Fibonacci Using DP in DSA C# - Time & Space Complexity

Time Complexity: Fibonacci Using DP

O(n)

Understanding Time Complexity

We want to understand how fast the Fibonacci sequence calculation runs when using dynamic programming.

Specifically, how the number of steps grows as the input number gets bigger.

Scenario Under Consideration

Analyze the time complexity of the following code snippet.

public int Fibonacci(int n) {
    int[] dp = new int[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 by building up from smaller results and storing them to avoid repeated work.

Identify Repeating Operations

Identify the loops, recursion, array traversals that repeat.

• Primary operation: The for-loop that runs from 2 to n, calculating each Fibonacci number once.
• How many times: Exactly n - 1 times (from 2 up to n).

How Execution Grows With Input

As n grows, the number of steps grows roughly the same as n because each Fibonacci number is calculated once.

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

Pattern observation: The steps increase linearly as input size increases.

Final Time Complexity

Time Complexity: O(n)

This means the time to compute grows in a straight line with the input size.

Common Mistake

[X] Wrong: "Calculating Fibonacci with DP is still slow because it uses recursion like the naive method."

[OK] Correct: This code uses a loop and stores results, so it avoids repeated calculations and runs much faster than naive recursion.

Interview Connect

Understanding this approach shows you can improve slow recursive solutions by saving work, a key skill in many coding problems.

Self-Check

"What if we changed the array to only store the last two Fibonacci numbers instead of all? How would the time complexity change?"

Practice

1. What is the main advantage of using Dynamic Programming (DP) to calculate Fibonacci numbers compared to simple recursion?

easy

A. It calculates Fibonacci numbers randomly

B. It uses less memory by not storing any results

C. It stores intermediate results to avoid repeated calculations

D. It only works for small Fibonacci numbers

Solution

1. Step 1: Understand recursion inefficiency

   Simple recursion recalculates the same Fibonacci numbers many times, causing exponential time complexity.

2. Step 2: Role of Dynamic Programming

   DP stores previously calculated Fibonacci numbers in a table (memoization or tabulation), so each number is computed once.

3. Final Answer:

   It stores intermediate results to avoid repeated calculations -> Option C

4. Quick Check:

   DP avoids repeated work = C [OK]

Hint: DP saves results to prevent repeated work [OK]

Common Mistakes:

• Thinking DP uses less memory than recursion
• Believing DP calculates numbers randomly
• Assuming DP only works for small inputs

2. Which of the following C# code snippets correctly initializes the DP array for Fibonacci calculation of size n?

easy

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

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

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

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

Solution

1. Step 1: Array size for DP

   We need an array of size n+1 to store Fibonacci numbers from 0 to n.

2. Step 2: Initial values for Fibonacci

   Fibonacci sequence starts with dp[0] = 0 and dp[1] = 1.

3. Final Answer:

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

4. Quick Check:

   Correct size and base values = A [OK]

Hint: DP array size = n+1, start with 0 and 1 [OK]

Common Mistakes:

• Using array size n instead of n+1
• Swapping dp[0] and dp[1] values
• Starting dp[0] or dp[1] with wrong numbers

3. What is the output of the following C# code snippet?

int n = 5;
int[] dp = new int[n + 1];
dp[0] = 0; dp[1] = 1;
for (int i = 2; i <= n; i++) {
    dp[i] = dp[i - 1] + dp[i - 2];
}
Console.WriteLine(dp[n]);

medium

A. 13

B. 8

C. 3

D. 5

Solution

1. Step 1: Calculate Fibonacci values up to n=5

   dp[0]=0, dp[1]=1 Then dp[2]=1, dp[3]=2, dp[4]=3, dp[5]=5.

2. Step 2: Output dp[5]

   The code prints dp[5], which is 5.

3. Final Answer:

   5 -> Option D

4. Quick Check:

   Fibonacci(5) = 5 [OK]

Hint: Fibonacci(5) = 5 by adding previous two numbers [OK]

Common Mistakes:

• Confusing Fibonacci index starting at 0 or 1
• Calculating dp[5] as 8 or 13 incorrectly
• Off-by-one errors in loop

4. Identify the error in this C# code for Fibonacci using DP:

int n = 6;
int[] dp = new int[n];
dp[0] = 0; dp[1] = 1;
for (int i = 2; i <= n; i++) {
    dp[i] = dp[i - 1] + dp[i - 2];
}
Console.WriteLine(dp[n]);

medium

A. Array size is too small causing IndexOutOfRangeException

B. dp[0] and dp[1] are initialized incorrectly

C. Loop condition should be i < n instead of i <= n

D. No error, code runs fine

Solution

1. Step 1: Check array size and indexing

   Array dp has size n=6, valid indices 0 to 5. Loop runs i=2 to 6 inclusive, accessing dp[6] which is out of bounds.

2. Step 2: Resulting error

   Accessing dp[6] causes IndexOutOfRangeException at runtime.

3. Final Answer:

   Array size is too small causing IndexOutOfRangeException -> Option A

4. Quick Check:

   Array size < loop max index = error [OK]

Hint: Array size must be n+1 for index n [OK]

Common Mistakes:

• Using array size n instead of n+1
• Ignoring loop boundary causing runtime error
• Assuming code runs without error

5. You want to find the 10th Fibonacci number using bottom-up DP but only want to use O(1) space instead of an array. Which approach below correctly implements this optimization?

hard

A. int[] dp = new int[10]; dp[0] = 0; dp[1] = 1; for (int i = 2; i <= 10; i++) dp[i] = dp[i - 1] + dp[i - 2]; Console.WriteLine(dp[10]);

B. int a = 0, b = 1, c = 0; for (int i = 2; i <= 10; i++) { c = a + b; a = b; b = c; } Console.WriteLine(c);

C. int a = 0, b = 1; for (int i = 2; i < 10; i++) { a = a + b; b = a; } Console.WriteLine(b);

D. int fib = 0; for (int i = 0; i <= 10; i++) fib += i; Console.WriteLine(fib);

Solution

1. Step 1: Understand space optimization

   We only need last two Fibonacci numbers to compute the next, so store them in variables a and b.

2. Step 2: Analyze the correct code logic

   int a = 0, b = 1, c = 0; for (int i = 2; i <= 10; i++) { c = a + b; a = b; b = c; } Console.WriteLine(c); correctly updates a and b in each iteration and stores current Fibonacci in c, printing c after loop.

3. Step 3: Check other options

   int[] dp = new int[10]; dp[0] = 0; dp[1] = 1; for (int i = 2; i <= 10; i++) dp[i] = dp[i - 1] + dp[i - 2]; Console.WriteLine(dp[10]); uses array (not O(1) space), the code with loop i < 10 updates variables incorrectly, the code with fib += i sums numbers instead of Fibonacci.

4. Final Answer:

   int a = 0, b = 1, c = 0; for (int i = 2; i <= 10; i++) { c = a + b; a = b; b = c; } Console.WriteLine(c); -> Option B

5. Quick Check:

   Use two vars to save space = B [OK]

Hint: Keep only last two Fibonacci numbers to save space [OK]

Common Mistakes:

• Using full array instead of variables
• Incorrect variable updates causing wrong output
• Summing indices instead of Fibonacci numbers