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. This sequence starts with 0 and 1, and each next number is the sum of the two before it. Calculating this directly can be slow for big numbers.We will use a method called Tabulation Bottom Up Dynamic Programming to find the Fibonacci number quickly by building up from the smallest cases.

🎯 Goal: Build a program that uses a vector to store Fibonacci numbers from the bottom up, then prints the nth Fibonacci number.

📋 What You'll Learn

Create a vector<int> called fib with size n+1

Initialize the first two Fibonacci numbers in fib[0] and fib[1]

Use a for loop with variable i from 2 to n to fill fib[i] as the sum of fib[i-1] and fib[i-2]

Print the value of fib[n]

💡 Why This Matters

🌍 Real World

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

💼 Career

Understanding bottom up dynamic programming is essential for software engineers to optimize recursive problems and improve performance.

1

Create the Fibonacci vector

Create an integer variable n and set it to 10. Then create a vector<int> called fib with size n + 1.

DSA C++

#include 
#include 

int main() {
    // Create integer n and set to 10
    // Create vector fib with size n + 1

    return 0;
}

Hint

Use int n = 10; and std::vector fib(n + 1); inside main().

2

Initialize the first two Fibonacci numbers

Set fib[0] to 0 and fib[1] to 1.

DSA C++

#include 
#include 

int main() {
    int n = 10;
    std::vector fib(n + 1);
    // Initialize fib[0] to 0
    // Initialize fib[1] to 1

    return 0;
}

Hint

Assign fib[0] = 0; and fib[1] = 1; after creating the vector.

3

Fill the Fibonacci vector using a for loop

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

DSA C++

#include 
#include 

int main() {
    int n = 10;
    std::vector fib(n + 1);
    fib[0] = 0;
    fib[1] = 1;
    // Use for loop from i = 2 to n
    // Set fib[i] = fib[i - 1] + fib[i - 2]

    return 0;
}

Hint

Write a for loop starting at 2 up to n. Inside, assign fib[i] as the sum of the two previous Fibonacci numbers.

4

Print the nth Fibonacci number

Print the value of fib[n] using std::cout.

DSA C++

#include 
#include 

int main() {
    int n = 10;
    std::vector fib(n + 1);
    fib[0] = 0;
    fib[1] = 1;
    for (int i = 2; i <= n; i++) {
        fib[i] = fib[i - 1] + fib[i - 2];
    }
    // Print fib[n]

    return 0;
}

Hint

Use std::cout << fib[n] << std::endl; to print the result.

Practice

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

easy

A. Build a table from the smallest subproblems up to the main problem

B. Solve the problem by recursion without storing results

C. Use random guesses to find the solution

D. Start solving from the largest problem and break it down

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 memory, bottom-up 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 A

4. Quick Check:

   Bottom-up DP = Build table from small to big [OK]

Hint: Think: solve small parts first, then bigger ones [OK]

Common Mistakes:

• Confusing bottom-up with recursion without memoization
• Thinking it starts from the largest problem
• Ignoring the use of a table to store results

2. Which of the following is the correct way to declare a DP table for Fibonacci numbers using tabulation in C++?

easy

A. int dp[]; dp[1] = 1;

B. int dp[100] = {0}; dp[1] = 1;

C. int dp[100]; dp[0] = 1;

D. int dp[100] = {1}; dp[2] = 1;

Solution

1. Step 1: Check array initialization

   int dp[100] = {0}; dp[1] = 1; initializes dp with zeros and sets dp[1] = 1, which is correct for Fibonacci base cases.

2. Step 2: Identify errors in other options

   int dp[]; dp[1] = 1; has no size for dp array, invalid in C++. int dp[100]; dp[0] = 1; sets dp[0] = 1 which is incorrect base case. int dp[100] = {1}; dp[2] = 1; initializes dp with 1 at dp[0], which is wrong.

3. Final Answer:

   int dp[100] = {0}; dp[1] = 1; -> Option B

4. Quick Check:

   Correct base initialization = int dp[100] = {0}; dp[1] = 1; [OK]

Hint: Initialize dp array with zeros and set base cases explicitly [OK]

Common Mistakes:

• Not specifying array size
• Wrong base case initialization
• Assuming default values without initialization

3. What will be the output of the following C++ code using bottom-up DP to find the 5th Fibonacci number?

int fib(int n) {
  int dp[6] = {0};
  dp[1] = 1;
  for (int i = 2; i <= n; i++) {
    dp[i] = dp[i-1] + dp[i-2];
  }
  return dp[n];
}

int main() {
  std::cout << fib(5);
  return 0;
}

medium

A. 5

B. 8

C. 3

D. 13

Solution

1. Step 1: Trace dp array filling

   dp[0]=0, dp[1]=1; then dp[2]=1, dp[3]=2, dp[4]=3, dp[5]=5 by adding previous two.

2. Step 2: Return dp[5]

   dp[5] is 5, so output is 5.

3. Final Answer:

   5 -> Option A

4. Quick Check:

   Fibonacci(5) = 5 [OK]

Hint: Remember Fibonacci sequence: 0,1,1,2,3,5... [OK]

Common Mistakes:

• Off-by-one errors in indexing
• Confusing dp[0] and dp[1] values
• Returning wrong dp index

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

int factorial(int n) {
  int dp[100];
  dp[0] = 1;
  for (int i = 1; i <= n; i++) {
    dp[i] = dp[i-1] * i;
  }
  return dp[n];
}

medium

A. dp array not initialized to zero

B. Loop should start from i = 0

C. dp array size might be too small for large n

D. No error, code is correct

Solution

1. Step 1: Check dp array size

   dp array size is fixed at 100, so if n > 99, dp[i] access is out of bounds causing error.

2. Step 2: Analyze other parts

   Initialization and loop logic are correct; loop starts at 1 and dp[0] = 1 is base case.

3. Final Answer:

   dp array size might be too small for large n -> Option C

4. Quick Check:

   Array size must fit n [OK]

Hint: Ensure dp array size >= n to avoid overflow [OK]

Common Mistakes:

• Assuming dp size is always enough
• Starting loop at wrong index
• Ignoring base case initialization

5. Given the problem to find the minimum number of coins to make amount 11 using coins {1, 3, 5}, which bottom-up DP table entry dp[11] will hold after running this code?

int coins[] = {1, 3, 5};
int n = 11;
int dp[12];
dp[0] = 0;
for (int i = 1; i <= n; i++) {
  dp[i] = 1000; // large number
  for (int c : coins) {
    if (i - c >= 0) {
      dp[i] = std::min(dp[i], dp[i-c] + 1);
    }
  }
}
std::cout << dp[n];

hard

A. 6

B. 5

C. 4

D. 3

Solution

1. Step 1: Understand DP logic for coin change

   dp[i] stores minimum coins for amount i. For 11, check coins 1,3,5 and pick minimum.

2. Step 2: Calculate dp values up to 11

   dp[0]=0; dp[1]=1; dp[2]=2; dp[3]=1; dp[4]=2; dp[5]=1; dp[6]=2; dp[7]=3; dp[8]=2; dp[9]=3; dp[10]=2; dp[11]=3 (e.g., 5+5+1 or 3+3+5)

3. Final Answer:

   3 -> Option D

4. Quick Check:

   Minimum coins for 11 = 3 [OK]

Hint: Build dp from 0 to n, pick min coins at each step [OK]

Common Mistakes:

• Not initializing dp[0] = 0
• Using max int instead of large number
• Forgetting to check i - coin >= 0