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793. Preimage Size of Factorial Zeroes Function

Description

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

 

Constraints:

• 0 <= k <= 109

Solutions

Binary search.

• Java
• C++
• Python
• Go

• class Solution {
      public int preimageSizeFZF(int k) {
          return g(k + 1) - g(k);
      }
  
      private int g(int k) {
          long left = 0, right = 5 * k;
          while (left < right) {
              long mid = (left + right) >> 1;
              if (f(mid) >= k) {
                  right = mid;
              } else {
                  left = mid + 1;
              }
          }
          return (int) left;
      }
  
      private int f(long x) {
          if (x == 0) {
              return 0;
          }
          return (int) (x / 5) + f(x / 5);
      }
  }
  

• class Solution {
  public:
      int preimageSizeFZF(int k) {
          return g(k + 1) - g(k);
      }
  
      int g(int k) {
          long long left = 0, right = 1ll * 5 * k;
          while (left < right) {
              long long mid = (left + right) >> 1;
              if (f(mid) >= k) {
                  right = mid;
              } else {
                  left = mid + 1;
              }
          }
          return (int) left;
      }
  
      int f(long x) {
          int res = 0;
          while (x) {
              x /= 5;
              res += x;
          }
          return res;
      }
  };
  

• class Solution:
      def preimageSizeFZF(self, k: int) -> int:
          def f(x):
              if x == 0:
                  return 0
              return x // 5 + f(x // 5)
  
          def g(k):
              return bisect_left(range(5 * k), k, key=f)
  
          return g(k + 1) - g(k)
  
  

• func preimageSizeFZF(k int) int {
  	f := func(x int) int {
  		res := 0
  		for x != 0 {
  			x /= 5
  			res += x
  		}
  		return res
  	}
  
  	g := func(k int) int {
  		left, right := 0, k*5
  		for left < right {
  			mid := (left + right) >> 1
  			if f(mid) >= k {
  				right = mid
  			} else {
  				left = mid + 1
  			}
  		}
  		return left
  	}
  
  	return g(k+1) - g(k)
  }
  

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