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feat: add solutions to lc problems: No.2605~2608
* No.2605.Form Smallest Number From Two Digit Arrays * No.2606.Find the Substring With Maximum Cost * No.2607.Make K-Subarray Sums Equal * No.2608.Shortest Cycle in a Graph
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‎solution/0700-0799/0789.Escape The Ghosts/README.md

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<p>如果你可以在任何阻碍者抓住你 <strong>之前</strong> 到达目的地（阻碍者可以采取任意行动方式），则被视为逃脱成功。如果你和阻碍者 <strong>同时</strong> 到达了一个位置（包括目的地）&nbsp;<strong>都不算</strong>&nbsp;是逃脱成功。</p>
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<p>只有在你有可能成功逃脱时，输出 <code>true</code> ；否则，输出 <code>false</code> 。</p>
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<p>如果不管阻碍者怎么移动都可以成功逃脱时，输出 <code>true</code> ；否则，输出 <code>false</code> 。</p>
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&nbsp;
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<p><strong>示例 1：</strong></p>

‎solution/1300-1399/1312.Minimum Insertion Steps to Make a String Palindrome/README_EN.md

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if s[i] == s[j]:
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return dfs(i + 1, j - 1)
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return 1 + min(dfs(i + 1, j), dfs(i, j - 1))
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return dfs(0, len(s) - 1)
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```
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‎solution/2600-2699/2605.Form Smallest Number From Two Digit Arrays/README.md

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# [2605. 从两个数字数组里生成最小数字](https://leetcode.cn/problems/form-smallest-number-from-two-digit-arrays)
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[English Version](/solution/2600-2699/2605.Form%20Smallest%20Number%20From%20Two%20Digit%20Arrays/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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给你两个只包含 1 到 9 之间数字的数组&nbsp;<code>nums1</code> 和&nbsp;<code>nums2</code>&nbsp;，每个数组中的元素 <strong>互不相同</strong>&nbsp;，请你返回 <strong>最小</strong> 的数字，两个数组都 <strong>至少</strong> 包含这个数字的某个数位。
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<p>&nbsp;</p>
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<p><strong>示例 1：</strong></p>
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<pre><b>输入：</b>nums1 = [4,1,3], nums2 = [5,7]
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<b>输出：</b>15
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<b>解释：</b>数字 15 的数位 1 在 nums1 中出现，数位 5 在 nums2 中出现。15 是我们能得到的最小数字。
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</pre>
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<p><strong>示例 2：</strong></p>
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<pre><b>输入：</b>nums1 = [3,5,2,6], nums2 = [3,1,7]
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<b>输出：</b>3
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<b>解释：</b>数字 3 的数位 3 在两个数组中都出现了。
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</pre>
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<p>&nbsp;</p>
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<p><strong>提示：</strong></p>
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<ul>
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<li><code>1 &lt;= nums1.length, nums2.length &lt;= 9</code></li>
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<li><code>1 &lt;= nums1[i], nums2[i] &lt;= 9</code></li>
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<li>每个数组中，元素 <strong>互不相同</strong>&nbsp;。</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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<!-- 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 minNumber(self, nums1: List[int], nums2: List[int]) -> int:
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ans = 100
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for a in nums1:
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for b in nums2:
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if a == b:
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ans = min(ans, a)
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else:
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ans = min(ans, 10 * a + b, 10 * b + a)
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return ans
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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 int minNumber(int[] nums1, int[] nums2) {
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int ans = 100;
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for (int a : nums1) {
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for (int b : nums2) {
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if (a == b) {
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ans = Math.min(ans, a);
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} else {
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ans = Math.min(ans, Math.min(a * 10 + b, b * 10 + a));
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}
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}
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}
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return 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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int minNumber(vector<int>& nums1, vector<int>& nums2) {
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int ans = 100;
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for (int a : nums1) {
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for (int b : nums2) {
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if (a == b) {
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ans = min(ans, a);
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} else {
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ans = min({ans, a * 10 + b, b * 10 + a});
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}
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}
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}
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return 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 minNumber(nums1 []int, nums2 []int) int {
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ans := 100
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for _, a := range nums1 {
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for _, b := range nums2 {
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if a == b {
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ans = min(ans, a)
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} else {
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ans = min(ans, min(a*10+b, b*10+a))
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}
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}
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}
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return ans
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}
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func min(a, b int) int {
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if a < b {
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return a
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}
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return b
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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/2600-2699/2605.Form Smallest Number From Two Digit Arrays/README_EN.md

