Tabulation Bottom Up DP in DSA Javascript - Build from Scratch

Tabulation Bottom Up DP: Fibonacci Sequence

📖 Scenario: Imagine you want to find the nth number in the Fibonacci sequence. This sequence starts with 0 and 1, and each next number is the sum of the two before it. Calculating this directly can be slow for big numbers.We will use a method called Tabulation Bottom Up Dynamic Programming to solve this quickly by building the answer step by step from the smallest cases.

🎯 Goal: Build a JavaScript program that uses tabulation (bottom-up dynamic programming) to find the nth Fibonacci number efficiently.

📋 What You'll Learn

Create an array to store Fibonacci numbers starting from 0 and 1

Create a variable n to represent which Fibonacci number to find

Use a for loop to fill the array from the bottom up

Print the nth Fibonacci number from the array

💡 Why This Matters

🌍 Real World

Calculating Fibonacci numbers quickly is useful in computer graphics, financial models, and nature simulations where patterns follow Fibonacci sequences.

💼 Career

Understanding dynamic programming and tabulation is important for software engineers to solve optimization problems efficiently.

1

Create the Fibonacci array with base cases

Create an array called fib with the first two Fibonacci numbers: 0 and 1.

DSA Javascript

// Create the fib array with first two numbers
// Your code here

Hint

The Fibonacci sequence always starts with 0 and 1. Store these in an array called fib.

2

Set the target Fibonacci number index

Create a variable called n and set it to 7 to find the 7th Fibonacci number.

DSA Javascript

const fib = [0, 1];
// Create variable n and set it to 7
// Your code here

Hint

We want to find the 7th Fibonacci number, so set n to 7.

3

Fill the Fibonacci array using a for loop

Use a for loop with variable i starting from 2 up to and including n. Inside the loop, set fib[i] to fib[i - 1] + fib[i - 2].

DSA Javascript

const fib = [0, 1];
const n = 7;
// Use a for loop from i = 2 to n to fill fib array
// Your code here

Hint

Start from 2 because the first two Fibonacci numbers are already known. Add the two previous numbers to get the current one.

4

Print the nth Fibonacci number

Print the value of fib[n] using console.log.

DSA Javascript

const fib = [0, 1];
const n = 7;
for (let i = 2; i <= n; i++) {
    fib[i] = fib[i - 1] + fib[i - 2];
}
// Print the nth Fibonacci number
// Your code here

Hint

The 7th Fibonacci number is 13. Use console.log(fib[n]) to print it.

Practice

1. What is the main idea behind Tabulation Bottom Up DP in dynamic programming?

easy

A. Build a table from the smallest subproblems up to the main problem

B. Solve the problem by recursion without storing results

C. Use random guesses to find the solution

D. Start solving from the biggest problem and break it down

Solution

1. Step 1: Understand bottom-up approach

   Tabulation means filling a table starting from the smallest subproblems to bigger ones.

2. Step 2: Compare with other methods

   Unlike recursion without storage, tabulation avoids repeated work by storing results in a table.

3. Final Answer:

   Build a table from the smallest subproblems up to the main problem -> Option A

4. Quick Check:

   Tabulation = Bottom-up table filling [OK]

Hint: Think: solve small parts first, then bigger [OK]

Common Mistakes:

• Confusing tabulation with recursion without memo
• Thinking tabulation starts from the biggest problem
• Ignoring the use of a table to store results

2. Which of the following is the correct way to initialize a tabulation array for Fibonacci numbers in JavaScript?

easy

A. const dp = [];

B. const dp = [0, 1];

C. const dp = [0, 0];

D. const dp = [1, 1];

Solution

1. Step 1: Recall Fibonacci base cases

   Fibonacci sequence starts with 0 and 1, so dp[0] = 0 and dp[1] = 1.

2. Step 2: Check initialization options

   Only const dp = [0, 1]; correctly sets dp = [0, 1] matching Fibonacci base values.

3. Final Answer:

   const dp = [0, 1]; -> Option B

4. Quick Check:

   Fibonacci base = [0,1] [OK]

Hint: Start dp with first two Fibonacci numbers [OK]

Common Mistakes:

• Starting dp with [1,1] which is incorrect base
• Initializing dp with zeros only
• Leaving dp empty without base values

3. What will be the output of the following JavaScript code using bottom-up DP to find Fibonacci number at position 5?

function fib(n) {
  const dp = [0, 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. 3

B. 8

C. 5

D. 13

Solution

1. Step 1: Trace dp array filling

   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 C

4. Quick Check:

   fib(5) = 5 [OK]

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

Common Mistakes:

• Confusing index with Fibonacci value
• Returning dp[n-1] instead of dp[n]
• Starting dp with wrong base values

4. Find the error in this bottom-up DP code for Fibonacci numbers:

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

medium

A. dp[i - 2] causes out-of-bound error

B. dp array is not initialized

C. Return statement is missing

D. Loop starts at 1 instead of 2

Solution

1. Step 1: Analyze loop start index

   The loop starts at i=1, but dp[1] is already initialized; calculation should start at i=2.

2. Step 2: Identify impact of wrong loop start

   At i=1, dp[1] = dp[0] + dp[-1] causes dp[-1] undefined, leading to NaN.

3. Final Answer:

   Loop starts at 1 instead of 2 -> Option D

4. Quick Check:

   Loop must start at 2 to avoid invalid index [OK]

Hint: Start loop at 2 to avoid negative index [OK]

Common Mistakes:

• Starting loop at 1 causing invalid dp[-1]
• Ignoring dp initialization
• Assuming dp[-1] is zero

5. Given the problem to find the minimum number of coins to make amount 7 using coins [1, 3, 4], which bottom-up DP table entry represents the minimum coins needed for amount 5?
Assuming dp[0] = 0 and dp[i] = Infinity initially for i > 0, what is dp[5] after filling the table?

hard

A. 2

B. 3

C. 1

D. Infinity

Solution

1. Step 1: Fill dp table for amounts up to 5

   dp[0]=0; dp[1]=1 (1 coin of 1); dp[2]=2 (1+1); dp[3]=1 (coin 3); dp[4]=1 (coin 4); dp[5]=min(dp[5-1], dp[5-3], dp[5-4]) +1 = min(dp[4], dp[2], dp[1]) +1 = min(1,2,1)+1=2.

2. Step 2: Confirm dp[5] value

   Minimum coins for amount 5 is 2.

3. Final Answer:

   2 -> Option A

4. Quick Check:

   dp[5] = 2 coins [OK]

Hint: Check all coin options and pick minimum +1 [OK]

Common Mistakes:

• Not considering all coin choices for dp[i]
• Using wrong initial dp values
• Confusing amount with coin count