Memoization to Optimize Recursion in DSA Typescript - Cheat Sheet & Quick Revision

Recall & Review

beginner

What is memoization in the context of recursion?

Memoization is a technique to store the results of expensive function calls and reuse them when the same inputs occur again, avoiding repeated calculations.

beginner

How does memoization improve recursive function performance?

By saving results of subproblems, memoization prevents the function from recalculating the same values multiple times, reducing time complexity from exponential to linear in many cases.

beginner

In TypeScript, what data structure is commonly used to implement memoization?

A JavaScript object or Map is commonly used to store computed results with input parameters as keys for quick lookup.

intermediate

Explain with a simple example how memoization works in a Fibonacci recursive function.

In a Fibonacci function, memoization stores the result of fib(n) after computing it once. When fib(n) is needed again, the stored value is returned immediately instead of recalculating fib(n-1) and fib(n-2).

intermediate

What is the difference between memoization and tabulation?

Memoization is a top-down approach that caches results during recursion, while tabulation is a bottom-up approach that builds a table iteratively without recursion.

What does memoization primarily help to reduce in recursive functions?

A Number of function parameters

B Repeated calculations of the same subproblems

C Memory usage by deleting variables

D Code length

Memoization stores results of subproblems to avoid recalculating them, reducing repeated work.

Which data structure is best suited for memoization in TypeScript?

A Map or Object

B Array

C Set

D String

Map or Object allows storing key-value pairs for quick lookup of computed results.

What is the time complexity improvement when using memoization on Fibonacci recursion?

A From O(n) to O(1)

B No improvement

C From O(2^n) to O(n)

D From O(n^2) to O(n^3)

Memoization reduces exponential time O(2^n) to linear time O(n) by caching results.

Memoization is an example of which programming technique?

A Greedy algorithm

B Brute force

C Divide and conquer

D Dynamic programming

Memoization is a dynamic programming technique that stores intermediate results.

Which of the following is NOT true about memoization?

A It always reduces space complexity

B It uses extra memory to store results

C It avoids repeated recursive calls

D It can be implemented using a cache

Memoization uses extra memory, so it does not reduce space complexity; it may increase it.

Describe how memoization optimizes a recursive Fibonacci function.

Think about how repeated calls to fib(n) are handled.

Explain the difference between memoization and tabulation in dynamic programming.

Consider the order of solving subproblems.

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