Given a number ‘n’, find the smallest number ‘p’ such that if we multiply all digits of ‘p’, we get ‘n’. The result ‘p’ should have minimum two digits.
Examples:
Input: n = 36 Output: p = 49 // Note that 4*9 = 36 and 49 is the smallest such number Input: n = 100 Output: p = 455 // Note that 4*5*5 = 100 and 455 is the smallest such number Input: n = 1 Output:p = 11 // Note that 1*1 = 1 Input: n = 13 Output: Not Possible
For a given n, following are the two cases to be considered.
Case 1: n < 10 When n is smaller than n, the output is always n+10. For example for n = 7, output is 17. For n = 9, output is 19.
Case 2: n >= 10 Find all factors of n which are between 2 and 9 (both inclusive). The idea is to start searching from 9 so that the number of digits in result are minimized. For example 9 is preferred over 33 and 8 is preferred over 24.
Store all found factors in an array. The array would contain digits in non-increasing order, so finally print the array in reverse order.
1 public static int findSmallest(int n){
2 if(n < 10) return 10 + n;
3 int out = 0;
4 int pos = 0;
5 for(int i = 9; i > 1; i --){
6 while(n % i == 0){
7 n /= i;
8 out += i * Math.pow(10, pos);
9 pos ++;
10 }
11 }
12 return n ==1 ? out : -1;
13 }
Find the smallest number whose digits multiply to a given number n