# Question

Write a program to find the n-th ugly number.

Ugly numbers are positive numbers whose prime factors only include 2, 3, 5.

Example:

Input: n = 10
Output: 12
Explanation: 1, 2, 3, 4, 5, 6, 8, 9, 10, 12 is the sequence of the first 10 ugly numbers.


Note:

1. 1 is typically treated as an ugly number.
2. n does not exceed 1690.

# Solution

The key is to realize each number can be and have to be generated by a former number multiplied by 2, 3 or 5. e.g.

1 2 3 4 5 6 8 9 10 12 15..

The next one must be x * 2 or y * 3 or z * 5, where x, y, z is an existing number.

How do we determine x, y, z then?

Apparently, you can just traverse the sequence generated by far from 1 … 15, until you find such x, y, z that x * 2, y * 3, z * 5 is just bigger than 15. In this case x=8, y=6, z=4. Then you compare x * 2, y * 3, z * 5 so you know next number will be x * 2 = 8 * 2 = 16. Now you have 1,2,3,4,….,15, 16, then what is next?

You wanna do the same process again to find the new x, y, z, but you realize, wait, do I have to traverse the sequence generated by far again?

NO! since you know last time, x=8, y=6, z=4 and x=8 was used to generate 16, so this time, you can immediately know the new_x = 9 (the next number after 8 is 9 in the generated sequence), y=6, z=4. Then you need to compare new_x * 2, y * 3, z * 5. You know next number is 9 * 2 = 18;

And you also know, the next x will be 10 since new_x = 9 was used this time. But what is next y? apparently, if y=6, 6*3 = 18, which is already generated in this round. So you also need to update next y from 6 to 8.

Based on the idea above, you can actually generated x,y,z from very beginning, and update x, y, z accordingly. It ends up with a O(n) solution.

class Solution(object):
def nthUglyNumber(self, n):
"""
:type n: int
:rtype: int
"""
if n == 1:
return 1
res = 
x, y, z = 0, 0, 0
while len(res) < n:
num = min(res[x] * 2, res[y] * 3, res[z] * 5)
if res[x] * 2 == num:
x += 1
if res[y] * 3 == num:
y += 1
if res[z] * 5 == num:
z += 1
res.append(num)

return res[-1]