Memoization to Optimize Recursion in DSA Typescript

Mental Model

Memoization remembers answers to small problems so we don't solve them again.

Analogy: Like writing down answers in a notebook while solving puzzles, so next time you see the same puzzle, you just look it up instead of solving it again.

fib(5)
  ↓
fib(4) + fib(3)
  ↓       ↓
fib(3)+fib(2) fib(2)+fib(1)
  ↓
[cache stores answers here]
fib(2) -> 1
fib(3) -> 2
fib(4) -> 3
fib(5) -> 5

Dry Run Walkthrough

Input: Calculate fib(5) using recursion with memoization

Goal: Find fib(5) efficiently by storing intermediate results to avoid repeated work

Step 1: Call fib(5), no cache yet

cache = {}
fib(5) -> fib(4) + fib(3)

Why: Start breaking down fib(5) into smaller problems

Step 2: Call fib(4), no cache yet

cache = {}
fib(4) -> fib(3) + fib(2)

Why: Need fib(4) to compute fib(5)

Step 3: Call fib(3), no cache yet

cache = {}
fib(3) -> fib(2) + fib(1)

Why: Need fib(3) to compute fib(4)

Step 4: Call fib(2), no cache yet

cache = {}
fib(2) -> fib(1) + fib(0)

Why: Need fib(2) to compute fib(3)

Step 5: fib(1) and fib(0) are base cases, return 1 and 0

cache = {}
fib(2) = 1 + 0 = 1

Why: Base cases stop recursion

Step 6: Store fib(2) = 1 in cache

cache = {2: 1}

Why: Remember fib(2) to avoid recomputing

Step 7: Call fib(1), base case returns 1

cache = {2: 1}

Why: fib(1) is base case

Step 8: Calculate fib(3) = fib(2) + fib(1) = 1 + 1 = 2 and store in cache

cache = {2: 1, 3: 2}

Why: Store fib(3) result for reuse

Step 9: Call fib(2) again, found in cache as 1

cache = {2: 1, 3: 2}

Why: Reuse cached fib(2) to save time

Step 10: Calculate fib(4) = fib(3) + fib(2) = 2 + 1 = 3 and store in cache

cache = {2: 1, 3: 2, 4: 3}

Why: Store fib(4) result for reuse

Step 11: Call fib(3) again, found in cache as 2

cache = {2: 1, 3: 2, 4: 3}

Why: Reuse cached fib(3) to save time

Step 12: Calculate fib(5) = fib(4) + fib(3) = 3 + 2 = 5 and store in cache

cache = {2: 1, 3: 2, 4: 3, 5: 5}

Why: Final answer stored for fib(5)

Result:

fib sequence cache: {2:1, 3:2, 4:3, 5:5}
Answer: 5

Annotated Code

DSA Typescript

class Fibonacci {
  private cache: Map<number, number> = new Map();

  fib(n: number): number {
    if (n <= 1) return n; // base case

    if (this.cache.has(n)) {
      return this.cache.get(n)!; // return cached result
    }

    const result = this.fib(n - 1) + this.fib(n - 2); // recursive calls
    this.cache.set(n, result); // store result in cache
    return result;
  }
}

const fibCalc = new Fibonacci();
console.log(fibCalc.fib(5));

if (n <= 1) return n; // base case

stop recursion when n is 0 or 1

if (this.cache.has(n)) { return this.cache.get(n)!; }

return cached answer if available to avoid recomputation

const result = this.fib(n - 1) + this.fib(n - 2);

compute fib(n) by recursive calls to smaller problems

this.cache.set(n, result);

store computed result in cache for future reuse

Output

5

Complexity Analysis

Time: O(n) because each fib number from 0 to n is computed once and then reused

Space: O(n) because the cache stores results for all numbers up to n and recursion stack can go up to n

vs Alternative: Without memoization, time is O(2^n) due to repeated calculations; memoization reduces it drastically to O(n)

Edge Cases

n = 0 (smallest input)

Returns 0 immediately as base case

DSA Typescript

if (n <= 1) return n; // base case

n = 1 (smallest positive input)

Returns 1 immediately as base case

DSA Typescript

if (n <= 1) return n; // base case

Repeated calls with same n

Returns cached result instantly without recursion

DSA Typescript

if (this.cache.has(n)) { return this.cache.get(n)!; }

When to Use This Pattern

When you see a problem with overlapping recursive calls and repeated subproblems, reach for memoization to save time by caching results.

Common Mistakes

Mistake: Not checking cache before recursive calls, causing repeated work
Fix: Always check if result is cached before computing recursively

Mistake: Not storing computed results in cache after recursion
Fix: Store the result in cache immediately after computing to reuse later

Mistake: Using global variables without encapsulation causing bugs
Fix: Use a class or closure to keep cache private and avoid side effects

Summary

Memoization stores answers to recursive calls to avoid repeating work.

Use it when recursion solves the same subproblems multiple times.

The key is to remember results so you never solve the same problem twice.

Practice

1. What is the main purpose of using memoization in recursive functions?

easy

A. To avoid using any variables inside the function

B. To make the function run slower by adding extra steps

C. To convert recursion into iteration automatically

D. To store results of expensive function calls and reuse them later

Solution

1. Step 1: Understand recursion inefficiency

   Recursive functions often repeat the same calculations multiple times, which wastes time.

2. Step 2: Role of memoization

   Memoization saves the results of these repeated calculations in a storage (like an object) so they can be reused without recalculating.

