Tabulation Bottom Up DP in DSA Java - Execution Trace

Concept Flow - Tabulation Bottom Up DP

Initialize DP table with base values

↓

Iterate over items

↓

Iterate over capacities

↓

Compute max value for current cell

↓

Fill DP table row by row

↓

Final answer at DP[last_item

Start by setting base cases in the table, then fill it row by row using previous results until the final answer is found.

Execution Sample

DSA Java

int[] weights = {1, 3, 4};
int[] values = {1, 4, 5};
int W = 7;
int[][] dp = new int[4][8];
// Fill dp bottom-up

This code sets up weights, values, and capacity, then creates a DP table to solve the knapsack problem using bottom-up tabulation.

Execution Table

Step	Operation	Item Index	Capacity	DP Cell Computed	Value Computed	DP Table State (Row)
1	Initialize base row	-	0 to 7	dp[0][0..7]	0	[0, 0, 0, 0, 0, 0, 0, 0]
2	Process item 1 (weight=1, value=1)	1	0	dp[1][0]	0	[0, _, _, _, _, _, _, _]
3	Process item 1	1	1	dp[1][1]	1	[0, 1, _, _, _, _, _, _]
4	Process item 1	1	2	dp[1][2]	1	[0, 1, 1, _, _, _, _, _]
5	Process item 1	1	3	dp[1][3]	1	[0, 1, 1, 1, _, _, _, _]
6	Process item 1	1	4	dp[1][4]	1	[0, 1, 1, 1, 1, _, _, _]
7	Process item 1	1	5	dp[1][5]	1	[0, 1, 1, 1, 1, 1, _, _]
8	Process item 1	1	6	dp[1][6]	1	[0, 1, 1, 1, 1, 1, 1, _]
9	Process item 1	1	7	dp[1][7]	1	[0, 1, 1, 1, 1, 1, 1, 1]
10	Process item 2 (weight=3, value=4)	2	0	dp[2][0]	0	[0, _, _, _, _, _, _, _]
11	Process item 2	2	1	dp[2][1]	1	[0, 1, _, _, _, _, _, _]
12	Process item 2	2	2	dp[2][2]	1	[0, 1, 1, _, _, _, _, _]
13	Process item 2	2	3	dp[2][3]	4	[0, 1, 1, 4, _, _, _, _]
14	Process item 2	2	4	dp[2][4]	5	[0, 1, 1, 4, 5, _, _, _]
15	Process item 2	2	5	dp[2][5]	5	[0, 1, 1, 4, 5, 5, _, _]
16	Process item 2	2	6	dp[2][6]	5	[0, 1, 1, 4, 5, 5, 5, _]
17	Process item 2	2	7	dp[2][7]	5	[0, 1, 1, 4, 5, 5, 5, 5]
18	Process item 3 (weight=4, value=5)	3	0	dp[3][0]	0	[0, _, _, _, _, _, _, _]
19	Process item 3	3	1	dp[3][1]	1	[0, 1, _, _, _, _, _, _]
20	Process item 3	3	2	dp[3][2]	1	[0, 1, 1, _, _, _, _, _]
21	Process item 3	3	3	dp[3][3]	4	[0, 1, 1, 4, _, _, _, _]
22	Process item 3	3	4	dp[3][4]	5	[0, 1, 1, 4, 5, _, _, _]
23	Process item 3	3	5	dp[3][5]	6	[0, 1, 1, 4, 5, 6, _, _]
24	Process item 3	3	6	dp[3][6]	6	[0, 1, 1, 4, 5, 6, 6, _]
25	Process item 3	3	7	dp[3][7]	9	[0, 1, 1, 4, 5, 6, 6, 9]
26	End	-	-	-	-	Final answer dp[3][7] = 9

💡 All items processed for all capacities; final answer found at dp[3][7]

Variable Tracker

Variable	Start	After Step 9	After Step 17	After Step 25	Final
dp[0]	[0,0,0,0,0,0,0,0]	[0,0,0,0,0,0,0,0]	[0,0,0,0,0,0,0,0]	[0,0,0,0,0,0,0,0]	[0,0,0,0,0,0,0,0]
dp[1]	[_,_,_,_,_,_,_,_]	[0,1,1,1,1,1,1,1]	[0,1,1,1,1,1,1,1]	[0,1,1,1,1,1,1,1]	[0,1,1,1,1,1,1,1]
dp[2]	[_,_,_,_,_,_,_,_]	[_,_,_,_,_,_,_,_]	[0,1,1,4,5,5,5,5]	[0,1,1,4,5,5,5,5]	[0,1,1,4,5,5,5,5]
dp[3]	[_,_,_,_,_,_,_,_]	[_,_,_,_,_,_,_,_]	[_,_,_,_,_,_,_,_]	[0,1,1,4,5,6,6,9]	[0,1,1,4,5,6,6,9]

Key Moments - 3 Insights

Why do we initialize the first row of the DP table with zeros?

The first row represents zero items, so no value can be taken regardless of capacity. This is shown in execution_table row 1 where dp[0][capacity] = 0.

How do we decide whether to include the current item or not in the DP cell?

We compare the value of not taking the item (dp from previous row) and taking it (value of current item plus dp of remaining capacity). This is seen in rows 13-25 where dp cells are computed by max of these two.

Why does the capacity loop start from 0 and go up to max capacity?

We fill the DP table for all capacities to build up solutions for smaller capacities first, enabling reuse for larger capacities. This is visible in the stepwise filling from capacity 0 to 7 in execution_table.

Visual Quiz - 3 Questions

Look at the execution_table at step 13, what value is computed for dp[2][3]?

A 1

B 5

C 4

D 0

💡 Hint

Check the 'Value Computed' column at step 13 in execution_table.

At which step does the DP table first show a value greater than 5?

A Step 23

B Step 14

C Step 25

D Step 17

💡 Hint

Look at the 'Value Computed' column and DP Table State rows for values >5.

If the weight of item 2 was increased to 5, how would dp[2][4] change at step 14?

A It would remain 5

B It would become 1

C It would become 4

D It would become 0

💡 Hint

Consider if item 2 can fit in capacity 4 with weight 5; check dp logic in execution_table steps.

Concept Snapshot

Tabulation Bottom Up DP:
- Create a 2D DP table with rows = items+1, cols = capacity+1
- Initialize first row with 0 (no items)
- Fill table row-wise: dp[i][w] = max(dp[i-1][w], value[i-1] + dp[i-1][w - weight[i-1]]) if weight[i-1] <= w
- Final answer at dp[items][capacity]
- Builds solution from smaller subproblems up

Full Transcript

Tabulation Bottom Up Dynamic Programming solves problems by building a table from the smallest subproblems up to the full problem. We start by initializing the first row with zeros because no items means no value. Then, for each item and capacity, we decide whether to include the item or not by comparing the value of including it plus the best value for the remaining capacity, or excluding it. This process fills the DP table row by row until the final answer is found at the bottom-right cell. This method avoids recursion and stores all intermediate results for efficient computation.

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