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DSA Javascriptprogramming~10 mins

Fibonacci Using DP in DSA Javascript - Interactive Practice

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Practice - 5 Tasks
Answer the questions below
1fill in blank
easy

Complete the code to initialize the memo array with base cases for Fibonacci.

DSA Javascript
const fib = (n, memo = [[1], 1]) => {
  if (memo[n] !== undefined) return memo[n];
  memo[n] = fib(n - 1, memo) + fib(n - 2, memo);
  return memo[n];
};
Drag options to blanks, or click blank then click option'
A-1
B1
C0
D2
Attempts:
3 left
💡 Hint
Common Mistakes
Starting memo with [1, 1] instead of [0, 1]
Using negative numbers in memo initialization
2fill in blank
medium

Complete the code to check if the Fibonacci value is already computed in memo.

DSA Javascript
if (memo[[1]] !== undefined) return memo[n];
Drag options to blanks, or click blank then click option'
An
B1
C0
Dn-1
Attempts:
3 left
💡 Hint
Common Mistakes
Checking memo[0] or memo[1] instead of memo[n]
Using n-1 or other indices incorrectly
3fill in blank
hard

Fix the error in the recursive call to compute Fibonacci using memoization.

DSA Javascript
memo[n] = fib(n [1] 1, memo) + fib(n - 2, memo);
Drag options to blanks, or click blank then click option'
A-
B+
C*
D/
Attempts:
3 left
💡 Hint
Common Mistakes
Using '+' instead of '-' causing wrong recursion
Using '*' or '/' which are invalid here
4fill in blank
hard

Fill both blanks to complete the base case check and return statement.

DSA Javascript
if (n [1] 0) return [2];
Drag options to blanks, or click blank then click option'
A<=
B==
C0
Dn
Attempts:
3 left
💡 Hint
Common Mistakes
Using '<=' instead of '==' causing incorrect base case
Returning n instead of 0 for base case
5fill in blank
hard

Fill all three blanks to create a memoized Fibonacci function with base cases and recursion.

DSA Javascript
const fib = (n, memo = [[1], 1]) => {
  if (n [2] 0) return [3];
  if (memo[n] !== undefined) return memo[n];
  memo[n] = fib(n - 1, memo) + fib(n - 2, memo);
  return memo[n];
};
Drag options to blanks, or click blank then click option'
A0
B==
D1
Attempts:
3 left
💡 Hint
Common Mistakes
Wrong memo initialization
Incorrect base case condition
Returning wrong value for base case

Practice

(1/5)
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 calculates Fibonacci numbers using only recursion without loops
C. It stores intermediate results to avoid repeated calculations
D. It always returns the Fibonacci sequence in reverse order

Solution

  1. Step 1: Understand recursion vs DP

    Simple recursion recalculates the same Fibonacci numbers many times, causing inefficiency.
  2. Step 2: Role of DP in Fibonacci

    DP stores previously calculated Fibonacci values in a table (memoization or tabulation) to reuse them, avoiding repeated work.
  3. Final Answer:

    It stores intermediate results to avoid repeated calculations -> Option C
  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 does not use loops or arrays
  • Confusing DP with simple recursion
2. Which of the following JavaScript code snippets correctly initializes an array for bottom-up DP to compute Fibonacci numbers up to n?
easy
A. const fib = new Array(n+1).fill(0); fib[0] = 0; fib[1] = 1;
B. const fib = []; fib[0] = 1; fib[1] = 1;
C. const fib = new Array(n).fill(0); fib[1] = 1;
D. const fib = new Array(n+1).fill(1); fib[0] = 0;

Solution

  1. Step 1: Array size for DP

    We need an array of size n+1 to store Fibonacci numbers from 0 to n.
  2. Step 2: Initial values for Fibonacci

    Fibonacci sequence starts with fib[0] = 0 and fib[1] = 1, so these must be set correctly.
  3. Final Answer:

    const fib = new Array(n+1).fill(0); fib[0] = 0; fib[1] = 1; -> Option A
  4. Quick Check:

