Fibonacci Using DP in DSA C - Execution Trace

Concept Flow - Fibonacci Using DP

Start with n

↓

Create DP array of size n+1

↓

Set base cases: dp[0

↓

For i from 2 to n

↓

Calculate dp[i

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Return dp[n

We build an array to store Fibonacci numbers from 0 to n, starting with base cases, then fill the array step-by-step using previous two values.

Execution Sample

DSA C

int fib(int n) {
  int dp[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]
0	Initialize dp array	-	[0, _, _, _, _, _]	-	-
1	Set base case	0	[0, _, _, _, _, _]	dp[0] = 0	0
2	Set base case	1	[0, 1, _, _, _, _]	dp[1] = 1	1
3	Calculate dp[i]	2	[0, 1, _, _, _, _]	dp[2] = dp[1] + dp[0] = 1 + 0	1
4	Calculate dp[i]	3	[0, 1, 1, _, _, _]	dp[3] = dp[2] + dp[1] = 1 + 1	2
5	Calculate dp[i]	4	[0, 1, 1, 2, _, _]	dp[4] = dp[3] + dp[2] = 2 + 1	3
6	Calculate dp[i]	5	[0, 1, 1, 2, 3, _]	dp[5] = dp[4] + dp[3] = 3 + 2	5
7	Return result	5	[0, 1, 1, 2, 3, 5]	-	5

💡 Reached i = n = 5, dp array fully computed, returned dp[5]

Variable Tracker

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

Key Moments - 3 Insights

Why do we start filling dp array from index 2, not 0 or 1?

Because dp[0] and dp[1] are base cases set before the loop (see steps 1 and 2 in execution_table). Starting from 2 avoids overwriting these known values.

Why do we add dp[i-1] and dp[i-2] to get dp[i]?

Fibonacci numbers are defined as sum of two previous numbers. The execution_table rows 3 to 6 show dp[i] calculated as dp[i-1] + dp[i-2].

What happens if n is 0 or 1?

The function returns dp[0] or dp[1] directly as base cases without entering the loop, as shown in steps 1 and 2.

Visual Quiz - 3 Questions

Look at the execution_table at Step 4, what is the value of dp[3]?

A 1

B 3

C 2

D 5

💡 Hint

Check the 'Result at dp[i]' column for Step 4 in execution_table.

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

A Step 5

B Step 4

C Step 6

D Step 7

💡 Hint

Look at the 'dp Array State' column in execution_table to see when 3 appears.

If we change the base case dp[1] to 2, what will be the value of dp[2] at Step 3?

A 1

B 2

C 3

D 4

💡 Hint

dp[2] = dp[1] + dp[0], so changing dp[1] affects dp[2] calculation at Step 3.

Concept Snapshot

Fibonacci Using DP:
- Use an array dp to store Fibonacci numbers up to n
- Set dp[0]=0 and dp[1]=1 as base cases
- For i from 2 to n, dp[i] = dp[i-1] + dp[i-2]
- Return dp[n] as the result
- Efficient: avoids repeated calculations by storing results

Full Transcript

This visualization shows how to calculate Fibonacci numbers using dynamic programming. We create an array dp to store Fibonacci values from 0 to n. We start by setting dp[0] to 0 and dp[1] to 1 as base cases. Then, for each index 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 the Fibonacci sequence step-by-step without repeating calculations. The execution table tracks each step, showing how the dp array fills up. The variable tracker shows how the index i and dp array values change over time. Key moments clarify why we start from index 2, why we add previous two values, and what happens for small n. The quiz tests understanding of dp values at specific steps and effects of changing base cases. This method is efficient and easy to understand for beginners.