Memoization Top Down DP in DSA C - Execution Trace

Concept Flow - Memoization Top Down DP

Start with problem input

↓

Check if result in memo?

Yes→Return memoized result

No↓

Compute result recursively

↓

Store result in memo

↓

Return computed result

The flow starts by checking if the answer is already saved. If yes, return it. If no, compute recursively, save it, then return.

Execution Sample

DSA C

int fib(int n, int memo[]) {
  if (n <= 1) return n;
  if (memo[n] != -1) return memo[n];
  memo[n] = fib(n-1, memo) + fib(n-2, memo);
  return memo[n];
}

This code computes Fibonacci numbers using memoization to avoid repeated calculations.

Execution Table

Step	Operation	n	Memo Before	Recursive Calls	Memo After	Returned Value
1	fib(4) called	4	[-1,-1,-1,-1,-1]	fib(3), fib(2)	[-1,-1,-1,-1,-1]	pending
2	fib(3) called	3	[-1,-1,-1,-1,-1]	fib(2), fib(1)	[-1,-1,-1,-1,-1]	pending
3	fib(2) called	2	[-1,-1,-1,-1,-1]	fib(1), fib(0)	[-1,-1,-1,-1,-1]	pending
4	fib(1) called	1	[-1,-1,-1,-1,-1]	none	[-1,-1,-1,-1,-1]	1
5	fib(0) called	0	[-1,-1,-1,-1,-1]	none	[-1,-1,-1,-1,-1]	0
6	fib(2) memoized	2	[-1,-1,-1,-1,-1]	none	[-1,-1,1,-1,-1]	1
7	fib(1) called again	1	[-1,-1,1,-1,-1]	none	[-1,-1,1,-1,-1]	1
8	fib(3) memoized	3	[-1,-1,1,-1,-1]	none	[-1,-1,1,2,-1]	2
9	fib(2) called again	2	[-1,-1,1,2,-1]	none	[-1,-1,1,2,-1]	1
10	fib(4) memoized	4	[-1,-1,1,2,-1]	none	[-1,-1,1,2,3]	3
11	Return final result	4	[-1,-1,1,2,3]	none	[-1,-1,1,2,3]	3

💡 fib(4) returns 3 after all recursive calls and memoization complete

Variable Tracker

Variable	Start	After Step 6	After Step 8	After Step 10	Final
n	4	2	3	4	4
memo	[-1,-1,-1,-1,-1]	[-1,-1,1,-1,-1]	[-1,-1,1,2,-1]	[-1,-1,1,2,3]	[-1,-1,1,2,3]
Returned Value	pending	1	2	3	3

Key Moments - 3 Insights

Why do we check memo[n] before computing fib(n)?

Because if memo[n] is not -1, it means fib(n) was already computed and saved, so we can return it immediately without repeating work. See steps 7 and 9 where memoized values are reused.

What happens when n is 0 or 1?

These are base cases returning n directly without recursion, as shown in steps 4 and 5. This stops infinite recursion.

Why do recursive calls reduce n by 1 and 2?

Because Fibonacci number fib(n) depends on fib(n-1) and fib(n-2). The recursion breaks the problem into smaller subproblems, as seen in steps 1-3.

Visual Quiz - 3 Questions

Look at the execution_table, what is the value of memo after step 8?

A [-1,-1,1,2,-1]

B [-1,-1,-1,-1,-1]

C [-1,-1,1,-1,-1]

D [-1,-1,1,2,3]

💡 Hint

Check the 'Memo After' column in row with Step 8

At which step does fib(2) get its value memoized for the first time?

A Step 3

B Step 9

C Step 6

D Step 10

💡 Hint

Look for when memo[2] changes from -1 to 1 in the 'Memo After' column

If we remove the memo check, what would happen to the number of recursive calls?

A It would decrease

B It would increase drastically

C It would stay the same

D It would stop immediately

💡 Hint

Memoization avoids repeated calls; without it, calls multiply exponentially

Concept Snapshot

Memoization Top Down DP:
- Use recursion with a memo array to save results
- Check memo before computing to avoid repeats
- Base cases return directly
- Store computed results in memo
- Return memoized or computed value

Full Transcript

Memoization Top Down DP solves problems by breaking them into smaller parts recursively. Before computing a result, it checks if the answer is already saved in a memo array. If yes, it returns that saved answer immediately. If no, it computes the result recursively, saves it in memo, then returns it. This avoids repeated work and speeds up the program. Base cases stop recursion. For example, Fibonacci numbers use this method by saving fib(n) once computed. The execution table shows each call, memo state before and after, and returned values step by step.