Tabulation Bottom Up DP in DSA Python - 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 cell

↓

Repeat until table is full

↓

Answer at DP table bottom-right

Start by setting base cases in the table, then fill the table row by row using previous results until the final answer is found at the bottom-right cell.

Execution Sample

DSA Python

weights = [1, 3, 4]
values = [1, 4, 5]
W = 7
DP = [[0]*(W+1) for _ in range(len(weights)+1)]
for i in range(1, len(weights)+1):
    for w in range(W+1):
        if weights[i-1] <= w:
            DP[i][w] = max(DP[i-1][w], DP[i-1][w-weights[i-1]] + values[i-1])
        else:
            DP[i][w] = DP[i-1][w]

This code fills a DP table to solve the 0/1 Knapsack problem using bottom-up tabulation.

Execution Table

Step	Operation	Item Index (i)	Capacity (w)	DP[i][w] Computed	DP Table State
1	Initialize DP row 0 with zeros	0	0 to 7	0	[[0,0,0,0,0,0,0,0], ...]
2	Process item 1 (weight=1, value=1)	1	0	0	[[0,0,0,0,0,0,0,0], [0,0,0,0,0,0,0,0], ...]
3	Process item 1	1	1	max(DP[0][1], DP[0][0]+1)=1	[[0,0,0,0,0,0,0,0], [0,1,0,0,0,0,0,0], ...]
4	Process item 1	1	2	max(DP[0][2], DP[0][1]+1)=1	[[0,0,0,0,0,0,0,0], [0,1,1,0,0,0,0,0], ...]
5	Process item 1	1	3	1	[[0,0,0,0,0,0,0,0], [0,1,1,1,0,0,0,0], ...]
6	Process item 1	1	4	1	[[0,0,0,0,0,0,0,0], [0,1,1,1,1,0,0,0], ...]
7	Process item 1	1	5	1	[[0,0,0,0,0,0,0,0], [0,1,1,1,1,1,0,0], ...]
8	Process item 1	1	6	1	[[0,0,0,0,0,0,0,0], [0,1,1,1,1,1,1,0], ...]
9	Process item 1	1	7	1	[[0,0,0,0,0,0,0,0], [0,1,1,1,1,1,1,1], ...]
10	Process item 2 (weight=3, value=4)	2	0	0	[[0,0,0,0,0,0,0,0], [0,1,1,1,1,1,1,1], [0,0,0,0,0,0,0,0]]
11	Process item 2	2	1	0	[[0,0,0,0,0,0,0,0], [0,1,1,1,1,1,1,1], [0,0,0,0,0,0,0,0]]
12	Process item 2	2	2	1	[[0,0,0,0,0,0,0,0], [0,1,1,1,1,1,1,1], [0,0,1,0,0,0,0,0]]
13	Process item 2	2	3	max(DP[1][3], DP[1][0]+4)=max(1,4)=4	[[0,0,0,0,0,0,0,0], [0,1,1,1,1,1,1,1], [0,0,1,4,0,0,0,0]]
14	Process item 2	2	4	max(DP[1][4], DP[1][1]+4)=max(1,5)=5	[[0,0,0,0,0,0,0,0], [0,1,1,1,1,1,1,1], [0,0,1,4,5,0,0,0]]
15	Process item 2	2	5	max(DP[1][5], DP[1][2]+4)=max(1,5)=5	[[0,0,0,0,0,0,0,0], [0,1,1,1,1,1,1,1], [0,0,1,4,5,5,0,0]]
16	Process item 2	2	6	max(DP[1][6], DP[1][3]+4)=max(1,5)=5	[[0,0,0,0,0,0,0,0], [0,1,1,1,1,1,1,1], [0,0,1,4,5,5,5,0]]
17	Process item 2	2	7	max(DP[1][7], DP[1][4]+4)=max(1,5)=5	[[0,0,0,0,0,0,0,0], [0,1,1,1,1,1,1,1], [0,0,1,4,5,5,5,5]]
18	Process item 3 (weight=4, value=5)	3	0	0	[[0,0,0,0,0,0,0,0], [0,1,1,1,1,1,1,1], [0,0,1,4,5,5,5,5], [0,0,0,0,0,0,0,0]]
19	Process item 3	3	1	0	[[0,0,0,0,0,0,0,0], [0,1,1,1,1,1,1,1], [0,0,1,4,5,5,5,5], [0,0,0,0,0,0,0,0]]
20	Process item 3	3	2	1	[[0,0,0,0,0,0,0,0], [0,1,1,1,1,1,1,1], [0,0,1,4,5,5,5,5], [0,0,1,0,0,0,0,0]]
21	Process item 3	3	3	4	[[0,0,0,0,0,0,0,0], [0,1,1,1,1,1,1,1], [0,0,1,4,5,5,5,5], [0,0,1,4,0,0,0,0]]
22	Process item 3	3	4	max(DP[2][4], DP[2][0]+5)=max(5,5)=5	[[0,0,0,0,0,0,0,0], [0,1,1,1,1,1,1,1], [0,0,1,4,5,5,5,5], [0,0,1,4,5,0,0,0]]
23	Process item 3	3	5	max(DP[2][5], DP[2][1]+5)=max(5,5)=5	[[0,0,0,0,0,0,0,0], [0,1,1,1,1,1,1,1], [0,0,1,4,5,5,5,5], [0,0,1,4,5,5,0,0]]
24	Process item 3	3	6	max(DP[2][6], DP[2][2]+5)=max(5,6)=6	[[0,0,0,0,0,0,0,0], [0,1,1,1,1,1,1,1], [0,0,1,4,5,5,5,5], [0,0,1,4,5,5,6,0]]
25	Process item 3	3	7	max(DP[2][7], DP[2][3]+5)=max(5,9)=9	[[0,0,0,0,0,0,0,0], [0,1,1,1,1,1,1,1], [0,0,1,4,5,5,5,5], [0,0,1,4,5,5,6,9]]

