Tabulation Bottom Up DP in DSA C# - 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 bottom-up

↓

Final answer at DP[last item

Start by setting base cases in the table, then fill each cell by checking if including the current item improves the value, moving from smaller subproblems to bigger ones.

Execution Sample

DSA C#

int[,] dp = new int[n+1, W+1];
for (int i = 1; i <= n; i++) {
  for (int w = 1; w <= W; w++) {
    if (weight[i-1] <= w) {
      dp[i,w] = Math.Max(dp[i-1,w], dp[i-1,w-weight[i-1]] + value[i-1]);
    } else {
      dp[i,w] = dp[i-1,w];
    }
  }
}

This code fills a 2D table to solve the 0/1 Knapsack problem by building solutions from smaller subproblems.

Execution Table

Step	Item (i)	Capacity (w)	Condition (weight[i-1] <= w)	dp[i-1,w]	dp[i-1,w-weight[i-1]] + value[i-1]	dp[i,w]	DP Table State (row i)
1	1	1	1 <= 1 True	0	0 + 1 = 1	1	[0, 1, 1, 1, 1, 1, 1, 1]
2	1	2	1 <= 2 True	0	0 + 1 = 1	1	[0, 1, 1, 1, 1, 1, 1, 1]
3	1	3	1 <= 3 True	0	0 + 1 = 1	1	[0, 1, 1, 1, 1, 1, 1, 1]
4	1	4	1 <= 4 True	0	0 + 1 = 1	1	[0, 1, 1, 1, 1, 1, 1, 1]
5	1	5	1 <= 5 True	0	0 + 1 = 1	1	[0, 1, 1, 1, 1, 1, 1, 1]
6	1	6	1 <= 6 True	0	0 + 1 = 1	1	[0, 1, 1, 1, 1, 1, 1, 1]
7	1	7	1 <= 7 True	0	0 + 1 = 1	1	[0, 1, 1, 1, 1, 1, 1, 1]
8	2	1	3 <= 1 False	1	-	1	[0, 1, 1, 4, 5, 5, 5, 5]
9	2	2	3 <= 2 False	1	-	1	[0, 1, 1, 4, 5, 5, 5, 5]
10	2	3	3 <= 3 True	1	0 + 4 = 4	4	[0, 1, 1, 4, 5, 5, 5, 5]
11	2	4	3 <= 4 True	1	1 + 4 = 5	5	[0, 1, 1, 4, 5, 5, 5, 5]
12	2	5	3 <= 5 True	1	1 + 4 = 5	5	[0, 1, 1, 4, 5, 5, 5, 5]
13	2	6	3 <= 6 True	1	1 + 4 = 5	5	[0, 1, 1, 4, 5, 5, 5, 5]
14	2	7	3 <= 7 True	1	1 + 4 = 5	5	[0, 1, 1, 4, 5, 5, 5, 5]
15	3	1	4 <= 1 False	1	-	1	[0, 1, 1, 4, 5, 6, 6, 9]
16	3	2	4 <= 2 False	1	-	1	[0, 1, 1, 4, 5, 6, 6, 9]
17	3	3	4 <= 3 False	4	-	4	[0, 1, 1, 4, 5, 6, 6, 9]
18	3	4	4 <= 4 True	5	0 + 5 = 5	5	[0, 1, 1, 4, 5, 6, 6, 9]
19	3	5	4 <= 5 True	5	1 + 5 = 6	6	[0, 1, 1, 4, 5, 6, 6, 9]
20	3	6	4 <= 6 True	5	1 + 5 = 6	6	[0, 1, 1, 4, 5, 6, 6, 9]
21	3	7	4 <= 7 True	5	4 + 5 = 9	9	[0, 1, 1, 4, 5, 6, 6, 9]

💡 All items and capacities processed; final answer is dp[3,7] = 9

Variable Tracker

Variable	Start	After Step 7	After Step 14	After Step 21
i (item index)	0	1	2	3
w (capacity)	0	7	7	7
dp table row i	[all zeros]	[0, 1, 1, 1, 1, 1, 1, 1]	[0, 1, 1, 4, 5, 5, 5, 5]	[0, 1, 1, 4, 5, 6, 6, 9]

Key Moments - 3 Insights

Why do we check if weight[i-1] <= w before updating dp[i,w]?

