Tabulation Bottom Up DP in DSA Javascript - Interactive Practice

Practice - 5 Tasks

1fill in blank

easy

Complete the code to initialize the DP array with zeros.

DSA Javascript

const dp = new Array(n + 1).fill([1]);

Drag options to blanks, or click blank then click option'

Aundefined

Bnull

C0

Dfalse

Attempts:

3 left

💡 Hint

Common Mistakes

Using null or undefined causes errors when performing arithmetic later.

Filling with false is a boolean, not a number.

We fill the dp array with 0 because in bottom-up DP, initial states usually start at zero.

2fill in blank

medium

Complete the code to start the loop from the first index for tabulation.

DSA Javascript

for (let i = [1]; i <= n; i++) {

Drag options to blanks, or click blank then click option'

A0

B-1

Cn

D1

Attempts:

3 left

💡 Hint

Common Mistakes

Starting from 0 when the problem requires starting from 1.

Starting from n causes no iterations.

In bottom-up DP, we usually start from 1 to build up solutions from smaller subproblems.

3fill in blank

hard

Fix the error in the code to update dp[i] using dp[i - 1].

DSA Javascript

dp[i] = dp[i - 1] [1] 1;

Drag options to blanks, or click blank then click option'

A+

B/

C*

D-

Attempts:

3 left

💡 Hint

Common Mistakes

Using subtraction causes decreasing values incorrectly.

Multiplication or division is not suitable here.

We add 1 to dp[i - 1] to build the solution incrementally in bottom-up DP.

4fill in blank

hard

Fill both blanks to complete the condition and update in the DP loop.

DSA Javascript

if (arr[i - 1] [1] target) {
  dp[i] = dp[i - 1] [2] dp[i - 1 - arr[i - 1]];
} else {
  dp[i] = dp[i - 1];
}

Drag options to blanks, or click blank then click option'

A<=

B+

C-

D>=

Attempts:

3 left

💡 Hint

Common Mistakes

Using > instead of <= causes logic errors.

Using subtraction instead of addition in dp update.

We check if current element is less or equal to target and add ways from previous states.

5fill in blank

hard

Fill all three blanks to complete the nested loop for 2D DP table filling.

DSA Javascript

for (let i = 1; i <= n; i++) {
  for (let j = 0; j <= target; j++) {
    if (arr[i - 1] [1] j) {
      dp[i][j] = dp[i - 1][j] [2] dp[i - 1][j [3] arr[i - 1]];
    } else {
      dp[i][j] = dp[i - 1][j];
    }
  }
}

Drag options to blanks, or click blank then click option'

A<=

B+

C-

D>=

Attempts:

3 left

💡 Hint

Common Mistakes

Using > instead of <= in condition.

Using subtraction instead of addition for dp update.

Using addition instead of subtraction inside index calculation.

We check if element fits (<=), then add ways (plus) from excluding and including element (subtracting its value from j).

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