💡 DP table fully filled; answer is DP[3][7] = 9

Variable Tracker

Variable	Start	After Step 3	After Step 13	After Step 25	Final
DP Table Row 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 Table Row 1	[0,0,0,0,0,0,0,0]	[0,1,0,0,0,0,0,0]	[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 Table Row 2	[0,0,0,0,0,0,0,0]	[0,0,0,0,0,0,0,0]	[0,0,1,4,5,5,5,5]	[0,0,1,4,5,5,5,5]	[0,0,1,4,5,5,5,5]
DP Table Row 3	[0,0,0,0,0,0,0,0]	[0,0,0,0,0,0,0,0]	[0,0,0,0,0,0,0,0]	[0,0,1,4,5,5,6,9]	[0,0,1,4,5,5,6,9]

Key Moments - 3 Insights

Why do we start filling the DP table from row 1 and not row 0?

Row 0 represents the base case of having zero items, so all values are zero. We start from row 1 to consider the first item, as shown in execution_table rows 2-9.

Why do we compare DP[i-1][w] and DP[i-1][w-weights[i-1]] + values[i-1] when weights[i-1] <= w?

This compares not taking the current item (DP[i-1][w]) versus taking it (adding its value plus the best for remaining capacity). This is the core decision step seen in execution_table rows like 13 and 25.

Why is the answer found at DP[len(weights)][W] after filling the table?

Because DP[i][w] stores the best value using first i items and capacity w, so the bottom-right cell DP[3][7] holds the max value for all items and full capacity, as shown in the last execution_table row.

Visual Quiz - 3 Questions

Look at the execution_table at step 13, what is the value of DP[2][3]?

A 0

B 4

C 1

D 5

💡 Hint

Check the DP[i][w] Computed column at step 13 in execution_table.

At which step does the DP table first show a value of 9?

A Step 22

B Step 18

C Step 25

D Step 15

💡 Hint

Look at the DP Table State column for the highest value in execution_table rows.

If the weight of item 3 was 5 instead of 4, how would DP[3][7] change?

A It would decrease

B It would remain 9

C It would increase

D It would be zero

💡 Hint

Consider how the capacity limits affect inclusion of item 3 in the DP table.

Concept Snapshot

Tabulation Bottom Up DP:
- Create a 2D DP table with rows = items+1, cols = capacity+1
- Initialize first row and column with zeros (base cases)
- Fill table row-wise: DP[i][w] = max(not take, take item i)
- Use previous row results to build current
- Final answer at DP[last item][full capacity]

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 a table with zeros representing no items or zero capacity. Then, for each item and capacity, we decide whether to include the item or not by comparing values from previous computations. This process fills the table step-by-step until the answer is found at the bottom-right cell, representing the best value achievable with all items and full capacity. Key points include starting from base cases, using previous results, and understanding the decision to take or skip an item. The final DP table shows the maximum values for all subproblems, making it easy to find the optimal solution.

Practice

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

easy

A. Start solving from the largest problem and break it down

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, so we solve all smaller parts first.

2. Step 2: Compare with other methods

   Unlike recursion without storage, 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 ones [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 table for the Fibonacci sequence using tabulation in Python?

easy

A. dp = [0, 1] + [0] * (n - 2)

B. dp = [1, 1] * n

C. dp = [0] * n; dp[0] = 1; dp[1] = 1

D. dp = [0] * (n + 1); dp[0] = 1

Solution

1. Step 1: Understand Fibonacci base cases

   Fibonacci starts with dp[0] = 0 and dp[1] = 1, so initialization must reflect that.

2. Step 2: Check each option

   dp = [0, 1] + [0] * (n - 2) correctly sets dp[0]=0, dp[1]=1, and fills the rest with zeros for further calculation.

