Here, We see Preimage Size of Factorial Zeroes Function LeetCode Solution. This Leetcode problem is done in many programming languages like C++, Java, JavaScript, Python, etc. with different approaches.

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Preimage Size of Factorial Zeroes Function LeetCode Solution

Problem Statement

Let f(x) be the number of zeroes at the end of x!. Recall that x! = 1 * 2 * 3 * ... * x and by convention, 0! = 1.

• For example, f(3) = 0 because 3! = 6 has no zeroes at the end, while f(11) = 2 because 11! = 39916800 has two zeroes at the end.

Given an integer k, return the number of non-negative integers x have the property that f(x) = k.

Example 1:
Input: k = 0 Output: 5 Explanation: 0!, 1!, 2!, 3!, and 4! end with k = 0 zeroes.

Example 2:
Input: k = 5 Output: 0 Explanation: There is no x such that x! ends in k = 5 zeroes.

Example 3:
Input: k = 3 Output: 5

Preimage Size of Factorial Zeroes Function Leetcode Solution C++

class Solution {
 public:
  int preimageSizeFZF(int k) {
    long l = 0;
    long r = 5L * k;
    while (l < r) {
      const long m = (l + r) / 2;
      if (trailingZeroes(m) >= k)
        r = m;
      else
        l = m + 1;
    }
    return trailingZeroes(l) == k ? 5 : 0;
  }
 private:
  int trailingZeroes(long n) {
    return n == 0 ? 0 : n / 5 + trailingZeroes(n / 5);
  }
};Code language: PHP (php)

Preimage Size of Factorial Zeroes Function Leetcode Solution Java

class Solution {
    public int preimageSizeFZF(int k) {
        return (int)(rightBoundOfKTrailingZeros(k) - leftBoundOfKTrailingZeros(k) + 1);
    }
    private long leftBoundOfKTrailingZeros(int k) {
        long start = 0;
        long end = 5L * (k + 1);
        while (start + 1 < end) {
            long mid = start + (end - start) / 2;
            if (trailingZeroes(mid) < k) {
                start = mid;
            } else {
                end = mid;
            }
        }
        if (trailingZeroes(start) == k) {
            return start;
        } else {
            return end;
        }
    }
    private long rightBoundOfKTrailingZeros(int k) {
        long start = 0;
        long end = 5L * (k + 1);
        while (start + 1 < end) {
            long mid = start + (end - start) / 2;
            if (trailingZeroes(mid) <= k) {
                start = mid;
            } else {
                end = mid;
            }
        }
        if (trailingZeroes(end) == k) {
            return end;
        } else {
            return start;
        }
    }
    private int trailingZeroes(long n) {
        int result = 0;
        long divisor = 5;
        while (divisor <= n) {
            result += n / divisor;
            divisor *= 5;
        }
        return result;
    }
}Code language: PHP (php)

Preimage Size of Factorial Zeroes Function Leetcode Solution JavaScript

var helper = function(m) {
    let r = m;
    while(m > 0) {
        m = Math.floor(m / 5);
        r += m;
    }
    return r;
}
var preimageSizeFZF = function(k) {
    let min = 0, max = k+1;
    while(min < max) {
        let center = ~~((min+max)/2);
        let zero = helper(center);
        if (zero > k) {
            max = center;
        }
        else if (zero < k) {
            min = center + 1;
        }
        else {
            return 5;
        }
    }
    return 0;
};Code language: JavaScript (javascript)

Preimage Size of Factorial Zeroes Function Solution Python

class Solution(object):
    def preimageSizeFZF(self, k):
        m = floor(log(4 * k + 1, 5))
        while m:
            unit = (5 ** m - 1) // 4
            if k
                return 0
            k %= unit
            m -= 1
        return 5