Tabulation Bottom Up DP in DSA C#

Mental Model

Build the answer step by step from the smallest parts up, storing results in a table to avoid repeating work.

Analogy: Like filling a staircase from the bottom step to the top, each step depends on the steps below it already being filled.

Table: [0] [1] [2] [3] [4] ...
Values:  0   1   1   2   3  ...

Dry Run Walkthrough

Input: Calculate Fibonacci number for n=5 using bottom-up tabulation

Goal: Find the 5th Fibonacci number by filling a table from the base cases up

Step 1: Initialize table with base values fib[0]=0, fib[1]=1

fib: [0]=0, [1]=1, [2]=_, [3]=_, [4]=_, [5]=_

Why: Base cases are known and needed to build next values

Step 2: Calculate fib[2] = fib[1] + fib[0] = 1 + 0 = 1

fib: [0]=0, [1]=1, [2]=1, [3]=_, [4]=_, [5]=_

Why: Use previous two results to find next Fibonacci number

Step 3: Calculate fib[3] = fib[2] + fib[1] = 1 + 1 = 2

fib: [0]=0, [1]=1, [2]=1, [3]=2, [4]=_, [5]=_

Why: Continue building up using stored results

Step 4: Calculate fib[4] = fib[3] + fib[2] = 2 + 1 = 3

fib: [0]=0, [1]=1, [2]=1, [3]=2, [4]=3, [5]=_

Why: Each step depends on the two before it

Step 5: Calculate fib[5] = fib[4] + fib[3] = 3 + 2 = 5

fib: [0]=0, [1]=1, [2]=1, [3]=2, [4]=3, [5]=5

Why: Final value is built from all previous computations

Result:

fib: [0]=0 -> [1]=1 -> [2]=1 -> [3]=2 -> [4]=3 -> [5]=5
Answer: 5

Annotated Code

DSA C#

using System;

class BottomUpFibonacci
{
    public static int Fib(int n)
    {
        if (n <= 1) return n; // base cases

        int[] fib = new int[n + 1];
        fib[0] = 0; // fib(0)
        fib[1] = 1; // fib(1)

        for (int i = 2; i <= n; i++)
        {
            fib[i] = fib[i - 1] + fib[i - 2]; // build from smaller subproblems
        }

        return fib[n]; // final answer
    }

    static void Main()
    {
        int n = 5;
        int result = Fib(n);
        Console.WriteLine(result);
    }
}

if (n <= 1) return n; // base cases

Handle smallest inputs directly without table

fib[0] = 0; // fib(0) fib[1] = 1; // fib(1)

Set known base values to start building

for (int i = 2; i <= n; i++) { fib[i] = fib[i - 1] + fib[i - 2]; // build from smaller subproblems }

Fill table from bottom up using previous two results

return fib[n]; // final answer

Return the computed result after table is complete

Output

5

Complexity Analysis

Time: O(n) because we fill the table once from 2 to n

Space: O(n) because we store all Fibonacci values up to n

vs Alternative: Compared to naive recursion with exponential time, tabulation is much faster by avoiding repeated calculations

Edge Cases

n = 0 or n = 1

Returns n immediately without building table

DSA C#

if (n <= 1) return n; // base cases

Large n (e.g., n=50)

Computes efficiently in linear time using table

DSA C#

for (int i = 2; i <= n; i++)

When to Use This Pattern

When you see a problem asking for a value built from smaller subproblems and overlapping calculations, use bottom-up tabulation to build answers stepwise and store them.

Common Mistakes

Mistake: Trying to compute Fibonacci recursively without memoization, causing repeated work and slow performance.
Fix: Use a table to store results and fill it from the bottom up to avoid repeated calculations.

Mistake: Not initializing base cases in the table before the loop.
Fix: Always set fib[0] and fib[1] before starting the loop to build further values.

Summary

Calculates answers by filling a table from smallest subproblems up to the target.

Use when a problem can be broken into smaller overlapping subproblems and you want efficient computation.

The key is to build solutions step by step and store them to avoid repeating work.

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