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# [2605. Form Smallest Number From Two Digit Arrays](https://leetcode.com/problems/form-smallest-number-from-two-digit-arrays)
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[中文文档](/solution/2600-2699/2605.Form%20Smallest%20Number%20From%20Two%20Digit%20Arrays/README.md)
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## Description
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Given two arrays of <strong>unique</strong> digits <code>nums1</code> and <code>nums2</code>, return <em>the <strong>smallest</strong> number that contains <strong>at least</strong> one digit from each array</em>.
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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> nums1 = [4,1,3], nums2 = [5,7]
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<strong>Output:</strong> 15
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<strong>Explanation:</strong> The number 15 contains the digit 1 from nums1 and the digit 5 from nums2. It can be proven that 15 is the smallest number we can have.
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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> nums1 = [3,5,2,6], nums2 = [3,1,7]
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<strong>Output:</strong> 3
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<strong>Explanation:</strong> The number 3 contains the digit 3 which exists in both arrays.
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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;= nums1.length, nums2.length &lt;= 9</code></li>
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<li><code>1 &lt;= nums1[i], nums2[i] &lt;= 9</code></li>
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<li>All digits in each array are <strong>unique</strong>.</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 minNumber(self, nums1: List[int], nums2: List[int]) -> int:
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ans = 100
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for a in nums1:
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for b in nums2:
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if a == b:
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ans = min(ans, a)
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else:
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ans = min(ans, 10 * a + b, 10 * b + a)
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return ans
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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 int minNumber(int[] nums1, int[] nums2) {
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int ans = 100;
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for (int a : nums1) {
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for (int b : nums2) {
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if (a == b) {
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ans = Math.min(ans, a);
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} else {
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ans = Math.min(ans, Math.min(a * 10 + b, b * 10 + a));
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}
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}
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}
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return 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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int minNumber(vector<int>& nums1, vector<int>& nums2) {
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int ans = 100;
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for (int a : nums1) {
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for (int b : nums2) {
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if (a == b) {
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ans = min(ans, a);
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} else {
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ans = min({ans, a * 10 + b, b * 10 + a});
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}
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}
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}
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return 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 minNumber(nums1 []int, nums2 []int) int {
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ans := 100
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for _, a := range nums1 {
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for _, b := range nums2 {
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if a == b {
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ans = min(ans, a)
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} else {
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ans = min(ans, min(a*10+b, b*10+a))
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}
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}
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}
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return ans
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}
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func min(a, b int) int {
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if a < b {
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return a
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}
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return b
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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/2600-2699/2605.Form Smallest Number From Two Digit Arrays/Solution.cpp

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class Solution {
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public:
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int minNumber(vector<int>& nums1, vector<int>& nums2) {
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int ans = 100;
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for (int a : nums1) {
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for (int b : nums2) {
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if (a == b) {
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ans = min(ans, a);
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} else {
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ans = min({ans, a * 10 + b, b * 10 + a});
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}
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}
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}
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return ans;
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}
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};

‎solution/2600-2699/2605.Form Smallest Number From Two Digit Arrays/Solution.go

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func minNumber(nums1 []int, nums2 []int) int {
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ans := 100
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for _, a := range nums1 {
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for _, b := range nums2 {
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if a == b {
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ans = min(ans, a)
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} else {
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ans = min(ans, min(a*10+b, b*10+a))
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}
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}
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}
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return ans
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}
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func min(a, b int) int {
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if a < b {
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return a
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}
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return b
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}

‎solution/2600-2699/2605.Form Smallest Number From Two Digit Arrays/Solution.java

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class Solution {
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public int minNumber(int[] nums1, int[] nums2) {
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int ans = 100;
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for (int a : nums1) {
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for (int b : nums2) {
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if (a == b) {
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ans = Math.min(ans, a);
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} else {
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ans = Math.min(ans, Math.min(a * 10 + b, b * 10 + a));
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}
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}
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}
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return ans;
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}
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}

‎solution/2600-2699/2605.Form Smallest Number From Two Digit Arrays/Solution.py

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class Solution:
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def minNumber(self, nums1: List[int], nums2: List[int]) -> int:
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ans = 100
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for a in nums1:
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for b in nums2:
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if a == b:
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ans = min(ans, a)
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else:
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ans = min(ans, 10 * a + b, 10 * b + a)
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return ans

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