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 ;
}
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