Fibonacci Using DP in DSA Java - 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

This flow shows how we build a table (dp array) from the bottom up to find the nth Fibonacci number efficiently.

Execution Sample

DSA Java

int fibDP(int n) {
  if (n == 0) return 0;
  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 using a bottom-up dynamic programming approach.

Execution Table

Step	Operation	Index i	dp Array State	Calculation	Result at dp[i]
1	Initialize dp array	-	[0, 0, 0, 0, 0, 0]	-	-
2	Set base cases	-	[0, 1, 0, 0, 0, 0]	dp[0]=0, dp[1]=1	-
3	Calculate dp[2]	2	[0, 1, 1, 0, 0, 0]	dp[2] = dp[1] + dp[0] = 1 + 0	1
4	Calculate dp[3]	3	[0, 1, 1, 2, 0, 0]	dp[3] = dp[2] + dp[1] = 1 + 1	2
5	Calculate dp[4]	4	[0, 1, 1, 2, 3, 0]	dp[4] = dp[3] + dp[2] = 2 + 1	3
6	Calculate dp[5]	5	[0, 1, 1, 2, 3, 5]	dp[5] = dp[4] + dp[3] = 3 + 2	5
7	Return result	5	[0, 1, 1, 2, 3, 5]	-	5

💡 Loop ends after i=5, dp[5] holds the 5th Fibonacci number.

Variable Tracker

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

Key Moments - 3 Insights

Why do we start filling dp array from index 2?

Because dp[0] and dp[1] are base cases already set (see Step 2 in execution_table). We need these to calculate dp[2] onwards.

Why does dp[i] depend on dp[i-1] and dp[i-2]?

Because Fibonacci numbers are defined as sum of two previous numbers. This is shown in the calculation column in steps 3 to 6.

What happens if n is 0 or 1?

The function returns dp[0] or dp[1] directly without entering the loop, as base cases are already set (Step 2).

Visual Quiz - 3 Questions

Look at the execution_table, what is the dp array state after calculating dp[3]?

A [0, 1, 2, 3, 0, 0]

B [0, 1, 1, 2, 0, 0]

C [0, 1, 1, 1, 0, 0]

D [0, 1, 0, 2, 0, 0]

💡 Hint

Check Step 4 row under 'dp Array State' in execution_table.

At which step does the dp array first contain the value 3?

A Step 5

B Step 3

C Step 4

D Step 6

💡 Hint

Look at the 'Result at dp[i]' column and 'dp Array State' in execution_table.

If we change the base case dp[1] to 2, what will be the value of dp[5]?

A 5

B 6

C 10

D 7

💡 Hint

Changing dp[1] affects all subsequent sums. Trace dp[2] to dp[5] with dp[1]=2 in variable_tracker.

Concept Snapshot

Fibonacci Using DP:
- Use an array dp[] to store Fibonacci numbers up to n.
- Initialize dp[0]=0, dp[1]=1 (base cases).
- For i=2 to n, dp[i] = dp[i-1] + dp[i-2].
- Return dp[n] as the nth Fibonacci number.
- This avoids repeated calculations by storing results.

Full Transcript

This visualization shows how to calculate the nth Fibonacci number using dynamic programming in Java. We create an array dp to store Fibonacci values from 0 to n. We set the first two values dp[0]=0 and dp[1]=1 as base cases. Then, starting from index 2, we fill dp[i] by adding the two previous values dp[i-1] and dp[i-2]. This process continues until we reach dp[n], which is returned as the result. The execution table tracks each step, showing how the dp array changes and how each dp[i] is calculated. The variable tracker shows the values of i and dp array after each step. Key moments clarify why we start from index 2, why dp[i] depends on previous two values, and what happens for small n. The quiz tests understanding of dp array states and effects of changing base cases. This method efficiently computes Fibonacci numbers by avoiding repeated work.

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 uses less memory by not storing any intermediate results.

B. It avoids repeated calculations by storing results of subproblems.

C. It calculates Fibonacci numbers using a closed-form formula.

D. It uses randomization to speed up the calculation.

Solution

1. Step 1: Understand recursive Fibonacci drawbacks

   Simple recursion recalculates the same Fibonacci numbers many times, causing inefficiency.

