Tabulation Bottom Up DP in DSA Python

Mental Model

Start solving the smallest parts first and build up to the full problem by saving answers in a table.

Analogy: Like filling a staircase from the bottom step by step, so you can stand on each step to reach the top easily.

Table: [0] [1] [2] [3] [4] [5]
Values:  _   _   _   _   _   _
↑ start filling from left to right

Dry Run Walkthrough

Input: Find the 5th Fibonacci number using bottom-up tabulation

Goal: Calculate Fibonacci(5) by filling a table from the smallest Fibonacci numbers up

Step 1: Initialize table with base cases: fib[0] = 0, fib[1] = 1

Table: [0]=0 [1]=1 [2]=_ [3]=_ [4]=_ [5]=_

Why: Base cases are known and needed to build bigger answers

Step 2: Calculate fib[2] = fib[1] + fib[0] = 1 + 0 = 1

Table: [0]=0 [1]=1 [2]=1 [3]=_ [4]=_ [5]=_

Why: Use previous results to find next Fibonacci number

Step 3: Calculate fib[3] = fib[2] + fib[1] = 1 + 1 = 2

Table: [0]=0 [1]=1 [2]=1 [3]=2 [4]=_ [5]=_

Why: Build on already computed values

Step 4: Calculate fib[4] = fib[3] + fib[2] = 2 + 1 = 3

Table: [0]=0 [1]=1 [2]=1 [3]=2 [4]=3 [5]=_

Why: Continue building up the table

Step 5: Calculate fib[5] = fib[4] + fib[3] = 3 + 2 = 5

Table: [0]=0 [1]=1 [2]=1 [3]=2 [4]=3 [5]=5

Why: Final answer is ready at the top of the table

Result:

Table: [0]=0 [1]=1 [2]=1 [3]=2 [4]=3 [5]=5
Answer: Fibonacci(5) = 5

Annotated Code

DSA Python

class FibonacciTabulation:
    def fib(self, n: int) -> int:
        if n <= 1:
            return n
        table = [0] * (n + 1)  # create table to store results
        table[0], table[1] = 0, 1  # base cases
        for i in range(2, n + 1):
            table[i] = table[i - 1] + table[i - 2]  # build table bottom-up
        return table[n]

# Driver code
fib_calc = FibonacciTabulation()
result = fib_calc.fib(5)
print(f"Fibonacci(5) = {result}")

table = [0] * (n + 1) # create table to store results

Initialize table to hold Fibonacci numbers up to n

table[0], table[1] = 0, 1 # base cases

Set known Fibonacci values for 0 and 1

for i in range(2, n + 1):

Iterate from smallest unsolved subproblem to n

table[i] = table[i - 1] + table[i - 2] # build table bottom-up

Calculate current Fibonacci using two previous results

return table[n]

Return final answer from table

Output

Fibonacci(5) = 5

Complexity Analysis

Time: O(n) because we fill the table once from 2 to n

Space: O(n) because we store all Fibonacci numbers up to n in the table

vs Alternative: Compared to naive recursion which is exponential, tabulation is much faster and avoids repeated work

Edge Cases

n = 0

Returns 0 immediately without building table

DSA Python

if n <= 1:
            return n

n = 1

Returns 1 immediately without building table

DSA Python

if n <= 1:
            return n

When to Use This Pattern

When you see a problem that can be broken into smaller overlapping parts and you want to avoid repeated work, use bottom-up tabulation to build answers from the smallest parts up.

Common Mistakes

Mistake: Not initializing base cases in the table before the loop
Fix: Always set the known base values before starting the loop to build the table

Mistake: Starting loop from 0 or 1 instead of 2
Fix: Start from 2 because base cases 0 and 1 are already set

Summary

Calculates answers by filling a table from the smallest subproblems up to the full problem.

Use when a problem can be split into smaller overlapping parts and you want to avoid repeated calculations.

The key is to solve and store small parts first, then use them to build bigger answers.

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