Tabulation Bottom Up DP in DSA Javascript - Time & Space Complexity

Time Complexity: Tabulation Bottom Up DP

O(n)

Understanding Time Complexity

We want to understand how the time needed grows when using tabulation in bottom-up dynamic programming.

Specifically, how the number of steps changes as the input size increases.

Scenario Under Consideration

Analyze the time complexity of the following code snippet.

function fib(n) {
  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];
}
    

This code calculates the nth Fibonacci number using bottom-up tabulation.

Identify Repeating Operations

Identify the loops, recursion, array traversals that repeat.

• Primary operation: The for-loop that fills the dp array from 2 to n.
• How many times: It runs (n - 1) times, which grows as n grows.

How Execution Grows With Input

Each new input number requires one more step in the loop to fill the dp array.

Input Size (n)	Approx. Operations
10	About 9 steps
100	About 99 steps
1000	About 999 steps

Pattern observation: The number of steps grows roughly in a straight line with n.

Final Time Complexity

Time Complexity: O(n)

This means the time needed grows directly in proportion to the input size.

Common Mistake

[X] Wrong: "Dynamic programming always takes exponential time because it solves many subproblems."

[OK] Correct: Tabulation solves each subproblem once in order, so it only takes linear time here, not exponential.

Interview Connect

Understanding how tabulation reduces repeated work helps you explain efficient solutions clearly and confidently.

Self-Check

"What if we changed the loop to run from n down to 2 instead? How would the time complexity change?"

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