Tabulation Bottom Up DP in DSA Typescript - 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.Using a smart way called Tabulation Bottom Up Dynamic Programming, you will build the sequence step by step from the bottom up, saving results to avoid repeating work.

🎯 Goal: Build a TypeScript program that uses tabulation to find the Fibonacci number at position n. You will create a table (array) to store results and fill it from the start up to n.

📋 What You'll Learn

Create an array called fibTable with size n + 1 to store Fibonacci numbers

Set the first two Fibonacci numbers in fibTable as 0 and 1

Use a for loop with variable i from 2 to n to fill fibTable

Calculate each Fibonacci number as the sum of the two previous numbers in fibTable

Print the Fibonacci number at position n from fibTable

💡 Why This Matters

🌍 Real World

Tabulation bottom up dynamic programming is used in many real-world problems like calculating sequences, optimizing resource allocation, and solving complex puzzles efficiently.

💼 Career

Understanding tabulation helps in software engineering roles that require optimization and efficient algorithm design, such as game development, data analysis, and system design.

1

Create the Fibonacci table array

Create a variable called n and set it to 7. Then create an array called fibTable with size n + 1 filled with zeros.

DSA Typescript

const n = 7;
// Your code here

Hint

Use new Array(n + 1).fill(0) to create an array of zeros with length n + 1.

2

Initialize the first two Fibonacci numbers

Set fibTable[0] to 0 and fibTable[1] to 1.

DSA Typescript

const n = 7;
const fibTable: number[] = new Array(n + 1).fill(0);
// Your code here

Hint

Assign the first two Fibonacci numbers directly to the first two positions of the array.

3

Fill the Fibonacci table using a for loop

Use a for loop with variable i from 2 to n inclusive. Inside the loop, set fibTable[i] to the sum of fibTable[i - 1] and fibTable[i - 2].

DSA Typescript

const n = 7;
const fibTable: number[] = new Array(n + 1).fill(0);
fibTable[0] = 0;
fibTable[1] = 1;
// Your code here

Hint

Loop from 2 to n and fill each position by adding the two previous values in the array.

4

Print the Fibonacci number at position n

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

DSA Typescript

const n = 7;
const fibTable: number[] = new Array(n + 1).fill(0);
fibTable[0] = 0;
fibTable[1] = 1;
for (let i = 2; i <= n; i++) {
    fibTable[i] = fibTable[i - 1] + fibTable[i - 2];
}
// Your code here

Hint

Use console.log(fibTable[n]) to print the final Fibonacci number.

Practice

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

easy

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

B. Solve the problem by dividing it into two halves recursively

C. Use recursion with memoization to avoid repeated work

D. Randomly guess answers and check correctness

Solution

1. Step 1: Understand Tabulation approach

   Tabulation means filling a table starting from the smallest subproblems, moving upwards.

2. Step 2: Compare with other methods

   Unlike recursion with memoization, tabulation solves subproblems iteratively in order.

3. Final Answer:

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

4. Quick Check:

   Tabulation = Bottom Up DP [OK]

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

Common Mistakes:

• Confusing tabulation with recursion
• Thinking tabulation guesses answers
• Mixing memoization with tabulation

2. Which of the following is the correct TypeScript syntax to initialize a DP array of size n+1 with zeros for tabulation?

easy

A. const dp = new Array(n+1).fill(0);

B. const dp = Array().map(() => 0);

C. const dp = new Array(n).fill(null);

D. const dp = Array.fill(n+1, 0);

Solution

1. Step 1: Check correct array initialization

   Using new Array(n+1).fill(0) creates an array of length n+1 filled with zeros.

2. Step 2: Identify incorrect options

   const dp = Array().map(() => 0); uses map on empty array which won't fill correctly; const dp = new Array(n).fill(null); fills with nulls; const dp = Array.fill(n+1, 0); is invalid syntax.

3. Final Answer:

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

4. Quick Check:

   Array fill with zeros = new Array(n+1).fill(0) [OK]

Hint: Use new Array(size).fill(value) to initialize arrays [OK]

Common Mistakes:

• Using map on empty array without elements
• Wrong array size (n instead of n+1)
• Using invalid Array.fill syntax

3. What will be the output of the following TypeScript code using bottom-up DP to find the nth Fibonacci number?

function fib(n: number): number {
  const dp = 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. 3

B. 5

C. 8

D. 13

Solution

1. Step 1: Trace dp array filling 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 B

4. Quick Check:

   Fibonacci(5) = 5 [OK]

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

Common Mistakes:

• Starting dp[1] as 0 instead of 1
• Off-by-one errors in loop
• Returning dp[n-1] instead of dp[n]

4. Identify the bug in this bottom-up DP code for calculating factorial of n:

function factorial(n: number): number {
  const dp = new Array(n + 1).fill(0);
  for (let i = 1; i <= n; i++) {
    dp[i] = dp[i - 1] * i;
  }
  return dp[n];
}
console.log(factorial(4));

medium

A. The loop should start from i = 0

B. dp array should be initialized with zeros, not ones

C. The multiplication should be dp[i] = dp[i] * i

D. dp[0] should be set to 1 before the loop

Solution

1. Step 1: Check dp initialization and base case

   dp is filled with 0s, but dp[0] must be explicitly set to 1 as factorial(0)=1.

2. Step 2: Verify loop correctness

   Loop from i=1 to n is correct; dp[i] = dp[i-1] * i is correct formula.

3. Final Answer:

   dp[0] should be set to 1 before the loop -> Option D

4. Quick Check:

   Base case dp[0]=1 needed [OK]

Hint: Always set base case explicitly in DP arrays [OK]

Common Mistakes:

• Starting loop at 0 causing dp[-1] access
• Wrong initialization of dp array
• Incorrect multiplication formula

5. You want to find the minimum number of coins needed to make amount n using coins of denominations [1, 3, 4] with bottom-up DP. Which of these is the correct tabulation approach?

hard

A. Initialize dp array with zeros; for each coin, add 1 to dp[coin] only once

B. Use recursion to try all coin combinations without dp array

C. Initialize dp array with Infinity except dp[0]=0; for each amount i from 1 to n, update dp[i] = min(dp[i], dp[i - coin] + 1) for each coin if i >= coin

D. Initialize dp array with zeros; for each amount i, set dp[i] = dp[i-1] + 1 only

Solution

1. Step 1: Understand coin change tabulation

   We start dp with Infinity except dp[0]=0 because 0 coins needed for amount 0.

2. Step 2: Update dp for each amount and coin

   For each amount i, check all coins; if coin <= i, update dp[i] = min(dp[i], dp[i-coin] + 1).

3. Final Answer:

   Initialize dp array with Infinity except dp[0]=0; for each amount i from 1 to n, update dp[i] = min(dp[i], dp[i - coin] + 1) for each coin if i >= coin -> Option C

4. Quick Check:

   Coin change DP = min coins with dp and Infinity init [OK]

Hint: Use Infinity init and min update for coin change DP [OK]

Common Mistakes:

• Not initializing dp with Infinity
• Ignoring multiple coin options per amount
• Using recursion instead of tabulation