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feat: add solutions to lc problem: No.2847 (#1583)
No.2847.Smallest Number With Given Digit Product
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‎solution/2800-2899/2847.Smallest Number With Given Digit Product/README.md

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# [2847. Smallest Number With Given Digit Product](https://leetcode.cn/problems/smallest-number-with-given-digit-product)
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[English Version](/solution/2800-2899/2847.Smallest%20Number%20With%20Given%20Digit%20Product/README_EN.md)
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## 题目描述
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<!-- 这里写题目描述 -->
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<p>Given a <strong>positive</strong> integer <code>n</code>, return <em>a string representing the <strong>smallest positive</strong> integer such that the product of its digits is equal to</em> <code>n</code><em>, or </em><code>&quot;-1&quot;</code><em> if no such number exists</em>.</p>
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<p>&nbsp;</p>
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<p><strong class="example">Example 1:</strong></p>
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<pre>
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<strong>Input:</strong> n = 105
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<strong>Output:</strong> &quot;357&quot;
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<strong>Explanation:</strong> 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 &quot;105&quot;.
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</pre>
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<p><strong class="example">Example 2:</strong></p>
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<pre>
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<strong>Input:</strong> n = 7
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<strong>Output:</strong> &quot;7&quot;
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<strong>Explanation:</strong> 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 &quot;7&quot; would be the answer.
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</pre>
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<p><strong class="example">Example 3:</strong></p>
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<pre>
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<strong>Input:</strong> n = 44
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<strong>Output:</strong> &quot;-1&quot;
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<strong>Explanation:</strong> It can be shown that there is no number such that its product of digits is equal to 44. So the answer would be &quot;-1&quot;.
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</pre>
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<p>&nbsp;</p>
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<p><strong>Constraints:</strong></p>
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<ul>
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<li><code>1 &lt;= n &lt;= 10<sup>18</sup></code></li>
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</ul>
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## 解法
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<!-- 这里可写通用的实现逻辑 -->
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**方法一：质因数分解 + 贪心**
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我们考虑对数字 $n$ 进行质因数分解，如果 $n$ 的质因数中存在大于 $9$ 的质数，那么一定无法找到符合条件的数字，因为大于 $9$ 的质数无法通过 $1$ 到 $9$ 的数字相乘得到，例如 $11$ 无法通过 $1$ 到 $9$ 的数字相乘得到，因此我们只需要考虑 $n$ 的质因数中是否存在大于 $9$ 的质数即可，如果存在，直接返回 $-1$。
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否则，如果质因数中包含 $7$ 和 $5$，那么数字 $n$ 首先可以拆分为若干个 $7$ 和 $5$，两个数字 $3$ 可以合成一个数字 $9$，三个数字 $2$ 可以合成一个数字 $8$，数字 $2$ 和数字 $3$ 可以合成一个数字 $6$，因此我们只需要将数字拆分为 $[2,..9]$ 的数字即可，我们可以使用贪心的方法，优先拆分出数字 $9$，然后拆分出数字 $8$，依次类推。
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时间复杂度 $O(\log n)$，空间复杂度 $O(1)$。
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<!-- tabs:start -->
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### **Python3**
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<!-- 这里可写当前语言的特殊实现逻辑 -->
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```python
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class Solution:
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def smallestNumber(self, n: int) -> str:
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cnt = [0] * 10
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for i in range(9, 1, -1):
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while n % i == 0:
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n //= i
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cnt[i] += 1
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if n > 1:
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return "-1"
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ans = "".join(str(i) * cnt[i] for i in range(2, 10))
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return ans if ans else "1"
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```
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### **Java**
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<!-- 这里可写当前语言的特殊实现逻辑 -->
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```java
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class Solution {
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public String smallestNumber(long n) {
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int[] cnt = new int[10];
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for (int i = 9; i > 1; --i) {
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while (n % i == 0) {
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++cnt[i];
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n /= i;
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}
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}
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if (n > 1) {
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return "-1";
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}
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StringBuilder sb = new StringBuilder();
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for (int i = 2; i < 10; ++i) {
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while (cnt[i] > 0) {
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sb.append(i);
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--cnt[i];
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}
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}
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String ans = sb.toString();
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return ans.isEmpty() ? "1" : ans;
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}
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}
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```
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### **C++**
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```cpp
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class Solution {
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public:
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string smallestNumber(long long n) {
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int cnt[10]{};
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for (int i = 9; i > 1; --i) {
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while (n % i == 0) {
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n /= i;
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++cnt[i];
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}
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}
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if (n > 1) {
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return "-1";
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}
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string ans;
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for (int i = 2; i < 10; ++i) {
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ans += string(cnt[i], '0' + i);
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}
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return ans == "" ? "1" : ans;
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}
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};
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```
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### **Go**
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```go
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func smallestNumber(n int64) string {
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cnt := [10]int{}
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for i := 9; i > 1; i-- {
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for n%int64(i) == 0 {
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cnt[i]++
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n /= int64(i)
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}
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}
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if n != 1 {
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return "-1"
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}
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sb := &strings.Builder{}
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for i := 2; i < 10; i++ {
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for j := 0; j < cnt[i]; j++ {
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sb.WriteByte(byte(i) + '0')
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}
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}
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ans := sb.String()
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if len(ans) > 0 {
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return ans
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}
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return "1"
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}
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```
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### **...**
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```
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```
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<!-- tabs:end -->

