# Java Programming – Ugly Numbers

Java Programming - Ugly Numbers - Dynamic Programming Ugly numbers are numbers whose only prime factors are 2, 3 or 5. Here we use similar merge

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.

Given a number n, the task is to find n’th Ugly number.

```Input  : n = 7
Output : 8

Input  : n = 10
Output : 12

Input  : n = 15
Output : 24

Input  : n = 150
Output : 5832```

Method 1 (Simple)

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.

Implementation:

Java
``````// Java program to find nth ugly number
class UglyNumber
{
/*This function divides a by greatest divisible
power of b*/
int maxDivide(int a, int b)
{
while(a % b == 0)
a = a/b;
return a;
}

/* Function to check if a number is ugly or not */
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*/
int getNthUglyNo(int n)
{
int i = 1;
int count = 1; // ugly number count

// check for all integers until count becomes n */
while(n > count)
{
i++;
if(isUgly(i) == 1)
count++;
}
return i;
}

/* Driver program to test above functions */
public static void main(String args[])
{
UglyNumber obj = new UglyNumber();
int no = obj.getNthUglyNo(150);
System.out.println("150th ugly no. is "+ no);
}
}
``````

Output:

`150th ugly no. is 5832`

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)

READ  Java Programming - Count number of binary strings without consecutive 1’s

Method 2 (Use Dynamic Programming)

Here is a time efficient solution with O(n) extra space. The ugly-number sequence is 1, 2, 3, 4, 5, 6, 8, 9, 10, 12, 15, …
because every number can only be divided by 2, 3, 5, one way to look at the sequence is to split the sequence to three groups as below:
(1) 1×2, 2×2, 3×2, 4×2, 5×2, …
(2) 1×3, 2×3, 3×3, 4×3, 5×3, …
(3) 1×5, 2×5, 3×5, 4×5, 5×5, …

We can find that every subsequence is the ugly-sequence itself (1, 2, 3, 4, 5, …) multiply 2, 3, 5. Then we use similar merge method as merge sort, to get every ugly number from the three subsequence. Every step we choose the smallest one, and move one step after.

```1 Declare an array for ugly numbers:  ugly[n]
2 Initialize first ugly no:  ugly = 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:
Let us see how it works

```initialize
ugly[] =  | 1 |
i2 =  i3 = i5 = 0;

First iteration
ugly = 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 = 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 = 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 = 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 = 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```
Java
``````// Java program to find nth ugly number
import java.lang.Math;

class UglyNumber
{
/* Function to get the nth ugly number*/
int getNthUglyNo(int n)
{
int ugly[] = new int[n];  // To store ugly numbers
int i2 = 0, i3 = 0, i5 = 0;
int next_multiple_of_2 = 2;
int next_multiple_of_3 = 3;
int next_multiple_of_5 = 5;
int next_ugly_no = 1;

ugly = 1;

for(int i = 1; i < n; i++)
{
next_ugly_no = Math.min(next_multiple_of_2,
Math.min(next_multiple_of_3,
next_multiple_of_5));

ugly[i] = next_ugly_no;
if (next_ugly_no == next_multiple_of_2)
{
i2 = i2+1;
next_multiple_of_2 = ugly[i2]*2;
}
if (next_ugly_no == next_multiple_of_3)
{
i3 = i3+1;
next_multiple_of_3 = ugly[i3]*3;
}
if (next_ugly_no == next_multiple_of_5)
{
i5 = i5+1;
next_multiple_of_5 = ugly[i5]*5;
}
} /*End of for loop (i=1; i<n; i++) */
return next_ugly_no;
}

/* Driver program to test above functions */
public static void main(String args[])
{
int n = 150;
UglyNumber obj = new UglyNumber();
System.out.println(obj.getNthUglyNo(n));
}
}
``````

Output :

```5832
```

Time Complexity: O(n)
Auxiliary Space: O(n) #### Venkatesan Prabu

Wikitechy Founder, Author, International Speaker, and Job Consultant. My role as the CEO of Wikitechy, I help businesses build their next generation digital platforms and help with their product innovation and growth strategy. I'm a frequent speaker at tech conferences and events.

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