2. Step 2: How DP improves calculation

   DP stores results of smaller Fibonacci numbers to reuse them, avoiding repeated work.

3. Final Answer:

   It avoids repeated calculations by storing results of subproblems. -> Option B

4. Quick Check:

   DP stores subproblem results = avoids repeats [OK]

Hint: DP saves previous results to avoid repeated work [OK]

Common Mistakes:

• Thinking DP uses less memory than recursion
• Confusing DP with closed-form formulas
• Believing DP uses randomization

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

easy

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

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

Solution

1. Step 1: Determine array size for Fibonacci DP

   To store Fibonacci numbers up to n, array size must be n+1 to include index n.

2. Step 2: Check base cases initialization

   Fibonacci starts with dp[0] = 0 and dp[1] = 1, so these must be set correctly.

3. Final Answer:

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

4. Quick Check:

   Array size n+1 with dp[0]=0, dp[1]=1 = correct initialization [OK]

Hint: Array size must be n+1 to hold Fibonacci up to n [OK]

Common Mistakes:

• Using array size n instead of n+1
• Initializing dp[0] as 1 instead of 0
• Using array size n-1 causing index errors

3. What is the output of the following Java 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];
}
System.out.println(dp[n]);

medium

A. 3

B. 8

C. 5

D. 13

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 value at dp[5] is 5, so the output is 5.

3. Final Answer:

   5 -> Option C

4. Quick Check:

   Fibonacci(5) = 5 [OK]

Hint: Fibonacci(5) is 5 by definition [OK]

Common Mistakes:

• Confusing index with Fibonacci value
• Calculating dp[5] as 8 (Fibonacci(6))
• Off-by-one errors in loop

4. Identify the error in the following Java code for Fibonacci using DP and select the correct fix:

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];
}
System.out.println(dp[n]);

medium

A. Change array size to n+1 to avoid index out of bounds error.

B. Initialize dp[2] = 1 before the loop.

C. Change loop condition to i < n instead of i <= n.

D. Remove dp[0] initialization.

Solution

1. Step 1: Analyze array size and loop bounds

   Array dp has size n (6), but loop runs i from 2 to 6 inclusive, accessing dp[6] which is out of bounds.

2. Step 2: Fix array size to n+1

   Increasing array size to n+1 (7) allows dp[6] access without error.

3. Final Answer:

   Change array size to n+1 to avoid index out of bounds error. -> Option A

4. Quick Check:

   Array size n+1 fixes index error [OK]

Hint: Array size must cover max index used in loop [OK]

Common Mistakes:

• Ignoring index out of bounds exception
• Changing loop condition without fixing array size
• Unnecessary initialization of dp[2]

5. You want to calculate the 10th Fibonacci number using DP but with space optimization to use only two variables instead of an array. Which code snippet correctly implements this?

hard

A. int a = 0, b = 1, c = 0; for (int i = 2; i <= 10; i++) { c = a + b; a = b; b = c; } System.out.println(c);

B. int a = 0, b = 1; for (int i = 2; i <= 10; i++) { a = a + b; b = a; } System.out.println(b);

C. int a = 0, b = 1; for (int i = 2; i < 10; i++) { int c = a + b; a = b; b = c; } System.out.println(b);

D. int a = 1, b = 1; for (int i = 2; i <= 10; i++) { int c = a + b; a = b; b = c; } System.out.println(b);

Solution

1. Step 1: Understand space optimization for Fibonacci

   We only need last two Fibonacci numbers to calculate the next, so two variables plus one temp variable suffice.

2. Step 2: Check loop and initialization correctness

   Start with a=0, b=1 (F0 and F1), loop from 2 to 10 inclusive, update c = a+b, then shift a=b, b=c.

3. Step 3: Verify output

   After loop, c holds Fibonacci(10) = 55, which is printed.

4. Final Answer:

   int a = 0, b = 1, c = 0; for (int i = 2; i <= 10; i++) { c = a + b; a = b; b = c; } System.out.println(c); -> Option A

5. Quick Check:

   Two vars + temp, correct loop = Fibonacci(10) [OK]

Hint: Use two vars and update in loop to save space [OK]

Common Mistakes:

• Incorrect loop bounds causing off-by-one
• Wrong initial values for a and b
• Not updating variables in correct order