Tabulation Bottom Up DP in DSA Typescript

Mental Model

Build the answer step-by-step from the smallest parts up, saving each result in a table so you don't repeat work.

Analogy: Like filling a staircase from the bottom step to the top, where each step depends on the steps below it.

Table: [0] [1] [2] [3] [4] [5]
Values:  0   1   1   2   3   5
↑ start filling from left to right

Dry Run Walkthrough

Input: Find the 5th Fibonacci number using bottom-up tabulation

Goal: Calculate Fibonacci(5) by filling a table from 0 to 5

Step 1: Initialize table with base cases: fib[0] = 0, fib[1] = 1

fib: [0, 1, _, _, _, _]

Why: We need starting points to build the rest

Step 2: Calculate fib[2] = fib[1] + fib[0] = 1 + 0 = 1

fib: [0, 1, 1, _, _, _]

Why: Each Fibonacci number depends on the two before it

Step 3: Calculate fib[3] = fib[2] + fib[1] = 1 + 1 = 2

fib: [0, 1, 1, 2, _, _]

Why: Build next value using previous results

Step 4: Calculate fib[4] = fib[3] + fib[2] = 2 + 1 = 3

fib: [0, 1, 1, 2, 3, _]

Why: Continue filling table step by step

Step 5: Calculate fib[5] = fib[4] + fib[3] = 3 + 2 = 5

fib: [0, 1, 1, 2, 3, 5]

Why: Final answer is at the last index

Result:

fib: [0, 1, 1, 2, 3, 5]
Answer: 5

Annotated Code

DSA Typescript

class Fibonacci {
  static fib(n: number): number {
    if (n <= 1) return n;
    const fibTable: number[] = new Array(n + 1).fill(0);
    fibTable[0] = 0; // base case
    fibTable[1] = 1; // base case
    for (let i = 2; i <= n; i++) {
      fibTable[i] = fibTable[i - 1] + fibTable[i - 2]; // build from smaller subproblems
    }
    return fibTable[n];
  }
}

// Driver code
const n = 5;
console.log(Fibonacci.fib(n));

if (n <= 1) return n;

handle smallest inputs directly

fibTable[0] = 0; // base case

initialize first base case

fibTable[1] = 1; // base case

initialize second base case

for (let i = 2; i <= n; i++) {

iterate from smallest subproblem to target

fibTable[i] = fibTable[i - 1] + fibTable[i - 2];

compute current value from previous two

return fibTable[n];

return final computed answer

Output

5

Complexity Analysis

Time: O(n) because we fill the table once from 0 to n

Space: O(n) because we store all Fibonacci values up to n

vs Alternative: Compared to naive recursion which is exponential, tabulation is much faster by avoiding repeated work

Edge Cases

n = 0

Returns 0 immediately as base case

DSA Typescript

if (n <= 1) return n;

n = 1

Returns 1 immediately as base case

DSA Typescript

if (n <= 1) return n;

When to Use This Pattern

When you see a problem asking for a value built from smaller subproblems with overlapping parts, use bottom-up tabulation to build answers from the smallest pieces up.

Common Mistakes

Mistake: Trying to compute the answer recursively without saving results, causing repeated work
Fix: Use a table to save intermediate results and fill it iteratively

Mistake: Not initializing base cases in the table before the loop
Fix: Always set base cases explicitly before building the rest

Summary

It builds the solution from the smallest subproblems up using a table.

Use it when a problem can be broken into smaller overlapping subproblems.

The key is to fill a table step-by-step so you never repeat calculations.

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