Memoization to Optimize Recursion in DSA Typescript - Time & Space Complexity

Time Complexity: Memoization to Optimize Recursion

O(n)

Understanding Time Complexity

We want to see how using memoization changes the speed of recursive functions.

How does remembering past answers help us avoid repeating work?

Scenario Under Consideration

Analyze the time complexity of the following code snippet.

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];
}
    

This function calculates the nth Fibonacci number using recursion and stores results to avoid repeated calculations.

Identify Repeating Operations

Identify the loops, recursion, array traversals that repeat.

• Primary operation: Recursive calls to fib with smaller n.
• How many times: Each unique n from 0 to n is computed once due to memoization.

How Execution Grows With Input

Each number from 0 up to n is calculated once and stored, so work grows steadily with n.

Input Size (n)	Approx. Operations
10	About 10 calculations
100	About 100 calculations
1000	About 1000 calculations

Pattern observation: The work grows roughly in a straight line as n increases.

Final Time Complexity

Time Complexity: O(n)

This means the function does work proportional to the input size, calculating each needed value once.

Common Mistake

[X] Wrong: "Recursive Fibonacci always takes exponential time because it calls itself twice each time."

[OK] Correct: With memoization, each number is calculated once and saved, so repeated calls don't redo work, making it much faster.

Interview Connect

Understanding how memoization changes recursion speed is a key skill that shows you can improve algorithms by avoiding repeated work.

Self-Check

"What if we removed memoization and just used plain recursion? How would the time complexity change?"

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