I like that Python code, and for three digits numbers it is just fine.
If the number could be in a larger range, factoring out digits starting
with larger digits should give the correct sequence, in reverse. If the
number has a prime factor larger than 9, there is no solution.
// Returns the smallest number such that the product of the digits is n.
// Returns -1 if no such number exists.
int productOfDigits(int n)
{
int result=0;
int mlt = 1;
if (n == 0) return 0;
if (n == 1) return 1;
for(int i = 9; i > 1; --i)
{
while(n % i == 0)
{
result += mlt*i;
mlt *= 10;
n /= i;
if (n == 1) return result;
}
}
return -1;
}
On Wednesday, February 12, 2014 10:23:52 AM UTC-5, Shashwat Anand wrote:
>
> Since the limit is so small, won't a simple bruteforce do ?
> Just run a loop from [100, 1000) and return the minimum.
>
> Here is a quick python code I wrote.
>
> >>> N = 24
> >>> min (i for i in range (100, 1001) if int (str (i) [0]) * int (str (i)
> [1]) * int (str (i) [2]) == N)
> 138
> >>> N = 36
> >>> min (i for i in range (100, 1001) if int (str (i) [0]) * int (str (i)
> [1]) * int (str (i) [2]) == N)
> 149
> >>> N = 100
> >>> min (i for i in range (100, 1001) if int (str (i) [0]) * int (str (i)
> [1]) * int (str (i) [2]) == N)
> 455
>
>
> On Wed, Feb 12, 2014 at 8:45 PM, atul anand <[email protected]<javascript:>
> > wrote:
>
>> Given a number N, find the smallest 3 digits number such that product of
>> its digits is equal to N.
>>
>> For Eg:-
>>
>> for input 24 , output :138 (1*3*8 = 24)
>> for input 36 , output :149 (1*4*9 = 36)
>> for input 100 , output : 455 (4*5*5 = 100)
>>
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