Why Tabulation Bottom Up DP in DSA Python?

The Big Idea

What if you could solve big problems by solving tiny pieces first and never repeating yourself?

The Scenario

Imagine you want to find the number of ways to climb stairs, but you try to count each path by writing them down one by one.

It quickly becomes confusing and takes forever as the stairs grow higher.

The Problem

Manually listing all possibilities is slow and easy to mess up.

You repeat the same counting many times, wasting effort and making mistakes.

The Solution

Tabulation Bottom Up DP builds the answer step by step from the smallest cases up.

It stores results in a table so you never repeat work, making it fast and reliable.

Before vs After

✗ Before

def climb_stairs(n):
    if n <= 2:
        return n
    return climb_stairs(n-1) + climb_stairs(n-2)

✓ After

def climb_stairs(n):
    if n <= 2:
        return n
    ways = [0] * (n+1)
    ways[1], ways[2] = 1, 2
    for i in range(3, n+1):
        ways[i] = ways[i-1] + ways[i-2]
    return ways[n]

What It Enables

You can solve complex problems quickly by breaking them into simple steps and remembering results.

Real Life Example

Calculating the fastest way to reach a destination by considering all smaller routes first and building up the best path.

Key Takeaways

Manual counting repeats work and is slow.

Tabulation builds answers from the bottom up using a table.

This method saves time and avoids mistakes by remembering past results.

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