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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 <= 10^{18}
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" }