Tabulation Bottom Up DP in DSA Python - Build from Scratch

Tabulation Bottom Up DP: Climbing Stairs Problem

📖 Scenario: Imagine you want to climb a staircase with a certain number of steps. You can take either 1 step or 2 steps at a time. You want to find out how many different ways you can reach the top.

🎯 Goal: Build a program using tabulation (bottom-up dynamic programming) to find the number of ways to climb a staircase with n steps.

📋 What You'll Learn

Create a variable n with the number of steps (exactly 5)

Create a list dp of size n+1 initialized with zeros

Set the base case values in dp for step 0 and step 1

Use a for loop with variable i from 2 to n to fill dp

Print the value of dp[n] which is the total ways to climb

💡 Why This Matters

🌍 Real World

This problem models situations where you want to count the number of ways to reach a goal by taking small steps, like climbing stairs or moving through stages.

💼 Career

Dynamic programming is a key technique in software engineering and data science for solving optimization and counting problems efficiently.

1

Set the number of steps

Create a variable called n and set it to 5 to represent the total steps in the staircase.

DSA Python

# Create a variable called n and set it to 5
# Your code here

Hint

Use n = 5 to set the number of steps.

2

Create the DP table and set base cases

Create a list called dp of size n+1 filled with zeros. Then set dp[0] to 1 and dp[1] to 1 as base cases.

DSA Python

n = 5
# Create a list dp of size n+1 filled with zeros
# Set dp[0] = 1 and dp[1] = 1
# Your code here

Hint

Use dp = [0] * (n + 1) to create the list. Then assign dp[0] = 1 and dp[1] = 1.

3

Fill the DP table using a for loop

Use a for loop with variable i from 2 to n (inclusive). Inside the loop, set dp[i] to dp[i-1] + dp[i-2].

DSA Python

n = 5
dp = [0] * (n + 1)
dp[0] = 1
dp[1] = 1
# Use a for loop from 2 to n to fill dp
# Your code here

Hint

Use for i in range(2, n + 1): and update dp[i] inside the loop.

4

Print the total number of ways to climb

Print the value of dp[n] which holds the total number of ways to climb the staircase.

DSA Python

n = 5
dp = [0] * (n + 1)
dp[0] = 1
dp[1] = 1
for i in range(2, n + 1):
    dp[i] = dp[i - 1] + dp[i - 2]
# Print the total ways to climb
# Your code here

Hint

Use print(dp[n]) to show the result.

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