Memoization Top Down DP in DSA Javascript - Time & Space Complexity

Time Complexity: Memoization Top Down DP

O(n)

Understanding Time Complexity

We want to understand how using memoization changes the time it takes to solve problems with repeated calculations.

How does saving results help reduce repeated work?

Scenario Under Consideration

Analyze the time complexity of the following memoized Fibonacci function.

function fib(n, memo = {}) {
  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 memoization to save results of smaller problems.

Identify Repeating Operations

Look at the recursive calls and how many times each happens.

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

How Execution Grows With Input

Each number from 0 up to n is calculated once, so the work grows roughly in a straight line with n.

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

Pattern observation: Doubling n roughly doubles the work, showing linear growth.

Final Time Complexity

Time Complexity: O(n)

This means the time to compute grows in a straight line with the input size because each subproblem is solved once.

Common Mistake

[X] Wrong: "Memoization still causes exponential time because of recursion."

[OK] Correct: Memoization saves results so each input is computed once, avoiding repeated work and reducing time to linear.

Interview Connect

Understanding memoization shows you how to turn slow recursive solutions into fast ones, a key skill for many coding challenges.

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 memoization in top-down dynamic programming?

easy

A. To store results of subproblems to avoid repeated calculations

B. To sort the input data before processing

C. To convert recursive code into iterative code

D. To reduce the memory usage by deleting variables

Solution

1. Step 1: Understand memoization concept

   Memoization saves the results of expensive function calls and returns the cached result when the same inputs occur again.

2. Step 2: Identify its role in top-down DP

   In top-down dynamic programming, memoization prevents repeated calculations of the same subproblems by storing their results.

3. Final Answer:

   To store results of subproblems to avoid repeated calculations -> Option A

4. Quick Check:

   Memoization = Store subproblem results [OK]

Hint: Memoization caches answers to reuse later [OK]

Common Mistakes:

• Confusing memoization with sorting
• Thinking memoization reduces memory usage
• Believing memoization converts recursion to iteration

2. Which of the following is the correct way to declare a memo object in JavaScript for top-down DP?

easy

A. const memo = 0;

B. let memo = [];

C. const memo = new Map();

D. var memo = function() {};

Solution

1. Step 1: Identify memo storage type

   Memoization requires a data structure to store key-value pairs of subproblem results. Map is ideal for this in JavaScript.

2. Step 2: Check options for correct syntax

   const memo = new Map(); uses new Map() which is correct. let memo = []; is an array, which can work but less flexible. var memo = function() {}; is a function, incorrect. const memo = 0; is a number, incorrect.

3. Final Answer:

   const memo = new Map(); -> Option C

4. Quick Check:

   Memo object = Map() [OK]

Hint: Use Map() for flexible memo keys [OK]

Common Mistakes:

• Using a function instead of an object
• Initializing memo as a number
• Confusing array with Map for memo

3. What will be the output of the following JavaScript code using memoization for Fibonacci?

const memo = new Map();
function fib(n) {
  if (n <= 1) return n;
  if (memo.has(n)) return memo.get(n);
  const result = fib(n - 1) + fib(n - 2);
  memo.set(n, result);
  return result;
}
console.log(fib(5));

medium

A. 5

B. 8

C. 3

D. 13

Solution

1. Step 1: Understand Fibonacci sequence

   Fibonacci sequence starts as 0,1,1,2,3,5,8,... So fib(5) = 5.

2. Step 2: Trace the code execution

   The function uses memoization to avoid repeated calls. fib(5) calculates fib(4)+fib(3), and so on, finally returning 5.

3. Final Answer:

   5 -> Option A

4. Quick Check:

   fib(5) = 5 [OK]

Hint: Recall Fibonacci sequence values [OK]

Common Mistakes:

• Confusing fib(5) with fib(6)
• Ignoring base cases
• Not understanding memo usage

4. Identify the bug in this memoized factorial function and select the fix:

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

medium

A. Change base case to 'if (n === 1) return 1;'

B. Change 'if (memo[n])' to 'if (memo.hasOwnProperty(n))'

C. Initialize memo as 'const memo = new Map();'

D. Remove memo and use plain recursion

Solution

1. Step 1: Understand the bug in memo check

   Using 'if (memo[n])' fails when memo[n] is 0 or falsy. It should check if the key exists.

2. Step 2: Fix by checking key existence

   Using 'memo.hasOwnProperty(n)' correctly checks if the result is stored, avoiding false negatives.

3. Final Answer:

   Change 'if (memo[n])' to 'if (memo.hasOwnProperty(n))' -> Option B

4. Quick Check:

   Check key existence, not truthiness [OK]

Hint: Check key presence, not value truthiness [OK]

Common Mistakes:

• Checking memo[n] directly causes bugs for 0 or falsy values
• Changing base case unnecessarily
• Removing memo defeats purpose

5. Given the following top-down memoized function to count ways to climb stairs, what is the output of countWays(4)?

const memo = {};
function countWays(n) {
  if (n === 0) return 1;
  if (n < 0) return 0;
  if (memo.hasOwnProperty(n)) return memo[n];
  memo[n] = countWays(n - 1) + countWays(n - 2);
  return memo[n];
}
console.log(countWays(4));

hard

A. 7

B. 6

C. 8

D. 5

Solution

1. Step 1: Understand the problem

   This function counts ways to climb stairs taking 1 or 2 steps. Base cases: countWays(0)=1, countWays(negative)=0.

2. Step 2: Calculate countWays(4) step-by-step

   countWays(4) = countWays(3) + countWays(2) countWays(3) = countWays(2) + countWays(1) countWays(2) = countWays(1) + countWays(0) countWays(1) = countWays(0) + countWays(-1) countWays(0) = 1, countWays(-1) = 0 So countWays(1) = 1 + 0 = 1 countWays(2) = 1 + 1 = 2 countWays(3) = 2 + 1 = 3 countWays(4) = 3 + 2 = 5

3. Final Answer:

   5 -> Option D

4. Quick Check:

   Ways to climb 4 stairs = 5 [OK]

Hint: Count ways as sum of previous two counts [OK]

Common Mistakes:

• Miscounting base cases
• Forgetting negative input returns 0
• Mixing up addition order