# 625. Minimum Factorization

## Description

Given a positive integer num, return the smallest positive integer x whose multiplication of each digit equals num. If there is no answer or the answer is not fit in 32-bit signed integer, return 0.

Example 1:

Input: num = 48
Output: 68


Example 2:

Input: num = 15
Output: 35


Constraints:

• 1 <= num <= 231 - 1

## Solutions

• class Solution {
public int smallestFactorization(int num) {
if (num < 2) {
return num;
}
long ans = 0, mul = 1;
for (int i = 9; i >= 2; --i) {
if (num % i == 0) {
while (num % i == 0) {
num /= i;
ans = mul * i + ans;
mul *= 10;
}
}
}
return num < 2 && ans <= Integer.MAX_VALUE ? (int) ans : 0;
}
}

• class Solution {
public:
int smallestFactorization(int num) {
if (num < 2) {
return num;
}
long long ans = 0, mul = 1;
for (int i = 9; i >= 2; --i) {
if (num % i == 0) {
while (num % i == 0) {
num /= i;
ans = mul * i + ans;
mul *= 10;
}
}
}
return num < 2 && ans <= INT_MAX ? ans : 0;
}
};

• class Solution:
def smallestFactorization(self, num: int) -> int:
if num < 2:
return num
ans, mul = 0, 1
for i in range(9, 1, -1):
while num % i == 0:
num //= i
ans = mul * i + ans
mul *= 10
return ans if num < 2 and ans <= 2**31 - 1 else 0


• func smallestFactorization(num int) int {
if num < 2 {
return num
}
ans, mul := 0, 1
for i := 9; i >= 2; i-- {
if num%i == 0 {
for num%i == 0 {
num /= i
ans = mul*i + ans
mul *= 10
}
}
}
if num < 2 && ans <= math.MaxInt32 {
return ans
}
return 0
}