Tabulation Bottom Up DP in DSA C# - Time & Space Complexity

Time Complexity: Tabulation Bottom Up DP

O(n²)

Understanding Time Complexity

We want to understand how fast a bottom-up dynamic programming solution runs.

Specifically, how the steps grow as the input size increases.

Scenario Under Consideration

Analyze the time complexity of the following code snippet.

int[] dp = new int[n + 1];
dp[0] = 0;
for (int i = 1; i <= n; i++) {
    dp[i] = int.MaxValue;
    for (int j = 1; j <= i; j++) {
        dp[i] = Math.Min(dp[i], dp[i - j] + cost[j - 1]);
    }
}
return dp[n];
    

This code fills a table from the bottom up to find the minimum cost for a problem of size n.

Identify Repeating Operations

Identify the loops, recursion, array traversals that repeat.

• Primary operation: Nested loops where the outer loop runs from 1 to n, and the inner loop runs from 1 to i.
• How many times: The inner loop runs i times for each i, so total operations add up as 1 + 2 + 3 + ... + n.

How Execution Grows With Input

As n grows, the total steps grow roughly like the sum of numbers from 1 to n.

Input Size (n)	Approx. Operations
10	55
100	5,050
1000	500,500

Pattern observation: The operations grow roughly with the square of n, so doubling n roughly quadruples the work.

Final Time Complexity

Time Complexity: O(n²)

This means the time needed grows roughly with the square of the input size.

Common Mistake

[X] Wrong: "Since we use a table, the solution must be very fast, like O(n)."

[OK] Correct: The nested loops cause the time to grow faster than linear, so it is actually quadratic, not linear.

Interview Connect

Understanding this time complexity helps you explain how dynamic programming improves over naive recursion.

It shows you can solve problems efficiently by building solutions step-by-step.

Self-Check

"What if we could reduce the inner loop to run only a fixed number of times? How would the time complexity change?"

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