feat: add solutions to lc problem: No.2847

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)
+
+[English Version](/solution/2800-2899/2847.Smallest%20Number%20With%20Given%20Digit%20Product/README_EN.md)
+
+## 题目描述
+
+<!-- 这里写题目描述 -->
+
+<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>
+
+<p>&nbsp;</p>
+<p><strong class="example">Example 1:</strong></p>
+
+<pre>
+<strong>Input:</strong> n = 105
+<strong>Output:</strong> &quot;357&quot;
+<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;.
+</pre>
+
+<p><strong class="example">Example 2:</strong></p>
+
+<pre>
+<strong>Input:</strong> n = 7
+<strong>Output:</strong> &quot;7&quot;
+<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.
+</pre>
+
+<p><strong class="example">Example 3:</strong></p>
+
+<pre>
+<strong>Input:</strong> n = 44
+<strong>Output:</strong> &quot;-1&quot;
+<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;.
+</pre>
+
+<p>&nbsp;</p>
+<p><strong>Constraints:</strong></p>
+
+<ul>
+	<li><code>1 &lt;= n &lt;= 10<sup>18</sup></code></li>
+</ul>
+
+## 解法
+
+<!-- 这里可写通用的实现逻辑 -->
+
+**方法一：质因数分解 + 贪心**
+
+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 $-1$。
+
+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。
+
+时间复杂度 $O(\log n)$，空间复杂度 $O(1)$。
+
+<!-- tabs:start -->
+
+### **Python3**
+
+<!-- 这里可写当前语言的特殊实现逻辑 -->
+
+```python
+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"
+```
+
+### **Java**
+
+<!-- 这里可写当前语言的特殊实现逻辑 -->
+
+```java
+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;
+    }
+}
+```
+
+### **C++**
+
+```cpp
+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;
+    }
+};
+```
+
+### **Go**
+
+```go
+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"
+}
+```
+
+### **...**
+
+```
+
+```
+
+<!-- tabs:end -->

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

@@ -0,0 +1,148 @@
+# [2847. Smallest Number With Given Digit Product](https://leetcode.com/problems/smallest-number-with-given-digit-product)
+
+[中文文档](/solution/2800-2899/2847.Smallest%20Number%20With%20Given%20Digit%20Product/README.md)
+
+## Description
+
+<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>
+
+<p>&nbsp;</p>
+<p><strong class="example">Example 1:</strong></p>
+
+<pre>
+<strong>Input:</strong> n = 105
+<strong>Output:</strong> &quot;357&quot;
+<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;.
+</pre>
+
+<p><strong class="example">Example 2:</strong></p>
+
+<pre>
+<strong>Input:</strong> n = 7
+<strong>Output:</strong> &quot;7&quot;
+<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.
+</pre>
+
+<p><strong class="example">Example 3:</strong></p>
+
+<pre>
+<strong>Input:</strong> n = 44
+<strong>Output:</strong> &quot;-1&quot;
+<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;.
+</pre>
+
+<p>&nbsp;</p>
+<p><strong>Constraints:</strong></p>
+
+<ul>
+	<li><code>1 &lt;= n &lt;= 10<sup>18</sup></code></li>
+</ul>
+
+## Solutions
+
+<!-- tabs:start -->
+
+### **Python3**
+
+```python
+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"
+```
+
+### **Java**
+
+```java
+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;
+    }
+}
+```
+
+### **C++**
+
+```cpp
+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;
+    }
+};
+```
+
+### **Go**
+
+```go
+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"
+}
+```
+
+### **...**
+
+```
+
+```
+
+<!-- tabs:end -->

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

@@ -0,0 +1,20 @@
+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;
+    }
+};

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

@@ -0,0 +1,23 @@
+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"
+}

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

@@ -0,0 +1,23 @@
+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;
+    }
+}

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

@@ -0,0 +1,11 @@
+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"