Fibonacci Using DP in DSA Python - Time & Space Complexity

Time Complexity: Fibonacci Using DP

O(n)

Understanding Time Complexity

We want to know how fast the Fibonacci numbers can be calculated using dynamic programming.

How does the time needed grow as the input number gets bigger?

Scenario Under Consideration

Analyze the time complexity of the following code snippet.

def fibonacci_dp(n):
    fib = [0, 1]
    for i in range(2, n + 1):
        fib.append(fib[i - 1] + fib[i - 2])
    return fib[n]

This code calculates the nth Fibonacci number by building up from the bottom using a list to store results.

Identify Repeating Operations

Identify the loops, recursion, array traversals that repeat.

• Primary operation: The for-loop that adds two previous Fibonacci numbers to get the next one.
• How many times: It runs once for each number from 2 up to n, so about n-1 times.

How Execution Grows With Input

Each new Fibonacci number is found by adding two earlier numbers once, so the work grows steadily with n.

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

Pattern observation: The number of operations grows roughly in a straight line as n increases.

Final Time Complexity

Time Complexity: O(n)

This means the time to find the nth Fibonacci number grows directly with n, so doubling n roughly doubles the work.

Common Mistake

[X] Wrong: "Calculating Fibonacci with recursion is always fast because it just calls itself."

[OK] Correct: Simple recursion repeats many calculations, making it much slower than this dynamic programming approach.

Interview Connect

Understanding how dynamic programming improves speed by remembering past results is a key skill that shows you can write efficient code.

Self-Check

"What if we used recursion with memoization instead of a loop? How would the time complexity change?"

Practice

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

easy

A. It avoids repeated calculations by storing intermediate results.

B. It uses less memory by not storing any results.

C. It calculates Fibonacci numbers using only addition without recursion.

D. It uses randomization to speed up calculations.

Solution

1. Step 1: Understand recursive Fibonacci inefficiency

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

2. Step 2: How DP improves calculation

   DP stores previously calculated Fibonacci numbers in a table (memoization or tabulation), avoiding repeated work.

3. Final Answer:

   It avoids repeated calculations by storing intermediate results. -> Option A

4. Quick Check:

   DP stores results to avoid repeats [OK]

Hint: DP saves results to avoid repeating work [OK]

Common Mistakes:

• Thinking DP uses less memory than recursion
• Believing DP removes addition operations
• Confusing DP with randomization

2. Which of the following Python code snippets correctly initializes a list to store Fibonacci numbers for DP with n = 5?

easy

A. fib = [0] * 5

B. fib = [0] * 6

C. fib = [1] * 5

D. fib = []

Solution

1. Step 1: Determine list size for Fibonacci DP

   For n=5, we need indices 0 to 5 inclusive, so list size must be 6.

2. Step 2: Check initialization correctness

   fib = [0] * 6 creates a list of six zeros, ready to store fib[0] to fib[5].

3. Final Answer:

   fib = [0] * 6 -> Option B

4. Quick Check:

   List size = n+1 for fib numbers [OK]

Hint: List size must be n+1 for fib up to n [OK]

Common Mistakes:

• Using list size equal to n instead of n+1
• Initializing with ones instead of zeros
• Leaving list empty without size

3. What will be the output of the following Python code?

def fib_dp(n):
    fib = [0] * (n+1)
    fib[0], fib[1] = 0, 1
    for i in range(2, n+1):
        fib[i] = fib[i-1] + fib[i-2]
    return fib[n]

print(fib_dp(6))

medium

A. 5

B. 13

C. 8

D. 21

Solution

1. Step 1: Trace Fibonacci values up to n=6

   fib[0]=0, fib[1]=1; then fib[2]=1, fib[3]=2, fib[4]=3, fib[5]=5, fib[6]=8.

2. Step 2: Return fib[6]

   The function returns fib[6], which is 8.

3. Final Answer:

   8 -> Option C

4. Quick Check:

   fib(6) = 8 [OK]

Hint: Calculate fib step-by-step from base cases [OK]

Common Mistakes:

• Confusing fib(6) with fib(7)
• Starting fib from 1 instead of 0
• Off-by-one errors in loop range

4. Identify the error in this Fibonacci DP code and select the correct fix:

def fib_dp(n):
    fib = [0] * n
    fib[0], fib[1] = 0, 1
    for i in range(2, n+1):
        fib[i] = fib[i-1] + fib[i-2]
    return fib[n]

medium

A. Change for loop to range(1, n)

B. Change fib[0], fib[1] = 0, 1 to fib[1], fib[2] = 0, 1

C. Return fib[n-1] instead of fib[n]

D. Change fib = [0] * n to fib = [0] * (n+1)

Solution

1. Step 1: Check list size for indexing

   fib list size is n, but code accesses fib[n], which is out of range.

2. Step 2: Fix list size

   Changing fib = [0] * n to fib = [0] * (n+1) allows indexing up to fib[n].

3. Final Answer:

   Change fib = [0] * n to fib = [0] * (n+1) -> Option D

4. Quick Check:

   List size must cover fib[n] index [OK]

Hint: List size must be n+1 for fib[n] access [OK]

Common Mistakes:

• Not increasing list size causes IndexError
• Changing base cases incorrectly
• Altering loop range wrongly

5. You want to find the sum of the first n Fibonacci numbers efficiently using DP. Which modification to the standard Fibonacci DP code below will correctly compute this sum?

def fib_sum(n):
    fib = [0] * (n+1)
    fib[0], fib[1] = 0, 1
    for i in range(2, n+1):
        fib[i] = fib[i-1] + fib[i-2]
    # What to add here to get sum of fib[0] to fib[n]?
    return ???

hard

A. return fib[n+2] - 1

B. return fib[n] * 2

C. return fib[n] + fib[n-1]

D. return sum(fib)

Solution

1. Step 1: Understand sum of Fibonacci numbers

   The sum of first n Fibonacci numbers is fib[0] + fib[1] + ... + fib[n].

2. Step 2: Use known Fibonacci sum formula

   The sum of first n Fibonacci numbers equals fib[n+2] - 1.

3. Final Answer:

   return fib[n+2] - 1 -> Option A

4. Quick Check:

   Sum of fib numbers = fib[n+2] - 1 [OK]

Hint: Sum of first n fib numbers = fib[n+2] - 1 [OK]

Common Mistakes:

• Using fib[n]*2 which is incorrect
• Using sum(fib) which is inefficient
• Adding only last two fib numbers