Why Tabulation Bottom Up DP in DSA Typescript?

The Big Idea

What if you could solve tough problems by building answers step-by-step without repeating yourself?

The Scenario

Imagine you want to find the fastest way to climb stairs where each step depends on the previous ones. You try to calculate the answer by repeating the same steps over and over manually.

The Problem

Doing this by hand means you keep solving the same smaller problems again and again, wasting time and making mistakes. It's like repeatedly counting the same stairs without remembering your progress.

The Solution

Tabulation Bottom Up DP solves this by starting from the smallest steps and building up the answer step by step, storing results in a table so you never repeat work. It's like writing down each stair's best time as you climb, so you always know the fastest path.

Before vs After

✗ Before

function climbStairs(n: number): number {
  if (n <= 2) return n;
  return climbStairs(n - 1) + climbStairs(n - 2);
}

✓ After

function climbStairs(n: number): number {
  const dp: number[] = [0, 1, 2];
  for (let step = 3; step <= n; step++) {
    dp[step] = dp[step - 1] + dp[step - 2];
  }
  return dp[n];
}

What It Enables

This method lets you solve complex problems quickly by building answers from the ground up without repeating work.

Real Life Example

Planning the fastest route in a video game where each move depends on previous moves, tabulation helps you find the best path efficiently.

Key Takeaways

Manual repeated calculations waste time and cause errors.

Tabulation stores results in a table from smallest to largest subproblems.

This approach speeds up problem solving by avoiding repeated work.

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