freeCodeCamp/curriculum/challenges/english/10-coding-interview-prep/rosetta-code/left-factorials.english.md

id, title, challengeType, forumTopicId

id	title	challengeType	forumTopicId
5a23c84252665b21eecc7ee0	Left factorials	5	302302

Description

Left factorials, $ !n $, may refer to either subfactorials or to factorial sums. The same notation can be confusingly seen used for the two different definitions. Sometimes, subfactorials (also known as derangements) may use any of the notations:

• $!n`$
• $!n$
• $n¡$

(It may not be visually obvious, but the last example uses an upside-down exclamation mark.) This task will be using this formula for left factorial: $ !n = \sum_{k=0}^{n-1} k! $ where $!0 = 0$

Instructions

Write a function to calculate the left factorial of a given number.

Tests

tests:
  - text: <code>leftFactorial</code> should be a function.
    testString: assert(typeof leftFactorial == 'function');
  - text: <code>leftFactorial(0)</code> should return a number.
    testString: assert(typeof leftFactorial(0) == 'number');
  - text: <code>leftFactorial(0)</code> should return <code>0</code>.
    testString: assert.equal(leftFactorial(0), 0);
  - text: <code>leftFactorial(1)</code> should return <code>1</code>.
    testString: assert.equal(leftFactorial(1), 1);
  - text: <code>leftFactorial(2)</code> should return <code>2</code>.
    testString: assert.equal(leftFactorial(2), 2);
  - text: <code>leftFactorial(3)</code> should return <code>4</code>.
    testString: assert.equal(leftFactorial(3), 4);
  - text: <code>leftFactorial(10)</code> should return <code>409114</code>.
    testString: assert.equal(leftFactorial(10), 409114);
  - text: <code>leftFactorial(17)</code> should return <code>22324392524314</code>.
    testString: assert.equal(leftFactorial(17), 22324392524314);
  - text: <code>leftFactorial(19)</code> should return <code>6780385526348314</code>.
    testString: assert.equal(leftFactorial(19), 6780385526348314);

Challenge Seed

function leftFactorial(n) {
  // Good luck!
}

Solution

function leftFactorial(n) {
  if (n == 0) return 0;
  if (n == 1) return 1;

  // Note: for n>=20, the result may not be correct.
  // This is because JavaScript uses 53 bit integers and
  // for n>=20 result becomes too large.

  let res = 2,
    fact = 2;
  for (var i = 2; i < n; i++) {
    res += fact;
    fact *= i + 1;
  }

  return res;
}