## Amazon Interview Question

**Country:**United States

Factor n into primes (assuming n is not an RSA kind of number). The simplest and most obvious number whose digits multiply to give n is simply these primes concatenated.

However, we need to find the smallest number. So, for that we will need to multiply all pairs of 2's, 2's and 3's and pairs of 3's. This is because only these are less than 10: 2*2=4, 2*3=6, 3*3=9. Once, we have multiplied all such pairs, we just need to check for any primes > 7. If no such prime, we just arrange the digits in ascending order and get the smallest number.

example: lets say n=2520=2*2*2*3*5*7*3,

then we can group this as (2*2), (2*3), 5,7,3. Sort the digits to give 34567.

for n= 2520, the smallest no whose digit's sum will be n is 5879 not 34567 as you suggested. Here is the explanation:

To get the smallest number, we divide the number with the greatest factor of the number which is less than 10, in this case it is 9, hence 9 is the last digit of the number. Remaining number is 2520/9=280. Now we divide 280 with the greatest factor of the number which is less than 10, which is 7. so second last digit of the required number is 7. Remaining number is 40. Again we divide 40 with 8, so the 3rd last digit is 8, remaining number is 5. As it is less than 10 so 5 is the first digit of the number. Hence the number comes out to be 5879.

*product(for n= 2520, the smallest no whose digit's product will be n is 5879 not 34567 )

@Isha, thanks for pointing that out. I missed a case there which is, 2*2*2=8.

So, my algorithm still works if I pick groups of factors using 2^3, 2^2, 2*3 or 3^2. Another plus point in using my algorithm is that I do not have to look for the largest factor of n which is less than 10. I can simply use the count of primes 2,3,5,7 to form the groups.

Instead of factoring the number, all we need to do, is to check whether the number can be formed by a combination of 2,3,5,7

(since no other prime number like 11,13 can be represented as a single digit)

So we go ahead , get the maximum power of 2 in the number, let's say w.

similarly we go ahead with 3,5 ,7 , and let's say we get x,y,z

and then if (2^w )*(3^x)*(5*y)(7^z) == number itself,

the smallest number would be

(2222....) w times ...(3333...) x times (5555....)y times (777...)z times

else print -1

Worst case time complexity - O(logn)

Wonder if we can optimize it further ....

and then we should multiply all 2's to make 4's and all 3's to make 9

if we have even of both(2's and 3's)---> do nothing

both odd----> multiply the remaining 2 and 3

and then sort to give the minimum number

#include <iostream>

#include <stdio.h>

int main()

{

int test;

scanf("%d",&test);

for(int i=0;i<test;i++)

{

long long int n;

scanf("%lld",&n);

int A[10]={0};

for(int j=9;j>0;j--)

{

if(n==1)

break;

while(n%j==0)

{

A[j]++;

n/=j;

}

}

if(n>1)

{

printf("-1\n");

continue;

}

for(int j=2;j<10;j++)

{

while(A[j]>0)

{

A[j]--;

printf("%d",j);

}

}

printf("\n");

}

return 0;

}

```
#include <iostream>
#include <stdio.h>
int main()
{
int test;
scanf("%d",&test);
for(int i=0;i<test;i++)
{
long long int n;
scanf("%lld",&n);
int A[10]={0};
for(int j=9;j>0;j--)
{
if(n==1)
break;
while(n%j==0)
{
A[j]++;
n/=j;
}
}
if(n>1)
{
printf("-1\n");
continue;
}
for(int j=2;j<10;j++)
{
while(A[j]>0)
{
A[j]--;
printf("%d",j);
}
}
printf("\n");
}
return 0;
}
```

int product()

{

int i,j,k, test, temp,temp1 = 1, remain, factor=0, found = 0;

int A[10]={0};

printf("enter the number \n");

scanf("%d",&test);

temp = test;

i = 9;

j = 0;

while (i>= 2)

{

factor = 0;

while((temp%i == 0) && (temp > 0))

{

factor = 1;

A[j] = i;

printf("filling is A[%d] is %d \n", j,A[j]);

j++;

temp /= i;

}

temp1 = 1;

if (factor)

{

k = j-1;

for (; k>=0; k--)

{

printf("factor at A[%d] is %d \n", k,A[k]);

temp1 *= A[k];

if (temp1 == test)

{

printf("factor found %d \n", temp1);

found = 1;

break;

}

}

}

if (found)

break;

temp = test;

i--;

}

j--;

if (found)

{

for (; j>=0; j--)

{

printf("factor is A[%d] is %d \n", j,A[j]);

}

}else

{

printf("factor is not found \n");

return -1;

}

return 0;

}

This is the code I came up with, to print the smallest number with the digits product equal to the number.

```
int prod(int n, int res)
{
if(res>n)
return 0;
else if(res==n)
{
return 1;
}
for(int i=9;i>2;i--) // we take i from 9 -> 2, since the o/p will be printed from last digit onwards so it will be the minimum no.
{
res*=i; // multiply a digit
if(prod(n,res)){
cout<<i;
return 1;
}
else
res/=i; // backtrack
}
return 0;
}
int main()
{
prod(2520,1);
}
```

The number printed on the screen will be the result. For above case the o/p is 5789

- Nitin February 20, 2014