Tabulation Bottom Up DP in DSA C++ - 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 input size increases.

Scenario Under Consideration

Analyze the time complexity of the following code snippet.

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

Identify the loops, recursion, array traversals that repeat.

• Primary operation: The for-loop that fills the dp array from 2 to n.
• How many times: Exactly n-1 times, once for each i from 2 to n.

How Execution Grows With Input

Each new input number adds one more step to fill the dp array.

Input Size (n)	Approx. Operations
10	About 9 steps
100	About 99 steps
1000	About 999 steps

Pattern observation: The number of steps grows roughly in a straight line with input size.

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 takes more time than recursion because it uses loops."

[OK] Correct: Although tabulation uses loops, it avoids repeated calculations, so it actually runs faster than naive recursion.

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 for only the last two Fibonacci numbers instead of all? How would the time complexity change?"

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