Memoization to Optimize Recursion in DSA Typescript - Execution Trace

Concept Flow - Memoization to Optimize Recursion

Start Recursive Call

↓

Check if result in cache?

Yes→Return cached result

No↓

Compute result recursively

↓

Store result in cache

↓

Return result

↓

End Recursive Call

The flow shows how recursion checks a cache before computing, stores results to avoid repeated work, and returns cached results when available.

Execution Sample

DSA Typescript

function fib(n: number, memo: Record<number, number> = {}): number {
  if (n <= 1) return n;
  if (memo[n] !== undefined) return memo[n];
  memo[n] = fib(n - 1, memo) + fib(n - 2, memo);
  return memo[n];
}

Calculates Fibonacci numbers using recursion with memoization to save and reuse results.

Execution Table

Step	Operation	Current n	Cache Before	Cache After	Returned Value	Visual State
1	Call fib(4)	4	{}	{}	pending	fib(4) called, cache empty
2	Call fib(3)	3	{}	{}	pending	fib(3) called inside fib(4)
3	Call fib(2)	2	{}	{}	pending	fib(2) called inside fib(3)
4	Call fib(1)	1	{}	{}	1	fib(1) base case returns 1
5	Call fib(0)	0	{}	{}	0	fib(0) base case returns 0
6	Compute fib(2)	2	{}	{"2": 1}	1	fib(2) = 1 + 0 = 1 stored in cache
7	Call fib(1)	1	{"2":1}	{"2":1}	1	fib(1) base case returns 1
8	Compute fib(3)	3	{"2":1}	{"2":1,"3":2}	2	fib(3) = 1 + 1 = 2 stored in cache
9	Call fib(2)	2	{"2":1,"3":2}	{"2":1,"3":2}	1	fib(2) result found in cache
10	Compute fib(4)	4	{"2":1,"3":2}	{"2":1,"3":2,"4":3}	3	fib(4) = 2 + 1 = 3 stored in cache
11	Return fib(4)	4	{"2":1,"3":2,"4":3}	{"2":1,"3":2,"4":3}	3	Final result returned

💡 All recursive calls resolved, final fib(4) value computed and cached.

Variable Tracker

Variable	Start	After Step 6	After Step 8	After Step 10	Final
n	4	2	3	4	4
memo	{}	{"2":1}	{"2":1,"3":2}	{"2":1,"3":2,"4":3}	{"2":1,"3":2,"4":3}
Returned Value	pending	1	2	3	3

Key Moments - 3 Insights

Why does the function check the cache before computing?

Because as shown in steps 9 and 10, the function returns cached results immediately to avoid repeating calculations, speeding up execution.

What happens when the base case is reached?

At steps 4 and 5, the function returns the base values (0 or 1) without recursion, stopping further calls and starting to build results back up.

How does memoization reduce the number of recursive calls?

By storing results in the cache (memo), repeated calls like fib(2) at step 9 return instantly from cache instead of recalculating, reducing total calls.

Visual Quiz - 3 Questions

Look at the execution table, what is the cache content after step 8?

A {"2":1,"3":2}

B {"2":1}

C {"3":2}

D {}

💡 Hint

Check the 'Cache After' column at step 8 in the execution table.

At which step does fib(2) return a cached value instead of computing?

A Step 4

B Step 6

C Step 9

D Step 10

💡 Hint

Look for when fib(2) is called and cache already has its value.

If memoization was removed, what would happen to the number of calls?

A Calls would stay the same

B Calls would increase due to repeated calculations

C Calls would decrease

D Calls would stop immediately

💡 Hint

Refer to how memoization avoids repeated calls in the execution table.

Concept Snapshot

Memoization stores results of recursive calls in a cache.
Before computing, it checks if result exists to avoid repeated work.
This optimizes recursion by reducing calls and speeding up execution.
Use a dictionary or object to save and retrieve results.
Base cases return directly without caching.
Memoization is key for efficient recursive algorithms.

Full Transcript

This visualization shows how memoization optimizes recursion by storing results in a cache object. When the recursive function fib is called with a number n, it first checks if the result is already in the cache. If yes, it returns that cached value immediately, avoiding repeated calculations. If not, it computes the value recursively, stores it in the cache, and then returns it. Base cases for fib(0) and fib(1) return immediately without recursion. The execution table tracks each step, showing calls, cache state before and after, and returned values. The variable tracker shows how the cache (memo) grows and how returned values build up. Key moments clarify why caching is checked first, how base cases stop recursion, and how memoization reduces calls. The quiz tests understanding of cache contents, when cached values are used, and the effect of removing memoization. The snapshot summarizes memoization as a technique to speed up recursion by saving and reusing results.