3. Final Answer:

   dp = [0, 1] + [0] * (n - 2) -> Option A

4. Quick Check:

   Correct base cases = dp = [0, 1] + [0] * (n - 2) [OK]

Hint: Start dp with base Fibonacci values 0 and 1 [OK]

Common Mistakes:

• Setting wrong base values
• Using multiplication incorrectly for initialization
• Not allocating enough space for dp

3. What will be the output of the following code snippet?

def fib(n):
    dp = [0, 1] + [0] * (n - 2)
    for i in range(2, n):
        dp[i] = dp[i-1] + dp[i-2]
    return dp[n-1]

print(fib(6))

medium

A. 5

B. 8

C. 13

D. 3

Solution

1. Step 1: Trace the DP table filling

   dp starts as [0,1,0,0,0,0]. Iterations fill dp as: dp[2]=1, dp[3]=2, dp[4]=3, dp[5]=5.

2. Step 2: Return the last element dp[5]

   dp[5] = 5, so fib(6) returns 5.

3. Final Answer:

   5 -> Option A

4. Quick Check:

   Fibonacci(6) = 5 [OK]

Hint: Remember Fibonacci sequence: 0,1,1,2,3,5... [OK]

Common Mistakes:

• Off-by-one errors in indexing
• Confusing dp[n-1] with dp[n]
• Incorrect loop range

4. Find the error in the following tabulation code for Fibonacci and select the fix:

def fib(n):
    dp = [0] * n
    dp[0] = 0
    dp[1] = 1
    for i in range(2, n):
        dp[i] = dp[i-1] + dp[i-2]
    return dp[n]

print(fib(5))

medium

A. Start loop from i=1 instead of i=2

B. Initialize dp with size n+1

C. Change return dp[n] to return dp[n-1]

D. Set dp[0] = 1 and dp[1] = 1

Solution

1. Step 1: Identify index error in return statement

   dp has indices 0 to n-1, so dp[n] is out of range causing error.

2. Step 2: Fix by returning dp[n-1]

   Returning dp[n-1] gives the last Fibonacci number correctly.

3. Final Answer:

   Change return dp[n] to return dp[n-1] -> Option C

4. Quick Check:

   Index error fixed by dp[n-1] return [OK]

Hint: Return dp[n-1] for nth Fibonacci (0-based index) [OK]

Common Mistakes:

• Returning dp[n] causing index error
• Wrong loop start index
• Incorrect base values

5. Given a list of coin denominations coins = [1, 2, 5] and an amount amount = 5, which tabulation code correctly computes the number of ways to make the amount using these coins?

hard

A. dp = [0]*(amount+1)
dp[0] = 1
for x in range(1, amount+1):
for coin in coins:
if coin <= x:
dp[x] += dp[x-coin]
print(dp[amount])

B. dp = [1] + [0]*amount
for coin in coins:
for x in range(coin, amount+1):
dp[x] += dp[x-coin]
print(dp[amount])

C. dp = [0]*(amount+1)
dp[0] = 1
for coin in coins:
for x in range(amount, coin-1, -1):
dp[x] += dp[x-coin]
print(dp[amount])

D. dp = [0]*(amount+1)
dp[0] = 1
for coin in coins:
for x in range(coin, amount+1):
dp[x] = dp[x-coin]
print(dp[amount])

Solution

1. Step 1: Understand coin change tabulation

   We initialize dp[0] = 1 (one way to make zero). For each coin, update dp[x] by adding dp[x-coin].

2. Step 2: Check each option's correctness

   dp = [1] + [0]*amount for coin in coins: for x in range(coin, amount+1): dp[x] += dp[x-coin] print(dp[amount]) correctly uses nested loops with dp[x] += dp[x-coin]. dp = [0]*(amount+1) dp[0] = 1 for x in range(1, amount+1): for coin in coins: if coin <= x: dp[x] += dp[x-coin] print(dp[amount]) loops coins inside amount, which counts duplicates. dp = [0]*(amount+1) dp[0] = 1 for coin in coins: for x in range(amount+1): if x >= coin: dp[x] += dp[x-coin] print(dp[amount]) loops x from 0 causing extra checks. dp = [0]*(amount+1) dp[0] = 1 for coin in coins: for x in range(coin, amount+1): dp[x] = dp[x-coin] print(dp[amount]) overwrites dp[x] instead of adding.

3. Final Answer:

   dp = [1] + [0]*amount for coin in coins: for x in range(coin, amount+1): dp[x] += dp[x-coin] print(dp[amount]) -> Option B

4. Quick Check:

   Correct coin change tabulation = dp = [1] + [0]*amount for coin in coins: for x in range(coin, amount+1): dp[x] += dp[x-coin] print(dp[amount]) [OK]

Hint: Add ways, don't overwrite dp[x] when counting combinations [OK]

Common Mistakes:

• Overwriting dp[x] instead of adding
• Wrong loop order causing duplicates
• Not initializing dp[0] = 1