Find the nth Ugly Number

Problem

Ugly numbers are numbers whose only prime factors are 2, 3 or 5. The sequence 1, 2, 3, 4, 5, 6, 8, 9, 10, 12, 15, ... shows the first 11 ugly numbers. ...By convention, 1 is included. Write a program to find and print the 1500'th ugly number. 

 Solution
Method 1 - Brute force
Loop for all positive integers until ugly number count is smaller than n, if an integer is ugly than increment ugly number count.
To check if a number is ugly, divide the number by greatest divisible powers of 2, 3 and 5, if the number becomes 1 then it is an ugly number otherwise not.
For example, let us see how to check for 300 is ugly or not. Greatest divisible power of 2 is 4, after dividing 300 by 4 we get 75. Greatest divisible power of 3 is 3, after dividing 75 by 3 we get 25. Greatest divisible power of 5 is 25, after dividing 25 by 25 we get 1. Since we get 1 finally, 300 is ugly number.

private static int maxDivide(int a, int b)
{
  while (a%b == 0)
   a = a/b; 
  return a;
}   

private static int isUgly(int no)
{
  no = maxDivide(no, 2);
  no = maxDivide(no, 3);
  no = maxDivide(no, 5);
   
  return (no == 1)? 1 : 0;
}    
 
/* Function to get the nth ugly number*/
public static int getNthUglyNo(int n)
{
  int i = 1; 
  int count = 1;   /* ugly number count */
 
  while (n > count)
  {
    i++;      
    if (isUgly(i))
      count++; 
  }
  return i;
}

This method is not time efficient as it checks for all integers until ugly number count becomes n, but space complexity of this method is O(1)

Method 2 - Using dynamic programming

1 Declare an array for ugly numbers:  ugly[150]
2 Initialize first ugly no:  ugly[0] = 1
3 Initialize three array index variables i2, i3, i5 to point to 
   1st element of the ugly array: 
        i2 = i3 = i5 =0; 
4 Initialize 3 choices for the next ugly no:
         next_mulitple_of_2 = ugly[i2]*2;
         next_mulitple_of_3 = ugly[i3]*3
         next_mulitple_of_5 = ugly[i5]*5;
5 Now go in a loop to fill all ugly numbers till 150:
For (i = 1; i < 150; i++ ) 
{
    /* These small steps are not optimized for good 
      readability. Will optimize them in C program */
    next_ugly_no  = Min(next_mulitple_of_2,
                                  next_mulitple_of_3,
                                  next_mulitple_of_5); 
    if  (next_ugly_no  == next_mulitple_of_2) 
    {             
        i2 = i2 + 1;        
        next_mulitple_of_2 = ugly[i2]*2;
    } 
    if  (next_ugly_no  == next_mulitple_of_3) 
    {             
        i3 = i3 + 1;        
        next_mulitple_of_3 = ugly[i3]*3;
     }            
     if  (next_ugly_no  == next_mulitple_of_5)
     {    
        i5 = i5 + 1;        
        next_mulitple_of_5 = ugly[i5]*5;
     } 
     ugly[i] =  next_ugly_no       
}/* end of for loop */ 
6.return next_ugly_no

Example
Suppose N = 50

initialize
   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 )

Fifth iteration
    ugly[4] = Min(ugly[i2]*2, ugly[i3]*3, ugly[i5]*5)
              = Min(6, 6, 10)
              = 6
    ugly[] =  | 1 | 2 | 3 |  4 | 5 | 6 |
    i2 = 3,  i3 =  2, i5 = 1  (i2 and i3 got incremented )

Will continue same way till I < 150

Here is the code in java

public static int getKthMagicNumber(int k) {
    if (k <= 0)
        return 0;
    int val = 1;
    Queue<Integer> Q2 = new LinkedList<Integer>();
    Queue<Integer> Q3 = new LinkedList<Integer>();
    Queue<Integer> Q5 = new LinkedList<Integer>();
    Q2.add(2);
    Q3.add(3);
    Q5.add(5);
    for (--k; k > 0; --k) { // We’ve done one iteration already.
        val = Math.min(Q2.peek().intValue(),
                Math.min(Q3.peek().intValue(), Q5.peek().intValue()));
        if (val == Q5.peek()) {
            Q5.remove();
        } else {
            if (val == Q3.peek()) {
                Q3.remove();
            } else { // must be from Q2
                Q2.remove();
                Q2.add(val * 2);
            }
            Q3.add(val * 3);
        }
        Q5.add(val * 5);
    }
    return val;
}

References
http://www.geeksforgeeks.org/ugly-numbers/
http://stackoverflow.com/questions/4600048/nth-ugly-number
http://tianrunhe.wordpress.com/2012/04/03/find-the-kth-number-with-prime-factors-3-5-and-7/