Factorial Trailing Zeroes
https://leetcode.com/problems/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.
Approach :
Lets see some examples :
5! = 1 * 2 * 3 * 4 * 5 = 1 trailing zero
7! = 1 * 2 * 3 * 4 * 5 * 6 * 7 = 1 trailing zero
10! = 1 * 2 * 3 * 4 * 5 * 6 * 7 * 8 * 9 * 10 = 2 trailing zeroes
15! = 1 * ... * 5 * .... * 10 * ... * 15 = 3 trailing zeroes
20! = 1 * ... * 5 * .... * 10 * ... * 15 * ... * 20 = 4 trailing zeroes
25! = 1 * ... * 5 * .... * 10 * ... * 15 * ... * 20 * .... * 25 = 6 trailing zeroes (25 = 5 * 5)
So basically we have to count how many times 5 will come, while calculating factorial of a number. Number of trailing zeroes will be equal to number of 5's.
Implementation 1 : Naive (Time Limit Exceeded 😀 )
import java.math.BigInteger;
class Solution {
public int trailingZeroes(int n) {
// Calculate n!
BigInteger nFactorial = BigInteger.ONE;
for (int i = 2; i <= n; i++) {
nFactorial = nFactorial.multiply(BigInteger.valueOf(i));
}
// Count how many 0's are on the end.
int trailingZeroes = 0;
while (nFactorial.mod(BigInteger.TEN).equals(BigInteger.ZERO)) {
nFactorial = nFactorial.divide(BigInteger.TEN);
trailingZeroes++;
}
return trailingZeroes;
}
}Implementation 2 :
class Solution {
public int trailingZeroes(int n) {
int trailingZeroes = 0;
while(n >= 5) {
n = n /5;
trailingZeroes += n;
}
return trailingZeroes;
}
}