LeetCode Solutions
625. Minimum Factorization
Time: $O(8\log \texttt{a}) = O(\log \texttt{a})$ Space: $O(1)$
class Solution {
public:
int smallestFactorization(int num) {
if (num == 1)
return 1;
long ans = 0;
long base = 1;
for (int i = 9; i > 1; --i)
while (num % i == 0) {
num /= i;
ans = base * i + ans;
base *= 10;
}
return num == 1 && ans <= INT_MAX ? ans : 0;
}
};
class Solution {
public int smallestFactorization(int num) {
if (num == 1)
return 1;
long ans = 0;
long base = 1;
for (int i = 9; i > 1; --i)
while (num % i == 0) {
num /= i;
ans = base * i + ans;
base *= 10;
}
return num == 1 && ans <= Integer.MAX_VALUE ? (int) ans : 0;
}
}
class Solution:
def smallestFactorization(self, num: int) -> int:
if num == 1:
return 1
ans = 0
base = 1
for i in range(9, 1, -1):
while num % i == 0:
num //= i
ans = base * i + ans
base *= 10
return ans if num == 1 and ans < 2**31 - 1 else 0