Geeks Interview Question： Ugly Numbers

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 150′th ugly number.

METHOD 1 (Simple)
Thanks to Nedylko Draganov for suggesting this solution.

Algorithm:
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.

Below is the simple method, with printing programme, which can print the ugly numbers:

int maxDivide(int num, int div)
{
    while (num % div == )
    {
        num /= div;
    }
    return num;
}

bool isUgly(int num)
{
    num = maxDivide(num, );
    num = maxDivide(num, );
    num = maxDivide(num, );
    return num == ? true:false;
}

int getNthUglyNo(int n)
{
    int c = ;
    int i = ;
    while (c < n)
    {
        if (isUgly(++i)) c++;
    }
    return i;
}
#include <vector>
using std::vector;
vector<int> getAllUglyNo(int n)
{
    vector<int> rs;
    for (int i = ; i <= n; i++)
    {
        if (isUgly(i)) rs.push_back(i);
    }
    return rs;
}

Dynamic programming:

Watch out:  We need to skip some repeated numbers, as commented out below.

Think about this algorithm, conclude as:

We caculate ugly numbers from button up, every new ugly number multiply 2,3,5 respectly would be a new ugly number.

class UglyNumbers
{
public:
    int getNthUglyNo(int n, vector<int> &rs)
    {
        if (n < ) return n;
        int n2 = , n3 = , n5 = ;
        int i2 = , i3 = , i5 = ;
        rs.resize(n, );
        for (int i = ; i < n; i++)
        {
            int t = min(n2, min(n3,n5));
            if (t == n2)
            {
                rs[i] = n2;
                n2 = rs[++i2]*;
            }
            if (t == n3) //Watch out, maybe repeated numbers
            {
                rs[i] = n3;
                n3 = rs[++i3]*;
            }
            if (t == n5) //Watch out, no else!
            {
                rs[i] = n5;
                n5 = rs[++i5]*;
            }
        }
        return rs.back();
    }
};

Testing：

int main()
    {
        unsigned no = getNthUglyNo();
        printf("ugly no. is %d \n",  no);
        vector<int> rs = getAllUglyNo();
        for (auto x:rs) cout<<x<<" ";
        cout<<endl;

        UglyNumbers un;
        printf("Ugly no. is %d \n", un.getNthUglyNo(, rs));
        for (auto x:rs) cout<<x<<" ";
        cout<<endl;

        system("pause");
        return ;
    }