Because we can only include the current item if it fits in the current capacity w. This is shown in execution_table rows like 10 where the condition is true and we update dp[i,w], and rows like 8 where it's false and we skip including the item.

Why does dp[i,w] take the max of dp[i-1,w] and dp[i-1,w-weight[i-1]] + value[i-1]?

dp[i-1,w] is the best value without the current item, dp[i-1,w-weight[i-1]] + value[i-1] is the value if we include it. We pick the better one to maximize value, as seen in steps 10 and 11.

Why do we start i and w from 1 instead of 0?

Because dp[0,w] and dp[i,0] represent base cases with no items or zero capacity, initialized to 0. The loops start from 1 to build on these base cases, as shown by the initial dp table rows in the execution_table.

Visual Quiz - 3 Questions

Look at the execution_table at step 10, what is the value of dp[2,3]?

A 1

B 4

C 5

D 0

💡 Hint

Check the dp[i,w] column at step 10 in the execution_table.

At which step does the condition weight[i-1] <= w become false for the first time?

A Step 8

B Step 1

C Step 15

D Step 10

💡 Hint

Look at the Condition column in execution_table rows; step 8 shows the first false.

If the value of the third item changed from 5 to 6, how would dp[3,7] change at step 21?

A It would remain 9

B It would become 11

C It would become 10

D It would become 12

💡 Hint

Look at the calculation dp[i-1,w-weight[i-1]] + value[i-1] at step 21 and add 1 to the value.

Concept Snapshot

Tabulation Bottom Up DP builds a table from smallest subproblems up.
Initialize dp table with zeros for base cases.
For each item i and capacity w:
  If item fits (weight[i-1] <= w), dp[i,w] = max(include, exclude).
  Else dp[i,w] = dp[i-1,w].
Answer is dp[n,W] after filling table.

Full Transcript

Tabulation Bottom Up Dynamic Programming solves problems by filling a table from the smallest cases to the full problem. We start with a table dp where dp[i,w] means the best value using first i items with capacity w. We initialize dp with zeros for no items or zero capacity. Then for each item and capacity, we check if the item fits. If yes, we choose the best between including or excluding it. If no, we exclude it. This process fills the table row by row. The final answer is at dp[n,W]. The execution table shows each step with conditions and updates. The variable tracker shows how dp table rows change after processing items. Key moments clarify why conditions and max choices are made. The quiz tests understanding of dp values and conditions. This method ensures all subproblems are solved before the big one, avoiding repeated work.

Practice

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

easy

A. Start solving from the main problem and break it down recursively

B. Solve the problem by recursion without storing results

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 bottom-up approach

   Tabulation means filling a table starting from the smallest subproblems, building up to the final answer.

2. Step 2: Compare with other methods

   Unlike recursion without memoization, 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 table filling [OK]

Hint: Think: solve small parts first, then bigger [OK]

Common Mistakes:

• Confusing tabulation with recursion
• Ignoring the order of filling the table
• Not storing intermediate results

2. Which of the following is the correct way to initialize a tabulation array for Fibonacci numbers in C#?

easy

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

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

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

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

Solution

1. Step 1: Check array size and base cases

   For Fibonacci, dp array size should be n+1 to include dp[n]. Base cases are dp[0]=0 and dp[1]=1.

2. Step 2: Verify initialization correctness

   int[] dp = new int[n + 1]; dp[0] = 0; dp[1] = 1; correctly initializes dp with size n+1 and sets dp[0]=0, dp[1]=1.

