# Python Programming – Ugly Numbers

Python Programming - Ugly Numbers - Dynamic Programming Ugly numbers are numbers whose only prime factors are 2, 3 or 5.

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 (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[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:
Let us see how it works

```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```
Python
``````# Python program to find n'th Ugly number

# Function to get the nth ugly number
def getNthUglyNo(n):

ugly = [0] * n # To store ugly numbers

# 1 is the first ugly number
ugly[0] = 1

# i2, i3, i5 will indicate indices for 2,3,5 respectively
i2 = i3 =i5 = 0

# set initial multiple value
next_multiple_of_2 = 2
next_multiple_of_3 = 3
next_multiple_of_5 = 5

# start loop to find value from ugly[1] to ugly[n]
for l in range(1, n):

# choose the min value of all available multiples
ugly[l] = min(next_multiple_of_2, next_multiple_of_3, next_multiple_of_5)

# increment the value of index accordingly
if ugly[l] == next_multiple_of_2:
i2 += 1
next_multiple_of_2 = ugly[i2] * 2

if ugly[l] == next_multiple_of_3:
i3 += 1
next_multiple_of_3 = ugly[i3] * 3

if ugly[l] == next_multiple_of_5:
i5 += 1
next_multiple_of_5 = ugly[i5] * 5

# return ugly[n] value
return ugly[-1]

def main():

n = 150

print getNthUglyNo(n)

if __name__ == '__main__':
main()
``````

Output :

```5832
```

Algorithmic Paradigm: Dynamic Programming
Time Complexity: O(n)
Auxiliary Space: O(n)

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#### 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.