Ugly Numbers

Ugly numbers are those number whose prime factors are 2, 3 or 5. From 1 to 15, there are 11 ugly numbers 1, 2, 3, 4, 5, 6, 8, 9, 10, 12, 15. The numbers 7, 11, 13 are not ugly because they are prime. The number 14 is not ugly because in its prime factor the 7 will come.

In this program, we will try to find the nth ugly number.

Input and Output

Input:
Take the term number. Say it is 10
Output:
The 10th ugly number is 12

Algorithm

getUglyNumbers(n)

Input: The number of terms.

Output: Find nth Ugly numbers.

Begin
   define array named uglyNum of size n
   i2 := 0, i3 := 0, i5 := 0
   next2mul := 2, next3mul := 3, next5Mul := 5
   next := 1
   ugluNum[0] := 1

   for i := 1 to n, do
      next := minimum of next2Mul, next3Mul and next5Mul
      uglyNum[i] := next
      if next = next2Mul, then
         i2 := i2 + 1
         next2mul := uglyNum[i2] * 2
      if next = next3Mul, then
         i3 := i3 + 1
         next3mul := uglyNum[i3] * 3
      if next = next5Mul, then
         i5 := i5 + 1
         next5mul := uglyNum[i5] * 5
   done
   return next
End

Example

# include<iostream>
using namespace std;

int min(int x, int y, int z) {            //find smallest among three numbers
   if(x < y) {
      if(x < z)
         return x;
      else
         return z;
   }else {
      if(y < z)
         return y;
      else
         return z;
   }
}

int getUglyNum(int n) {
   int uglyNum[n];          // To store ugly numbers
   int i2 = 0, i3 = 0, i5 = 0;

   //find next multiple as 1*2, 1*3, 1*5

   int next2mul = 2;
   int next3mul = 3;
   int next5mul = 5;
   int next = 1;              //initially the ugly number is 1

   uglyNum[0] = 1;

   for (int i=1; i<n; i++) {
      next = min(next2mul, next3mul, next5mul);       //find next ugly number
      uglyNum[i] = next;

      if (next == next2mul) {
         i2++;             //increase iterator of ugly numbers whose factor is 2
         next2mul = uglyNum[i2]*2;
      }

      if (next == next3mul) {
         i3++;             //increase iterator of ugly numbers whose factor is 3
         next3mul = uglyNum[i3]*3;
      }

      if (next == next5mul) {
         i5++;              //increase iterator of ugly numbers whose factor is 5
         next5mul = uglyNum[i5]*5;
      }
   }
   return next;        //the nth ugly number
}

int main() {
   int n;
   cout << "Enter term: "; cin >> n;
   cout << n << "th Ugly number is: " << getUglyNum(n) << endl;
}

Output

Enter term: 10
10th Ugly number is: 12