Ugly Number II

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.
 class Solution {
        public int nthUglyNumber(int n) {
            int[] ugly = new int[n];
            ugly[0] = 1;
            int indexOf2 = 0, indexOf3 = 0, indexOf5 = 0;
            int factorOf2 = 2, factorOf3 = 3, factorOf5 = 5;
            for(int i=1;i<n;i++){
                int min = Math.min(Math.min(factorOf2,factorOf3),factorOf5);
                ugly[i] = min;
                if(factorOf2 == min)
                    factorOf2 = 2*ugly[++indexOf2];
                if(factorOf3 == min)
                    factorOf3 = 3*ugly[++indexOf3];
                if(factorOf5 == min)
                    factorOf5 = 5*ugly[++indexOf5];
            }
            return ugly[n-1];
        }
    }


Here, we are going to solve using efficient solution but having O(n) extra space. Ugly number sequence are 1, 2, 3, 4, 5, 8, 9, 10, 12, 15... A number should be divided only by 2, 3, 5 so that only it can become a ugly number. So for that, a number can be formed by upcoming split sequences (solution can be come up based on Hints given on Leetcode)

1×2, 2×2, 3×2, 4×2, 5×2, …
1×3, 2×3, 3×3, 4×3, 5×3, …
1×5, 2×5, 3×5, 4×5, 5×5, …

Below you find the detailed explanation of working of above code...,

Initially
 ugly[] =  | 1 |
   i2 =  i3 = i5 = 0;

First iteration
   ugly[1] = Min(ugly[i2]*2, ugly[i3]*3, ugly[i5]*5)
            = Min(2, 3, 5)
            = 2
   ugly[] =  | 1 | 2 |
   i2 = 1,  i3 = i5 = 0  (i2 got incremented ) 

Second iteration
    ugly[2] = Min(ugly[i2]*2, ugly[i3]*3, ugly[i5]*5)
             = Min(4, 3, 5)
             = 3
    ugly[] =  | 1 | 2 | 3 |
    i2 = 1,  i3 =  1, i5 = 0  (i3 got incremented ) 

Third iteration
    ugly[3] = Min(ugly[i2]*2, ugly[i3]*3, ugly[i5]*5)
             = Min(4, 6, 5)
             = 4
    ugly[] =  | 1 | 2 | 3 |  4 |
    i2 = 2,  i3 =  1, i5 = 0  (i2 got incremented )

Fourth iteration
    ugly[4] = Min(ugly[i2]*2, ugly[i3]*3, ugly[i5]*5)
              = Min(6, 6, 5)
              = 5
    ugly[] =  | 1 | 2 | 3 |  4 | 5 |
    i2 = 2,  i3 =  1, i5 = 1  (i5 got incremented )
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