3. Final Answer:

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

4. Quick Check:

   Correct array size and base values [OK]

Hint: Remember Fibonacci base: 0 and 1 at dp[0], dp[1] [OK]

Common Mistakes:

• Using wrong array size (n instead of n+1)
• Swapping base values dp[0] and dp[1]
• Not initializing base cases

3. What will be the output of the following C# code using bottom-up DP to find the number of ways to climb 3 steps if you can take 1 or 2 steps at a time?

int n = 3;
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];
}
Console.WriteLine(dp[n]);

medium

A. 2

B. 3

C. 4

D. 5

Solution

1. Step 1: Understand dp array meaning

   dp[i] stores number of ways to reach step i. Base: dp[0]=1 (one way to stand still), dp[1]=1 (one way to step 1).

2. Step 2: Calculate dp values up to n=3

   dp[2] = dp[1] + dp[0] = 1 + 1 = 2; dp[3] = dp[2] + dp[1] = 2 + 1 = 3.

3. Final Answer:

   3 -> Option B

4. Quick Check:

   Ways to climb 3 steps = 3 [OK]

Hint: Sum ways to previous two steps for each step [OK]

Common Mistakes:

• Confusing dp indices or base cases
• Off-by-one errors in loop
• Misunderstanding dp meaning

4. Identify the error in this bottom-up DP code snippet for calculating factorial of n:

int n = 5;
int[] dp = new int[n + 1];
dp[0] = 1;
for (int i = 1; i <= n; i++) {
    dp[i] = dp[i - 1] * i;
}
Console.WriteLine(dp[n]);

medium

A. No error, code works correctly

B. Array dp should be of size n, not n+1

C. dp[0] should be initialized to 0

D. Loop should start from i = 0

Solution

1. Step 1: Check array size and initialization

   dp size n+1 is correct to store factorials from 0 to n. dp[0] = 1 is correct as 0! = 1.

2. Step 2: Verify loop and calculation

   Loop from i=1 to n correctly calculates dp[i] = dp[i-1] * i, which is factorial definition.

3. Final Answer:

   No error, code works correctly -> Option A

4. Quick Check:

   Factorial DP code is correct [OK]

Hint: Check base case and loop range carefully [OK]

Common Mistakes:

• Wrong dp array size causing index errors
• Incorrect base case initialization
• Starting loop at wrong index

5. Given a list of coin denominations coins = {1, 2, 5} and a target amount amount = 5, which bottom-up DP approach correctly computes the minimum number of coins needed to make the amount?

int amount = 5;
int[] dp = new int[amount + 1];
for (int i = 1; i <= amount; i++) dp[i] = int.MaxValue;
int[] coins = {1, 2, 5};
for (int i = 1; i <= amount; i++) {
    foreach (int coin in coins) {
        if (coin <= i && dp[i - coin] != int.MaxValue) {
            dp[i] = Math.Min(dp[i], dp[i - coin] + 1);
        }
    }
}
Console.WriteLine(dp[amount]);

hard

A. 1

B. 2

C. 3

D. 5

Solution

1. Step 1: Understand dp meaning and initialization

   dp[i] stores minimum coins to make amount i. Initialized dp[0]=0 (default), others to MaxValue.

2. Step 2: Calculate dp values for amount=5

   Calculate dp[1]=1 (coin 1), dp[2]=1 (coin 2), dp[3]=2 (1+2), dp[4]=2 (2+2), dp[5]=1 (coin 5). Minimum coins for 5 is 1.

3. Step 3: Re-check code logic

   Code correctly updates dp[i] with minimum coins. Output dp[5] = 1.

4. Final Answer:

   1 -> Option A

5. Quick Check:

   Minimum coins for 5 = 1 (coin 5) [OK]

Hint: Check dp updates with all coins for each amount [OK]

Common Mistakes:

• Not initializing dp array properly
• Ignoring dp[i - coin] == int.MaxValue check
• Confusing minimum with maximum coin count