Task 84. UniLecs. Find the number of anagrams.

anagramNumber.js

// it can be optimised via lazy evaluation or cached values
function factorial(n) {
  if (n < 0) {
    throw new TypeError('n should be non-negative')
  }
  if (n < 2) {
    return 1
  }
  return n * factorial(n - 1)
}

// can be optimised as well
// k-permutations of n, link: https://en.wikipedia.org/wiki/Permutation
function permutations(n, k) {
  return factorial(n) / (factorial(n - k) * factorial(k))
}

// count the quantity of each letter in a word
function lettersIn(word) {
  const hash = {}
  for (let ch of word) {
    hash[ch] = (hash[ch] || 0) + 1
  }
  return Object.values(hash)
}

function anagramNumber(word) {
  const ks = lettersIn(word)
  const [_, quantity] = ks.reduce(
    // n is a number of free elements where we can place k-letters (initial value is word.length)
    // k is a number of letters which need to be placed (initial value is 1)
    // q is a quantity of pre-calculated placements which we already calculated
    ([n, q], k) => [n - k, q * permutations(n, k)],
    [word.length, 1]
  )
  return quantity
}

console.log(anagramNumber('линия'))         // 60
console.log(anagramNumber('вектор'))        // 720 
console.log(anagramNumber('парабола'))      // 6720 
console.log(anagramNumber('математика'))    // 151200
console.log(anagramNumber('биссектриса'))   // 3326400