which modification correctly reduces space complexity from O(n) to O(1)?

hard

A. Use a recursive function without memoization.

B. Store all Fibonacci numbers in a global array.

C. Use two variables to store only the last two Fibonacci numbers instead of an array.

D. Initialize the dp array with zeros and fill it completely every call.

Solution

1. Step 1: Understand space usage in DP Fibonacci

   DP uses an array of size n+1, so space is O(n).

2. Step 2: Optimize space by storing only needed values

   Since each Fibonacci number depends only on the previous two, storing just two variables suffices.

3. Step 3: Compare options

   Use two variables to store only the last two Fibonacci numbers instead of an array. uses two variables, reducing space to O(1). Others do not reduce space or are inefficient.

4. Final Answer:

   Use two variables to store only the last two Fibonacci numbers instead of an array. -> Option C

5. Quick Check:

   Two vars = O(1) space [OK]

Hint: Keep only last two Fibonacci numbers to save space [OK]

Common Mistakes:

• Using recursion without memoization increases time
• Storing all numbers globally wastes space
• Filling full dp array every call wastes memory