Tabulation Bottom Up DP in DSA Java - Interactive Practice

Practice - 5 Tasks

1fill in blank

easy

Complete the code to initialize the DP array for bottom-up tabulation.

DSA Java

int[] dp = new int[[1]];

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An - 1

Bn

Cn + 1

Dn * 2

Attempts:

3 left

💡 Hint

Common Mistakes

Using size n causes index out of bounds when accessing dp[n].

The DP array size should be n + 1 to include the base case at index 0.

2fill in blank

medium

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

DSA Java

dp[0] = [1];

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A1

Bn

C-1

D0

Attempts:

3 left

💡 Hint

Common Mistakes

Setting dp[0] to 1 or n causes incorrect base initialization.

The base case for many DP problems is dp[0] = 0, representing zero items or zero capacity.

3fill in blank

hard

Fix the error in the loop condition to iterate over all subproblems.

DSA Java

for (int i = 1; i [1] dp.length; i++) {

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A<=

B<

C>

D>=

Attempts:

3 left

💡 Hint

Common Mistakes

Using '<=' causes index out of bounds error.

The loop should run while i is less than dp.length to avoid index out of bounds.

4fill in blank

hard

Fill both blanks to complete the inner loop and condition for the knapsack problem.

DSA Java

for (int w = 1; w [1] W; w++) {
    if (weights[i - 1] [2] w) {

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A<=

B<

C>=

D>

Attempts:

3 left

💡 Hint

Common Mistakes

Using '<' excludes the max capacity W.

Using '>' reverses the condition logic.

We loop w from 1 to W inclusive, and check if the current item's weight is less than or equal to w.

5fill in blank

hard

Fill all three blanks to complete the DP state update for the knapsack problem.

DSA Java

dp[i][w] = Math.max(dp[i - 1][w], dp[i - 1][w - [1]] + [2]);
} else {
    dp[i][w] = [3];

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Aweights[i - 1]

Bvalues[i - 1]

Cdp[i - 1][w]

Dw

Attempts:

3 left

💡 Hint

Common Mistakes

Mixing up weights and values.

Using wrong DP indices.

The DP formula chooses the max between not taking the item and taking it (adding its value and reducing capacity).

Practice

1. What is the main idea behind Tabulation Bottom Up DP in dynamic programming?

easy

A. Solve the problem by recursion without storing results

B. Start solving from the biggest problem and break it down

C. Use random guesses to find the solution

D. Build a table from the smallest subproblems up to the main problem

Solution

1. Step 1: Understand Tabulation Approach

   Tabulation means filling a table starting from the smallest subproblems to bigger ones.

2. Step 2: Compare with other methods

   Unlike recursion, 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 D

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 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 DP array for bottom-up Fibonacci in Java?

easy

A. int[] dp = new int[n+1]; dp[0] = 0; dp[1] = 1;

B. int[] dp = new int[n]; dp[1] = 0; dp[2] = 1;

C. int[] dp = new int[n+1]; dp[0] = 1; dp[1] = 1;

D. int[] dp = new int[n]; dp[0] = 0; dp[1] = 1;

Solution

1. Step 1: Check array size for Fibonacci

   We need size n+1 to store from 0 to n indices.

2. Step 2: Set base cases correctly

   Fibonacci starts with dp[0]=0 and dp[1]=1.

3. Final Answer:

   int[] dp = new int[n+1]; dp[0] = 0; dp[1] = 1; -> Option A

4. Quick Check:

   Base cases dp[0]=0, dp[1]=1 [OK]

Hint: Array size n+1 and dp[0]=0, dp[1]=1 for Fibonacci [OK]

Common Mistakes:

• Using wrong array size causing IndexOutOfBounds
• Swapping base case values
• Starting indices from 1 instead of 0

3. What will be the output of the following Java code using bottom-up DP to find Fibonacci of 5?

int n = 5;
int[] dp = new int[n+1];
dp[0] = 0; dp[1] = 1;
for (int i = 2; i <= n; i++) {
    dp[i] = dp[i-1] + dp[i-2];
}
System.out.println(dp[n]);

medium

A. 13

B. 5

C. 3

D. 8

Solution

1. Step 1: Trace DP array filling

   dp[2]=1, dp[3]=2, dp[4]=3, dp[5]=5 by adding previous two values.

2. Step 2: Output dp[5]

   dp[5] is 5, so printed output 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:

• Confusing index with Fibonacci value
• Off-by-one errors in loop
• Printing wrong dp index

4. Identify the error in this bottom-up DP code for Fibonacci:

int n = 5;
int[] dp = new int[n];
dp[0] = 0; dp[1] = 1;
for (int i = 2; i <= n; i++) {
    dp[i] = dp[i-1] + dp[i-2];
}
System.out.println(dp[n]);

medium

A. Array size is too small, causing IndexOutOfBoundsException

B. Base cases dp[0] and dp[1] are incorrect

C. Loop condition should be i < n instead of i <= n

D. No error, code runs fine

Solution

1. Step 1: Check array size

   Array dp has size n=5, valid indices 0 to 4 only.

2. Step 2: Loop accesses dp[5]

   Loop runs i=2 to 5 inclusive, dp[5] access causes error.

3. Final Answer:

   Array size is too small, causing IndexOutOfBoundsException -> Option A

4. Quick Check:

   Array size must be n+1 for dp[n] access [OK]

Hint: Array size must cover all indices used in loop [OK]

Common Mistakes:

• Using array size n instead of n+1
• Ignoring loop boundary causing runtime error
• Assuming code runs without error

5. You want to find the number of ways to climb n stairs if you can take 1 or 2 steps at a time using bottom-up DP. Which code snippet correctly implements this?

hard

A. int[] dp = new int[n+1]; dp[0]=0; dp[1]=1; for(int i=2;i<=n;i++){ dp[i]=dp[i-1]+dp[i-2]; } return dp[n];

B. int[] dp = new int[n]; dp[0]=1; dp[1]=2; for(int i=2;i

C. int[] dp = new int[n+1]; dp[0]=1; dp[1]=1; for(int i=2;i<=n;i++){ dp[i]=dp[i-1]+dp[i-2]; } return dp[n];

D. int[] dp = new int[n+1]; dp[0]=1; dp[1]=1; for(int i=1;i<=n;i++){ dp[i]=dp[i-1]+dp[i-2]; } return dp[n];

Solution

1. Step 1: Initialize base cases for climbing stairs

   dp[0]=1 means 1 way to stand still; dp[1]=1 means 1 way to climb 1 step.

2. Step 2: Fill dp using sum of previous two steps

   For i≥2, dp[i] = dp[i-1] + dp[i-2] counts ways taking 1 or 2 steps.

3. Step 3: Check loop and return

   Loop runs i=2 to n inclusive; return dp[n] is correct.

4. Final Answer:

   int[] dp = new int[n+1]; dp[0]=1; dp[1]=1; for(int i=2;i<=n;i++){ dp[i]=dp[i-1]+dp[i-2]; } return dp[n]; -> Option C

5. Quick Check:

   Climbing stairs DP uses dp[i]=dp[i-1]+dp[i-2] [OK]

Hint: Base dp[0]=1, dp[1]=1; sum previous two for dp[i] [OK]

Common Mistakes:

• Wrong base case values
• Multiplying instead of adding
• Incorrect loop start or end indices