Fibonacci Using DP in DSA Python - Execution Trace

Concept Flow - Fibonacci Using DP

Start with n

↓

Create DP array of size n+1

↓

Initialize base cases: dp[0

↓

For i from 2 to n

↓

Calculate dp[i

↓

Return dp[n

We build a list to store Fibonacci numbers from 0 to n, starting with base cases, then fill each next value by adding the two before it.

Execution Sample

DSA Python

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

This code calculates the nth Fibonacci number using a list to store intermediate results.

Execution Table

Step	Operation	DP Array State	Calculation	Result at dp[i]
0	Initialize dp	[0, 1, 0, 0, 0, 0, 0]	Set dp[0]=0, dp[1]=1	dp = [0, 1, 0, 0, 0, 0, 0]
1	i=2	[0, 1, 0, 0, 0, 0, 0]	dp[2] = dp[1] + dp[0] = 1 + 0	dp[2] = 1
2	i=3	[0, 1, 1, 0, 0, 0, 0]	dp[3] = dp[2] + dp[1] = 1 + 1	dp[3] = 2
3	i=4	[0, 1, 1, 2, 0, 0, 0]	dp[4] = dp[3] + dp[2] = 2 + 1	dp[4] = 3
4	i=5	[0, 1, 1, 2, 3, 0, 0]	dp[5] = dp[4] + dp[3] = 3 + 2	dp[5] = 5
5	i=6	[0, 1, 1, 2, 3, 5, 0]	dp[6] = dp[5] + dp[4] = 5 + 3	dp[6] = 8
6	Return dp[6]	[0, 1, 1, 2, 3, 5, 8]	Final result	8

💡 Loop ends when i > n (i=7), dp[n] is returned

Variable Tracker

Variable	Start	After 1	After 2	After 3	After 4	After 5	Final
i	-	2	3	4	5	6	-
dp	[0,1,0,0,0,0,0]	[0,1,1,0,0,0,0]	[0,1,1,2,0,0,0]	[0,1,1,2,3,0,0]	[0,1,1,2,3,5,0]	[0,1,1,2,3,5,8]	[0,1,1,2,3,5,8]

Key Moments - 3 Insights

Why do we start the loop from i=2 instead of i=0 or i=1?

Because dp[0] and dp[1] are base cases already set. The loop starts at 2 to build from these known values, as shown in execution_table rows 1 and 2.

Why do we use dp[i-1] + dp[i-2] to calculate dp[i]?

The Fibonacci number at position i is the sum of the two previous Fibonacci numbers. This is the core relation, visible in the 'Calculation' column of the execution_table.

What happens if n is 0 or 1?

The function returns dp[0] or dp[1] directly without entering the loop, since the loop starts at 2. This is why base cases are initialized first.

Visual Quiz - 3 Questions

Look at the execution_table at step 3 (i=4). What is the value of dp[4] after calculation?

A 3

B 5

C 2

D 8

💡 Hint

Check the 'Result at dp[i]' column in execution_table row with Step=3

At which step does the dp array first contain the value 5?

A Step 3

B Step 4

C Step 5

D Step 6

💡 Hint

Look at the 'DP Array State' column and find when 5 appears first

If we change the base case dp[1] from 1 to 2, what would dp[2] be at step 1?

A 1

B 2

C 3

D 4

💡 Hint

dp[2] = dp[1] + dp[0], so changing dp[1] affects dp[2] directly (see execution_table step 1)

Concept Snapshot

Fibonacci Using DP:
- Create dp array size n+1
- Set dp[0]=0, dp[1]=1 (base cases)
- For i in 2 to n: dp[i] = dp[i-1] + dp[i-2]
- Return dp[n]
- Efficient: avoids repeated calculations

Full Transcript

This visualization shows how to calculate Fibonacci numbers using dynamic programming. We start with a list dp that holds Fibonacci values from 0 to n. The first two values dp[0] and dp[1] are set to 0 and 1. Then, for each position i from 2 to n, we calculate dp[i] by adding the two previous values dp[i-1] and dp[i-2]. This way, we build up the Fibonacci sequence step by step without repeating calculations. The final answer is dp[n]. The execution table tracks each step, showing how dp changes and how each Fibonacci number is computed. Key moments clarify why the loop starts at 2, why the sum of two previous values is used, and what happens for small n. The quiz tests understanding of dp values at specific steps and effects of changing base cases.

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