3. Final Answer:

   To store results of expensive function calls and reuse them later -> Option D

4. Quick Check:

   Memoization = Store and reuse results [OK]

Hint: Memoization saves repeated results to speed up recursion [OK]

Common Mistakes:

• Thinking memoization slows down the function
• Confusing memoization with iteration
• Believing memoization removes recursion

2. Which of the following is the correct way to declare a memo object in TypeScript for memoization?

easy

A. let memo = []

B. const memo: Record = {}

C. var memo = 0

D. const memo: string = ''

Solution

1. Step 1: Identify memo storage type

   Memoization requires a storage that maps input (number) to output (number), so an object with number keys and number values is suitable.

2. Step 2: Check TypeScript syntax

   Record correctly types an object with number keys and number values.

3. Final Answer:

   const memo: Record = {} -> Option B

4. Quick Check:

   Memo object type = Record [OK]

Hint: Use Record for memo object in TypeScript [OK]

Common Mistakes:

• Using array instead of object for memo
• Declaring memo as a number or string
• Not initializing memo as an empty object

3. What will be the output of the following TypeScript code?

function fib(n: number, memo: Record = {}): number {
  if (n <= 1) return n;
  if (memo[n]) return memo[n];
  memo[n] = fib(n - 1, memo) + fib(n - 2, memo);
  return memo[n];
}
console.log(fib(5));

medium

A. 5

B. 8

C. 3

D. 13

Solution

1. Step 1: Understand Fibonacci sequence

   Fibonacci numbers start as 0,1,1,2,3,5,8,... The 5th Fibonacci number (0-based) is 5.

2. Step 2: Trace fib(5) call

   fib(5) = fib(4) + fib(3) = 3 + 2 = 5, memoization avoids repeated calls.

3. Final Answer:

   5 -> Option A

4. Quick Check:

   fib(5) = 5 [OK]

Hint: Remember Fibonacci sequence values for small n [OK]

Common Mistakes:

• Confusing index starting at 0 or 1
• Ignoring memoization effect on output
• Calculating fib(5) as fib(6)

4. Identify the error in this memoized factorial function and choose the correct fix:

function factorial(n: number, memo: Record = {}): number {
  if (n === 0) return 1;
  if (memo[n]) return memo[n];
  memo[n] = n * factorial(n - 1, memo);
  return memo[n];
}
console.log(factorial(5));

medium

A. Remove memo parameter from function

B. Add base case for n === 1

C. Change if (memo[n]) to if (memo[n] !== undefined)

D. Initialize memo as an array instead of object

Solution

1. Step 1: Understand memo check issue

   Checking if (memo[n]) fails when memo[n] is 0 or falsy, but factorial values are always positive, so 0 is not possible here, but better to check for undefined explicitly.

2. Step 2: Fix condition to check existence properly

   Use if (memo[n] !== undefined) to correctly detect cached values.

3. Final Answer:

   Change if (memo[n]) to if (memo[n] !== undefined) -> Option C

4. Quick Check:

   Check memo existence with !== undefined [OK]

Hint: Check memo keys with !== undefined to avoid false negatives [OK]

Common Mistakes:

• Using if(memo[n]) which fails for falsy cached values
• Adding unnecessary base case for n=1
• Removing memo parameter breaks memoization

5. You want to optimize a recursive function that calculates the number of ways to climb n stairs taking 1 or 2 steps at a time. Which memoization approach below correctly implements this in TypeScript?

hard

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

B. function climb(n: number): number {
if (n <= 1) return 1;
return climb(n - 1) + climb(n - 2);
}

C. function climb(n: number, memo: number[] = []): number {
if (n <= 1) return 1;
if (memo[n]) return memo[n];
memo[n] = climb(n - 1, memo) * climb(n - 2, memo);
return memo[n];
}

D. function climb(n: number, memo: Record = {}): number {
if (n === 0) return 0;
if (memo[n]) return memo[n];
memo[n] = climb(n - 1, memo) + climb(n - 2, memo);
return memo[n];
}

Solution

1. Step 1: Understand problem base cases

   Number of ways to climb 0 or 1 stairs is 1 (one way: stand still or one step).

2. Step 2: Check memoization and recursion logic

   function climb(n: number, memo: Record = {}): number { if (n <= 1) return 1; if (memo[n] !== undefined) return memo[n]; memo[n] = climb(n - 1, memo) + climb(n - 2, memo); return memo[n]; } correctly uses memo object, checks with !== undefined, and sums ways from n-1 and n-2 stairs.

3. Step 3: Identify errors in other options

   function climb(n: number): number { if (n <= 1) return 1; return climb(n - 1) + climb(n - 2); } lacks memoization, function climb(n: number, memo: number[] = []): number { if (n <= 1) return 1; if (memo[n]) return memo[n]; memo[n] = climb(n - 1, memo) * climb(n - 2, memo); return memo[n]; } multiplies instead of sums, function climb(n: number, memo: Record = {}): number { if (n === 0) return 0; if (memo[n]) return memo[n]; memo[n] = climb(n - 1, memo) + climb(n - 2, memo); return memo[n]; } returns 0 for n=0 which is incorrect.

4. Final Answer:

   Option A code correctly implements memoized climb function -> Option A

5. Quick Check:

   Memo + sum of previous two = correct climb ways [OK]

Hint: Sum ways for n-1 and n-2 with memo check !== undefined [OK]

Common Mistakes:

• Forgetting memoization causing slow recursion
• Multiplying instead of adding ways
• Wrong base case for n=0