Tabulation Bottom Up DP in DSA C# - Build from Scratch

Tabulation Bottom Up DP: Fibonacci Numbers

📖 Scenario: Imagine you want to find the nth number in the Fibonacci sequence quickly. The Fibonacci sequence starts with 0 and 1, and each next number is the sum of the two before it.Using a method called tabulation (bottom-up dynamic programming), you will build the solution step by step from the smallest cases up to the number you want.

🎯 Goal: You will create a program that calculates the nth Fibonacci number using tabulation. This means you will fill a table (array) from the bottom up to find the answer efficiently.

📋 What You'll Learn

Create an integer variable n with the value 7

Create an integer array fibTable of size n + 1

Initialize the first two Fibonacci numbers in fibTable

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

Print the nth Fibonacci number stored in fibTable[n]

💡 Why This Matters

🌍 Real World

Calculating Fibonacci numbers efficiently is useful in computer graphics, financial models, and algorithm design where repeated calculations happen.

💼 Career

Understanding dynamic programming and tabulation is important for software engineers to solve optimization problems efficiently.

1

Create the input variable n

Create an integer variable called n and set it to 7.

DSA C#

# Your code here

Hint

Use int n = 7; to create the variable.

2

Create the Fibonacci table array

Create an integer array called fibTable with size n + 1.

DSA C#

int n = 7;
# Your code here

Hint

Use new int[n + 1] to create the array.

3

Initialize base cases and fill the table

Set fibTable[0] to 0 and fibTable[1] to 1. Then use a for loop with variable i from 2 to n to fill fibTable[i] as the sum of fibTable[i - 1] and fibTable[i - 2].

DSA C#

int n = 7;
int[] fibTable = new int[n + 1];
# Your code here

Hint

Remember to set the first two Fibonacci numbers before the loop.

Inside the loop, add the two previous numbers to get the current one.

4

Print the nth Fibonacci number

Print the value stored in fibTable[n].

DSA C#

int n = 7;
int[] fibTable = new int[n + 1];
fibTable[0] = 0;
fibTable[1] = 1;
for (int i = 2; i <= n; i++)
{
    fibTable[i] = fibTable[i - 1] + fibTable[i - 2];
}
# Your code here

Hint

Use System.Console.WriteLine(fibTable[n]); to print the result.

Practice

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

easy

A. Start solving from the main problem and break it down recursively

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, building up to the final answer.

2. Step 2: Compare with other methods

   Unlike recursion without memoization, 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 [OK]

Common Mistakes:

• Confusing tabulation with recursion
• Ignoring the order of filling the table
• Not storing intermediate results

2. Which of the following is the correct way to initialize a tabulation array for Fibonacci numbers in C#?

easy

A. int[] dp = new int[n + 1]; dp[0] = 1; dp[1] = 0;

B. int[] dp = new int[n]; dp[0] = 1; dp[1] = 1;

C. int[] dp = new int[n + 1]; dp[0] = 0; dp[1] = 1;

D. int[] dp = new int[n]; dp[0] = 0; dp[1] = 1;

Solution

1. Step 1: Check array size and base cases

   For Fibonacci, dp array size should be n+1 to include dp[n]. Base cases are dp[0]=0 and dp[1]=1.

2. Step 2: Verify initialization correctness

   int[] dp = new int[n + 1]; dp[0] = 0; dp[1] = 1; correctly initializes dp with size n+1 and sets dp[0]=0, dp[1]=1.

3. Final Answer:

   int[] dp = new int[n + 1]; dp[0] = 0; dp[1] = 1; -> Option C

4. Quick Check:

   Correct array size and base values [OK]

Hint: Remember Fibonacci base: 0 and 1 at dp[0], dp[1] [OK]

Common Mistakes:

• Using wrong array size (n instead of n+1)
• Swapping base values dp[0] and dp[1]
• Not initializing base cases

3. What will be the output of the following C# code using bottom-up DP to find the number of ways to climb 3 steps if you can take 1 or 2 steps at a time?

int n = 3;
int[] dp = new int[n + 1];
dp[0] = 1;
dp[1] = 1;
for (int i = 2; i <= n; i++) {
    dp[i] = dp[i - 1] + dp[i - 2];
}
Console.WriteLine(dp[n]);

medium

A. 2

B. 3

C. 4

D. 5

Solution

1. Step 1: Understand dp array meaning

   dp[i] stores number of ways to reach step i. Base: dp[0]=1 (one way to stand still), dp[1]=1 (one way to step 1).

2. Step 2: Calculate dp values up to n=3

   dp[2] = dp[1] + dp[0] = 1 + 1 = 2; dp[3] = dp[2] + dp[1] = 2 + 1 = 3.

3. Final Answer:

   3 -> Option B

4. Quick Check:

   Ways to climb 3 steps = 3 [OK]

Hint: Sum ways to previous two steps for each step [OK]

Common Mistakes:

• Confusing dp indices or base cases
• Off-by-one errors in loop
• Misunderstanding dp meaning

4. Identify the error in this bottom-up DP code snippet for calculating factorial of n:

int n = 5;
int[] dp = new int[n + 1];
dp[0] = 1;
for (int i = 1; i <= n; i++) {
    dp[i] = dp[i - 1] * i;
}
Console.WriteLine(dp[n]);

medium

A. No error, code works correctly

B. Array dp should be of size n, not n+1

C. dp[0] should be initialized to 0

D. Loop should start from i = 0

Solution

1. Step 1: Check array size and initialization

   dp size n+1 is correct to store factorials from 0 to n. dp[0] = 1 is correct as 0! = 1.

2. Step 2: Verify loop and calculation

   Loop from i=1 to n correctly calculates dp[i] = dp[i-1] * i, which is factorial definition.

3. Final Answer:

   No error, code works correctly -> Option A

4. Quick Check:

   Factorial DP code is correct [OK]

Hint: Check base case and loop range carefully [OK]

Common Mistakes:

• Wrong dp array size causing index errors
• Incorrect base case initialization
• Starting loop at wrong index

5. Given a list of coin denominations coins = {1, 2, 5} and a target amount amount = 5, which bottom-up DP approach correctly computes the minimum number of coins needed to make the amount?

int amount = 5;
int[] dp = new int[amount + 1];
for (int i = 1; i <= amount; i++) dp[i] = int.MaxValue;
int[] coins = {1, 2, 5};
for (int i = 1; i <= amount; i++) {
    foreach (int coin in coins) {
        if (coin <= i && dp[i - coin] != int.MaxValue) {
            dp[i] = Math.Min(dp[i], dp[i - coin] + 1);
        }
    }
}
Console.WriteLine(dp[amount]);

hard

A. 1

B. 2

C. 3

D. 5

Solution

1. Step 1: Understand dp meaning and initialization

   dp[i] stores minimum coins to make amount i. Initialized dp[0]=0 (default), others to MaxValue.

2. Step 2: Calculate dp values for amount=5

   Calculate dp[1]=1 (coin 1), dp[2]=1 (coin 2), dp[3]=2 (1+2), dp[4]=2 (2+2), dp[5]=1 (coin 5). Minimum coins for 5 is 1.

3. Step 3: Re-check code logic

   Code correctly updates dp[i] with minimum coins. Output dp[5] = 1.

4. Final Answer:

   1 -> Option A

5. Quick Check:

   Minimum coins for 5 = 1 (coin 5) [OK]

Hint: Check dp updates with all coins for each amount [OK]

Common Mistakes:

• Not initializing dp array properly
• Ignoring dp[i - coin] == int.MaxValue check
• Confusing minimum with maximum coin count