# 2847. Smallest Number With Given Digit Product

## Description

Given a positive integer n, return a string representing the smallest positive integer such that the product of its digits is equal to n, or "-1" if no such number exists.

Example 1:

Input: n = 105
Output: "357"
Explanation: 3 * 5 * 7 = 105. It can be shown that 357 is the smallest number with a product of digits equal to 105. So the answer would be "105".


Example 2:

Input: n = 7
Output: "7"
Explanation: Since 7 has only one digit, its product of digits would be 7. We will show that 7 is the smallest number with a product of digits equal to 7. Since the product of numbers 1 to 6 is 1 to 6 respectively, so "7" would be the answer.


Example 3:

Input: n = 44
Output: "-1"
Explanation: It can be shown that there is no number such that its product of digits is equal to 44. So the answer would be "-1".


Constraints:

• 1 <= n <= 1018

## Solutions

Solution 1: Prime Factorization + Greedy

We consider prime factorizing the number $n$. If there are prime factors greater than $9$ in $n$, then it is impossible to find a number that meets the conditions, because prime factors greater than $9$ cannot be obtained by multiplying numbers from $1$ to $9$. For example, $11$ cannot be obtained by multiplying numbers from $1$ to $9$. Therefore, we only need to consider whether there are prime factors greater than $9$ in $n$. If there are, return $-1$ directly.

Otherwise, if the prime factors include $7$ and $5$, then the number $n$ can first be decomposed into several $7$s and $5$s. Two $3$s can be combined into a $9$, three $2$s can be combined into an $8$, and a $2$ and a $3$ can be combined into a $6$. Therefore, we only need to decompose the number into numbers from $2$ to $9$. We can use a greedy method, preferentially decomposing into $9$, then decomposing into $8$, and so on.

The time complexity is $O(\log n)$, and the space complexity is $O(1)$.

• class Solution {
public String smallestNumber(long n) {
int[] cnt = new int[10];
for (int i = 9; i > 1; --i) {
while (n % i == 0) {
++cnt[i];
n /= i;
}
}
if (n > 1) {
return "-1";
}
StringBuilder sb = new StringBuilder();
for (int i = 2; i < 10; ++i) {
while (cnt[i] > 0) {
sb.append(i);
--cnt[i];
}
}
String ans = sb.toString();
return ans.isEmpty() ? "1" : ans;
}
}

• class Solution {
public:
string smallestNumber(long long n) {
int cnt[10]{};
for (int i = 9; i > 1; --i) {
while (n % i == 0) {
n /= i;
++cnt[i];
}
}
if (n > 1) {
return "-1";
}
string ans;
for (int i = 2; i < 10; ++i) {
ans += string(cnt[i], '0' + i);
}
return ans == "" ? "1" : ans;
}
};

• class Solution:
def smallestNumber(self, n: int) -> str:
cnt = [0] * 10
for i in range(9, 1, -1):
while n % i == 0:
n //= i
cnt[i] += 1
if n > 1:
return "-1"
ans = "".join(str(i) * cnt[i] for i in range(2, 10))
return ans if ans else "1"


• func smallestNumber(n int64) string {
cnt := [10]int{}
for i := 9; i > 1; i-- {
for n%int64(i) == 0 {
cnt[i]++
n /= int64(i)
}
}
if n != 1 {
return "-1"
}
sb := &strings.Builder{}
for i := 2; i < 10; i++ {
for j := 0; j < cnt[i]; j++ {
sb.WriteByte(byte(i) + '0')
}
}
ans := sb.String()
if len(ans) > 0 {
return ans
}
return "1"
}