Using Closure to Calculate Large Fibonacci Numbers

The code defines a function createFib that returns another function fib, which calculates Fibonacci numbers. It uses a technique called memoization to optimize performance. Let’s go through it line by line.

const createFib = () => {
  // Create an empty object called memo to store calculated Fibonacci numbers.
  const memo = {};

  // Define the function fib, which calculates the nth Fibonacci number.
  const fib = (n) => {
    // Base case: If n is 0, the Fibonacci number is 0.
    if (n === 0) return 0;

    // Base case: If n is 1, the Fibonacci number is 1.
    if (n === 1) return 1;

    // Check if we've already calculated fib(n) and stored it in memo.
    // If so, return it to avoid redundant calculations.
    if (memo[n]) return memo[n];

    // Calculate fib(n) recursively.
    let fibn = fib(n - 1) + fib(n - 2);

    // Store the newly calculated fib(n) in memo for future use.
    memo[n] = fibn;

    // Return the calculated fib(n).
    return fibn;
  };

  // Return the fib function, which now has access to memo because of closure.
  return fib;
};

const createFib = () => { ... }: This is an arrow function that, when invoked, returns another function (fib).
const memo = {};: This object stores previously computed Fibonacci numbers. This is the memoization part.
const fib = (n) => { ... }: This is the function that actually calculates Fibonacci numbers.
if (n === 0) return 0; and if (n === 1) return 1;: These are the base cases for Fibonacci numbers. Fib(0) is 0, and Fib(1) is 1.
if (memo[n]) return memo[n];: Before calculating fib(n), the function checks if it’s already stored in memo. If it is, it returns the stored value to avoid redundant calculations.
let fibn = fib(n - 1) + fib(n - 2);: This is where the Fibonacci number is actually calculated using recursion.
memo[n] = fibn;: The newly calculated value is stored in memo for future use.
return fibn;: Finally, the function returns the calculated Fibonacci number.
return fib;: At the end of createFib, the fib function is returned. It retains access to memo thanks to closure, so it can use memo in future calls.

This code leverages memoization to avoid recalculating Fibonacci numbers, which makes it more efficient. In a typical recursive Fibonacci function without memoization, the time complexity would be exponential (O(2^n)). With memoization, the time complexity is linear (O(n)), which is a substantial optimization.

I hope this explanation helps clarify how the code works! Feel free to ask if you have more questions.