Cracking the Coding Interview - Fooo

Solutions to Chapter 19 | Moderate
19 3 Write an algorithm which computes the number of trailing zeros in n factorial
SOLUTION
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pg 89
Trailing zeros are contributed by pairs of 5 and 2, because 5*2 = 10 To count the number of
pairs, we just have to count the number of multiples of 5 Note that while 5 contributes to
one multiple of 10, 25 contributes two (because 25 = 5*5)
1 public static int numZeros(int num) {
2 int count = 0;
3 if (num < 0) {
4 System.out.println(“Factorial is not defined for < 0”);
5 return 0;
6 }
7 for (int i = 5; num / i > 0; i *= 5) {
8 count += num / i;
9 }
10 return count;
11 }
Let’s walk through an example to see how this works: Suppose num = 26 In the first loop, we
count how many multiples of five there are by doing 26 / 5 = 5 (these multiples are 5, 10, 15,
20, and 25) In the next loop, we count how many multiples of 25 there are: 26 / 25 = 1 (this
multiple is 25) Thus, we see that we get one zero from 5, 10, 15 and 20, and two zeros from
25 (note how it was counted twice in the loops) Therefore, 26! has six zeros
OBSERVATIONS AND SUGGESTIONS:
» This is a bit of a brain teaser, but it can be approached logically (as shown above) By
thinking through what exactly will contribute a zero, and what doesn’t matter, you can
come up with a solution Again, be very clear in your rules up front so that you can
implement this correctly
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