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 == 0)

    {

        num /= div;

    }

    return num;

}



bool isUgly(int num)

{

    num = maxDivide(num, 2);

    num = maxDivide(num, 3);

    num = maxDivide(num, 5);

    return num == 1? true:false;

}



int getNthUglyNo(int n)

{

    int c = 0;

    int i = 0;

    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 = 1; 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 < 1) return n; 

        int n2 = 2, n3 = 3, n5 = 5;

        int i2 = 0, i3 = 0, i5 = 0;

        rs.resize(n, 1);

        for (int i = 1; i < n; i++)

        {

            int t = min(n2, min(n3,n5));

            if (t == n2) 

            {

                rs[i] = n2;

                n2 = rs[++i2]*2;

            }

            if (t == n3) //Watch out, maybe repeated numbers

            {

                rs[i] = n3;

                n3 = rs[++i3]*3;

            }

            if (t == n5) //Watch out, no else!

            {

                rs[i] = n5;

                n5 = rs[++i5]*5;

            }

        }

        return rs.back();

    }

};

Testing：

int main()

    {

        unsigned no = getNthUglyNo(35);

        printf("ugly no. is %d \n",  no);

        vector<int> rs = getAllUglyNo(35);

        for (auto x:rs) cout<<x<<" ";

        cout<<endl;



        UglyNumbers un;

        printf("Ugly no. is %d \n", un.getNthUglyNo(35, rs));

        for (auto x:rs) cout<<x<<" ";

        cout<<endl;



        system("pause");    

        return 0;

    }