Maximize number segments of cutted in length x, y and z (Memoization)

Given a rod of length n, maximize the total number of segments can be cutted in length of x, y, and z.

Hint

This algorithm uses memoization to improve performance. It's same as the recursive version of Maximize number segments of cutted in length x, y and z, but with a special 'cache' to store results. The 'cache key' is crucial. It's how we identify and store past results. Sometimes we need one key, sometimes multiple keys. We add code to:

• Initialize cache to store results.
• Using the key to check cached result. If the result is not found in the cache, do the calculation and save the result in cache.
• Return the cached result.

• Python
• Javascript
•  Run

cache = None

def cut(n, x, y, z):
  if n <= 0:
    return 0
  
  global cache
  if not cache:
    cache = [float('-inf')] * (n + 1)

  if cache[n] == float('-inf'):
    cache[n] = 1 + max(cut(n - x, x, y, z), cut(n - y, x, y, z), cut(n - z, x, y, z))

  return cache[n]

n = 4
x = 2
y = 1
z = 1

print(cut(n, x, y, z))

let cache = null;

function cut(n, x, y, z) {
  if (n <= 0) {
    return 0;
  }

  if (!cache) {
    cache = Array(n + 1);
  }

  if (!cache[n]) {
    cache[n] = 1 + Math.max(cut(n - x, x, y, z), cut(n - y, x, y, z), cut(n - z, x, y, z));
  }

  return cache[n];
}

const n = 4;
const x = 2;
const y = 1;
const z = 1;

console.log(cut(n, x, y, z));