Tabulation Bottom Up DP in DSA Typescript - Interactive Practice

Practice - 5 Tasks

1fill in blank

easy

Complete the code to initialize the DP array with zeros.

DSA Typescript

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

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

A1

B-1

C0

Dnull

Attempts:

3 left

💡 Hint

Common Mistakes

Using 1 instead of 0 causes incorrect base values.

Using null or -1 can cause runtime errors in calculations.

We fill the dp array with 0 because initially no subproblems are solved.

2fill in blank

medium

Complete the code to set the base case for the DP array.

DSA Typescript

dp[0] = [1];

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

An

B1

C0

D-1

Attempts:

3 left

💡 Hint

Common Mistakes

Setting dp[0] to 0 means no ways to make sum zero, which is incorrect.

Setting dp[0] to n or -1 is logically wrong for base case.

Setting dp[0] = 1 means there is exactly one way to reach sum zero: by choosing nothing.

3fill in blank

hard

Fix the error in the loop condition to correctly iterate over sums.

DSA Typescript

for (let sum = 1; sum [1]= target; sum++) {

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

A<

B>=

C>

D<=

Attempts:

3 left

💡 Hint

Common Mistakes

Using < excludes the target sum from calculation.

Using > or >= causes the loop to never run or run incorrectly.

The loop must include the target sum, so use <= to cover all sums up to target.

4fill in blank

hard

Fill both blanks to update the DP array correctly inside the nested loop.

DSA Typescript

if (sum >= nums[i]) {
  dp[sum] = dp[sum] + dp[sum [1] nums[i]];
}

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

A-

B+

C*

D/

Attempts:

3 left

💡 Hint

Common Mistakes

Using + instead of - causes wrong indexing and errors.

Using * or / is not meaningful in this context.

We subtract nums[i] from sum to find the previous state to add ways from.

5fill in blank

hard

Fill all three blanks to complete the tabulation function for subset sum count.

DSA Typescript

function countSubsets(nums: number[], target: number): number {
  const dp: number[] = new Array(target + 1).fill([1]);
  dp[0] = [2];
  for (let i = 0; i < nums.length; i++) {
    for (let sum = target; sum >= [3]; sum--) {
      if (sum >= nums[i]) {
        dp[sum] += dp[sum - nums[i]];
      }
    }
  }
  return dp[target];
}

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

A0

B1

Cnums[i]

Dtarget

Attempts:

3 left

💡 Hint

Common Mistakes

Initializing dp with 1 instead of 0 causes wrong counts.

Setting dp[0] to 0 means no base case.

Iterating sums forward causes overcounting.

Initialize dp with 0, set dp[0] = 1 for base case, and iterate sums down from target to nums[i] to avoid reuse in the same iteration.

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