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# Ugly Number II LeetCode Solution

## Problem – Ugly Number II LeetCode Solution

An ugly number is a positive integer whose prime factors are limited to `2``3`, and `5`.

Given an integer `n`, return the `nth` ugly number.

Example 1:

``````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.
``````

Example 2:

``````Input: n = 1
Output: 1
Explanation: 1 has no prime factors, therefore all of its prime factors are limited to 2, 3, and 5.
``````

Constraints:

• `1 <= n <= 1690`

## Ugly Number II LeetCode Solution in C++

``````class Solution {
public:
int nthUglyNumber(int n) {
if(n <= 0) return false; // get rid of corner cases
if(n == 1) return true; // base case
int t2 = 0, t3 = 0, t5 = 0; //pointers for 2, 3, 5
vector<int> k(n);
k = 1;
for(int i  = 1; i < n ; i ++)
{
k[i] = min(k[t2]*2,min(k[t3]*3,k[t5]*5));
if(k[i] == k[t2]*2) t2++;
if(k[i] == k[t3]*3) t3++;
if(k[i] == k[t5]*5) t5++;
}
return k[n-1];
}
};
``````

## Ugly Number II LeetCode Solution in Java

``````public class Solution {
public int nthUglyNumber(int n) {
int[] ugly = new int[n];
ugly = 1;
int index2 = 0, index3 = 0, index5 = 0;
int factor2 = 2, factor3 = 3, factor5 = 5;
for(int i=1;i<n;i++){
int min = Math.min(Math.min(factor2,factor3),factor5);
ugly[i] = min;
if(factor2 == min)
factor2 = 2*ugly[++index2];
if(factor3 == min)
factor3 = 3*ugly[++index3];
if(factor5 == min)
factor5 = 5*ugly[++index5];
}
return ugly[n-1];
}
}
``````

## Ugly Number II LeetCode Solution in Python

``````def nthUglyNumber(self, n):
ugly = 
i2, i3, i5 = 0, 0, 0
while n > 1:
u2, u3, u5 = 2 * ugly[i2], 3 * ugly[i3], 5 * ugly[i5]
umin = min((u2, u3, u5))
if umin == u2:
i2 += 1
if umin == u3:
i3 += 1
if umin == u5:
i5 += 1
ugly.append(umin)
n -= 1
return ugly[-1]
``````
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