Why Tabulation Bottom Up DP in DSA Javascript?

The Big Idea

Discover how building answers step-by-step can save you from endless repeated work!

The Scenario

Imagine you want to find the number of ways to climb stairs, but you try to list all possibilities by hand or with simple loops.

As the number of stairs grows, this manual counting becomes impossible to keep track of.

The Problem

Manually counting or using simple recursion repeats the same calculations many times, making it very slow and confusing.

You waste time recalculating answers you already know, and it's easy to make mistakes.

The Solution

Tabulation Bottom Up DP builds the answer step by step from the smallest cases up, storing results in a table.

This way, each subproblem is solved once, and the final answer is quickly found without repeated work.

Before vs After

✗ Before

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

✓ After

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

What It Enables

It enables solving complex problems efficiently by breaking them into simple steps and remembering results.

Real Life Example

Calculating the fastest way to reach a destination by considering all possible routes without repeating the same path calculations.

Key Takeaways

Manual counting repeats work and is slow.

Tabulation stores answers from bottom up to avoid repeats.

This method speeds up solving problems with overlapping subproblems.

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