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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 <= 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"
}

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