Tabulation Bottom Up DP in DSA Java - Cheat Sheet & Quick Revision

Recall & Review

beginner

What is Tabulation Bottom Up DP?

It is a way to solve problems by filling a table from the smallest subproblems up to the main problem, avoiding repeated work.

beginner

How does Tabulation differ from Memoization?

Tabulation solves subproblems first and stores results in a table, while Memoization solves problems top-down and stores results when needed.

beginner

Why do we use a table in Bottom Up DP?

The table stores answers to smaller problems so we can build answers to bigger problems quickly without repeating work.

beginner

What is the first step in creating a Bottom Up DP solution?

Identify the smallest subproblems and fill their answers in the table as the base cases.

beginner

In Bottom Up DP, how do you find the answer to the main problem?

After filling the table from smallest to largest subproblems, the answer to the main problem is in the last cell of the table.

What does Bottom Up DP start with?

A Smallest subproblems

B The main problem

C Random guesses

D Largest subproblems

Bottom Up DP starts solving from the smallest subproblems and builds up.

Which data structure is commonly used in Tabulation?

A Queue

B Table or array

C Stack

D Linked list

Tabulation uses a table or array to store answers to subproblems.

What is the main advantage of Bottom Up DP over naive recursion?

A Avoids repeated calculations

B Uses less memory

C Runs slower

D Uses random values

Bottom Up DP avoids repeated calculations by storing results.

In Bottom Up DP, where is the final answer usually found?

A First cell of the table

B Middle of the table

C Last cell of the table

D Not stored in the table

The final answer is stored in the last cell after filling the table.

Which problem is a classic example solved by Bottom Up DP?

A Sorting an array

B Binary search

C Depth-first search

D Fibonacci numbers

Fibonacci numbers are a classic example solved efficiently by Bottom Up DP.

Explain how Tabulation Bottom Up DP works using a simple example like Fibonacci numbers.

Think about how you build the Fibonacci sequence from 0 and 1 upwards.

Describe the difference between Memoization and Tabulation in dynamic programming.

Consider how each method approaches solving subproblems.

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