Memoization to Optimize Recursion in DSA Typescript - Interactive Practice

Practice - 5 Tasks

1fill in blank

easy

Complete the code to declare the memo object for caching results.

DSA Typescript

const memo: [1] = {};

Drag options to blanks, or click blank then click option'

ASet<number>

BArray<number>

CRecord<number, number>

DMap<number, number>

Attempts:

3 left

💡 Hint

Common Mistakes

Using Array instead of an object for memoization.

Using Set which only stores unique values without keys.

We use a Record with number keys and number values to store computed results for each input number.

2fill in blank

medium

Complete the code to check if the result for n is already cached in memo.

DSA Typescript

if ([1] in memo) return memo[n];

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Acache

Bmemo

Cresult

Dn

Attempts:

3 left

💡 Hint

Common Mistakes

Checking if 'memo' is in memo (wrong variable).

Using 'result' or 'cache' which are not defined here.

We check if the input number n is a key in the memo object to reuse the cached result.

3fill in blank

hard

Fix the error in storing the computed Fibonacci value in memo.

DSA Typescript

memo[n] = [1] + [2];

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Afib(n - 1)

Bfib(n - 2)

Cmemo[n - 1]

Dmemo[n - 2]

Attempts:

3 left

💡 Hint

Common Mistakes

Calling fib recursively again instead of using memoized values.

Mixing up memo keys or using wrong indices.

We store the sum of previously computed Fibonacci values from memo to avoid repeated recursion.

4fill in blank

hard

Fill both blanks to complete the memoized Fibonacci function.

DSA Typescript

function fib(n: number): number {
  if (n <= 1) return n;
  if (n [1] memo) return memo[n];
  memo[n] = fib(n - 1) [2] fib(n - 2);
  return memo[n];
}

Drag options to blanks, or click blank then click option'

Ain

B+

C-

D*

Attempts:

3 left

💡 Hint

Common Mistakes

Using '-' or '*' instead of '+' for Fibonacci addition.

Using wrong operator for checking memo keys.

We check if n is in memo and then add the results of fib(n-1) and fib(n-2) to store in memo.

5fill in blank

hard

Fill all three blanks to create a memoized factorial function.

DSA Typescript

const factorialMemo: Record<number, number> = {};
function factorial(n: number): number {
  if (n <= 1) return 1;
  if (n [1] factorialMemo) return factorialMemo[n];
  factorialMemo[n] = n [2] factorial(n [3] 1);
  return factorialMemo[n];
}

Drag options to blanks, or click blank then click option'

Ain

B*

C-

D+

Attempts:

3 left

💡 Hint

Common Mistakes

Using '+' instead of '*' for multiplication.

Using '+' instead of '-' to decrement n.

Not checking cache correctly.

We check if n is in factorialMemo, then multiply n by factorial(n-1) and store the result.

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