‎solution/2800-2899/2847.Smallest Number With Given Digit Product/README_EN.md

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# [2847. Smallest Number With Given Digit Product](https://leetcode.com/problems/smallest-number-with-given-digit-product)
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[中文文档](/solution/2800-2899/2847.Smallest%20Number%20With%20Given%20Digit%20Product/README.md)
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## Description
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<p>Given a <strong>positive</strong> integer <code>n</code>, return <em>a string representing the <strong>smallest positive</strong> integer such that the product of its digits is equal to</em> <code>n</code><em>, or </em><code>&quot;-1&quot;</code><em> if no such number exists</em>.</p>
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<p>&nbsp;</p>
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<p><strong class="example">Example 1:</strong></p>
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<pre>
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<strong>Input:</strong> n = 105
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<strong>Output:</strong> &quot;357&quot;
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<strong>Explanation:</strong> 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 &quot;105&quot;.
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</pre>
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<p><strong class="example">Example 2:</strong></p>
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<pre>
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<strong>Input:</strong> n = 7
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<strong>Output:</strong> &quot;7&quot;
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<strong>Explanation:</strong> 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 &quot;7&quot; would be the answer.
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</pre>
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<p><strong class="example">Example 3:</strong></p>
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<pre>
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<strong>Input:</strong> n = 44
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<strong>Output:</strong> &quot;-1&quot;
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<strong>Explanation:</strong> It can be shown that there is no number such that its product of digits is equal to 44. So the answer would be &quot;-1&quot;.
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</pre>
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<p>&nbsp;</p>
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<p><strong>Constraints:</strong></p>
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<ul>
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<li><code>1 &lt;= n &lt;= 10<sup>18</sup></code></li>
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</ul>
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## Solutions
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<!-- tabs:start -->
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### **Python3**
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```python
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class Solution:
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def smallestNumber(self, n: int) -> str:
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cnt = [0] * 10
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for i in range(9, 1, -1):
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while n % i == 0:
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n //= i
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cnt[i] += 1
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if n > 1:
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return "-1"
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ans = "".join(str(i) * cnt[i] for i in range(2, 10))
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return ans if ans else "1"
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```
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### **Java**
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```java
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class Solution {
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public String smallestNumber(long n) {
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int[] cnt = new int[10];
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for (int i = 9; i > 1; --i) {
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while (n % i == 0) {
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++cnt[i];
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n /= i;
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}
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}
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if (n > 1) {
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return "-1";
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}
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StringBuilder sb = new StringBuilder();
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for (int i = 2; i < 10; ++i) {
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while (cnt[i] > 0) {
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sb.append(i);
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--cnt[i];
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}
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}
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String ans = sb.toString();
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return ans.isEmpty() ? "1" : ans;
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}
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}
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```
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### **C++**
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```cpp
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class Solution {
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public:
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string smallestNumber(long long n) {
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int cnt[10]{};
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for (int i = 9; i > 1; --i) {
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while (n % i == 0) {
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n /= i;
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++cnt[i];
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}
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}
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if (n > 1) {
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return "-1";
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}
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string ans;
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for (int i = 2; i < 10; ++i) {
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ans += string(cnt[i], '0' + i);
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}
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return ans == "" ? "1" : ans;
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}
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};
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```
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### **Go**
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```go
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func smallestNumber(n int64) string {
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cnt := [10]int{}
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for i := 9; i > 1; i-- {
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for n%int64(i) == 0 {
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cnt[i]++
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n /= int64(i)
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}
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}
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if n != 1 {
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return "-1"
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}
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sb := &strings.Builder{}
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for i := 2; i < 10; i++ {
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for j := 0; j < cnt[i]; j++ {
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sb.WriteByte(byte(i) + '0')
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}
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}
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ans := sb.String()
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if len(ans) > 0 {
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return ans
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}
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return "1"
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}
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```
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### **...**
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```
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```
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<!-- tabs:end -->

‎solution/2800-2899/2847.Smallest Number With Given Digit Product/Solution.cpp

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

‎solution/2800-2899/2847.Smallest Number With Given Digit Product/Solution.go

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

‎solution/2800-2899/2847.Smallest Number With Given Digit Product/Solution.java

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

‎solution/2800-2899/2847.Smallest Number With Given Digit Product/Solution.py

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

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