    Array size n+1, fib[0]=0, fib[1]=1 [OK]
Hint: Array size n+1 with fib[0]=0 and fib[1]=1 [OK]
Common Mistakes:
  • Using array size n instead of n+1
  • Setting fib[0] = 1 incorrectly
  • Filling array with 1 instead of 0
3. What will be the output of the following JavaScript code?
function fibDP(n) {
  const fib = [0, 1];
  for (let i = 2; i <= n; i++) {
    fib[i] = fib[i - 1] + fib[i - 2];
  }
  return fib[n];
}
console.log(fibDP(6));
medium
A. 8
B. 13
C. 21
D. 34

Solution

  1. Step 1: Trace Fibonacci values up to n=6

    fib[0]=0, fib[1]=1, fib[2]=1, fib[3]=2, fib[4]=3, fib[5]=5, fib[6]=8.
  2. Step 2: Return fib[6]

    The function returns fib[6], which is 8.
  3. Final Answer:

    8 -> Option A
  4. Quick Check:

    fib(6) = 8 [OK]
Hint: fib(6) = 8 by adding previous two numbers [OK]
Common Mistakes:
  • Confusing index and value (e.g., fib[6] = 13)
  • Starting array with wrong initial values
  • Off-by-one errors in loop
4. Identify the error in this JavaScript DP Fibonacci function and choose the correct fix:
function fibDP(n) {
  const fib = [0, 1];
  for (let i = 1; i <= n; i++) {
    fib[i] = fib[i - 1] + fib[i - 2];
  }
  return fib[n];
}
medium
A. Remove the loop and use recursion instead
B. Initialize fib array with zeros instead of [0,1]
C. Change return to fib[n-1] to fix off-by-one
D. Change loop start to i = 2 to avoid undefined fib[-1]

Solution

  1. Step 1: Analyze loop start index

    The loop starts at i=1, so fib[i-2] becomes fib[-1] which is undefined and causes error.
  2. Step 2: Correct loop start

    Starting loop at i=2 ensures fib[i-2] is fib[0], which is defined, fixing the error.
  3. Final Answer:

    Change loop start to i = 2 to avoid undefined fib[-1] -> Option D
  4. Quick Check:

    Loop from 2 avoids negative index [OK]
Hint: Start loop at 2 to avoid negative indices [OK]
Common Mistakes:
  • Starting loop at 1 causing undefined access
  • Changing return index incorrectly
  • Removing DP and using recursion unnecessarily
5. Given the following code to compute Fibonacci numbers using DP, what will be the output of fibDP(0) and fibDP(1)? How does the function handle these edge cases?
function fibDP(n) {
  if (n <= 1) return n;
  const fib = new Array(n + 1).fill(0);
  fib[1] = 1;
  for (let i = 2; i <= n; i++) {
    fib[i] = fib[i - 1] + fib[i - 2];
  }
  return fib[n];
}
hard
A. fibDP(0) = 1, fibDP(1) = 1; initializes fib array incorrectly
B. fibDP(0) = 0, fibDP(1) = 1; returns n directly for n ≤ 1
C. fibDP(0) = 0, fibDP(1) = 0; misses setting fib[1] properly
D. fibDP(0) and fibDP(1) cause errors due to array size

Solution

  1. Step 1: Check base case handling

    The function returns n directly if n ≤ 1, so fibDP(0) returns 0 and fibDP(1) returns 1 immediately.
  2. Step 2: Confirm array initialization and loop

    For n > 1, the array is initialized and filled correctly, but for n=0 or 1, the function returns early, handling edge cases properly.
  3. Final Answer:

    fibDP(0) = 0, fibDP(1) = 1; returns n directly for n ≤ 1 -> Option B
  4. Quick Check:

    Edge cases return n directly [OK]
Hint: Returns n directly for n ≤ 1 to handle edges [OK]
Common Mistakes:
  • Not handling n=0 or n=1 separately
  • Incorrect array initialization for small n
  • Assuming loop runs for n=0 or n=1
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