Tabulation Bottom Up DP in DSA Python - Time & Space Complexity

Time Complexity: Tabulation Bottom Up DP

O(n)

Understanding Time Complexity

We want to understand how the time needed grows when using tabulation in bottom-up dynamic programming.

Specifically, how the number of steps changes as the input size increases.

Scenario Under Consideration

Analyze the time complexity of the following code snippet.

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

This code calculates the nth Fibonacci number using bottom-up tabulation.

Identify Repeating Operations

• Primary operation: The for-loop that fills the dp array from 2 to n.
• How many times: It runs (n - 1) times, which grows as n grows.

How Execution Grows With Input

Each step in the loop does a simple addition and assignment, repeated for each number up to n.

Input Size (n)	Approx. Operations
10	About 9 additions and assignments
100	About 99 additions and assignments
1000	About 999 additions and assignments

Pattern observation: The number of operations grows roughly in a straight line as n increases.

Final Time Complexity

Time Complexity: O(n)

This means the time needed grows directly in proportion to the input size.

Common Mistake

[X] Wrong: "Tabulation is always slower than recursion because it uses loops."

[OK] Correct: Tabulation avoids repeated work by storing results, so it actually runs faster than naive recursion which repeats calculations many times.

Interview Connect

Understanding how tabulation scales helps you explain efficient solutions clearly and confidently in interviews.

Self-Check

"What if we changed the dp array to store results only for the last two numbers instead of all n? How would the time complexity change?"

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