factorial-trailing-zeroes.js

/**
 * Factorial Trailing Zeroes
 *
 * Given an integer n, return the number of trailing zeroes in n!.
 *
 * Example 1:
 *
 * Input: 3
 * Output: 0
 * Explanation: 3! = 6, no trailing zero.
 * Example 2:
 *
 * Input: 5
 * Output: 1
 * Explanation: 5! = 120, one trailing zero.
 * Note: Your solution should be in logarithmic time complexity.
 *
 * Solution: Because all trailing 0 is from factors 5 * 2.
 *
 * But sometimes one number may have several 5 factors, for example, 25 have two 5 factors,
 * 125 have three 5 factors. In the n! operation, factors 2 is always ample. So we just count
 * how many 5 factors in all number from 1 to n.
 */

/**
 * @param {number} n
 * @return {number}
 */
const trailingZeroes = n => {
  return n < 5 ? 0 : Math.floor(n / 5) + trailingZeroes(n / 5);
};

export { trailingZeroes };