Count number of derangements (Memoization)

Given a number n, find number of derangements in a set of n elements. A Derangement is a permutation with no element appears in its original position. For example, a derangement of [0, 1, 2] is [2, 0, 1].

Hint

This algorithm uses memoization to improve performance. It's same as the recursive version of Count number of derangements, 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

# Python implementation

cache = {}

def placement(n):
  if n == 1:
    return 0

  if n == 2:
    return 1
  
  if n not in cache:
    cache[n] = (n - 1) * (placement(n - 1) + placement(n - 2))

  return cache[n]

n = 4

print(placement(n))

// Javascript implementation

let cache = [];

function placement(n) {
  if (n == 1) {
    return 0;
  }

  if (n == 2) {
    return 1;
  }

  if (cache[n] === undefined) {
    cache[n] = (n - 1) * (placement(n - 1) + placement(n - 2));
  }

  return cache[n];
}

const n = 4;

console.log(placement(n));