# C Programming – Ugly Numbers

C Programming - Ugly Numbers - Dynamic Programming Ugly numbers are numbers whose only prime factors are 2, 3 or 5. Every step we choose the smallest one

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:

C
# include<stdio.h>
# include<stdlib.h>

/*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 untill ugly count
becomes n*/
while (n > count)
{
i++;
if (isUgly(i))
count++;
}
return i;
}

/* Driver program to test above functions */
int main()
{
unsigned no = getNthUglyNo(150);
printf("150th ugly no. is %d ",  no);
getchar();
return 0;
}

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  C++ Programming-Backtracking Set 6 (Hamiltonian Cycle)

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[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
C
# include<bits/stdc++.h>
using namespace std;

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

ugly[0] = 1;
for (int i=1; i<n; i++)
{
next_ugly_no = min(next_multiple_of_2,
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 */
int main()
{
int n = 150;
cout << getNthUglyNo(n);
return 0;
}

Output :

5832