feat: update lc problems

solution/0800-0899/0813.Largest Sum of Averages/README.md

@@ -31,9 +31,9 @@ tags:
 <pre>
 <strong>输入:</strong> nums = [9,1,2,3,9], k = 3
 <strong>输出:</strong> 20.00000
-<strong>解释:</strong>
-nums 的最优分组是[9], [1, 2, 3], [9]. 得到的分数是 9 + (1 + 2 + 3) / 3 + 9 = 20.
-我们也可以把 nums 分成[9, 1], [2], [3, 9].
+<strong>解释:</strong> 
+nums 的最优分组是[9], [1, 2, 3], [9]. 得到的分数是 9 + (1 + 2 + 3) / 3 + 9 = 20. 
+我们也可以把 nums 分成[9, 1], [2], [3, 9]. 
 这样的分组得到的分数为 5 + 2 + 6 = 13, 但不是最大值.
 </pre>
 

solution/0800-0899/0813.Largest Sum of Averages/README_EN.md

@@ -30,7 +30,7 @@ tags:
 <pre>
 <strong>Input:</strong> nums = [9,1,2,3,9], k = 3
 <strong>Output:</strong> 20.00000
-<strong>Explanation:</strong>
+<strong>Explanation:</strong> 
 The best choice is to partition nums into [9], [1, 2, 3], [9]. The answer is 9 + (1 + 2 + 3) / 3 + 9 = 20.
 We could have also partitioned nums into [9, 1], [2], [3, 9], for example.
 That partition would lead to a score of 5 + 2 + 6 = 13, which is worse.

solution/0800-0899/0814.Binary Tree Pruning/README_EN.md

@@ -28,7 +28,7 @@ tags:
 <pre>
 <strong>Input:</strong> root = [1,null,0,0,1]
 <strong>Output:</strong> [1,null,0,null,1]
-<strong>Explanation:</strong>
+<strong>Explanation:</strong> 
 Only the red nodes satisfy the property &quot;every subtree not containing a 1&quot;.
 The diagram on the right represents the answer.
 </pre>

solution/0800-0899/0849.Maximize Distance to Closest Person/README.md

@@ -34,7 +34,7 @@ tags:
 <strong>解释：
 </strong>如果亚历克斯坐在第二个空位（seats[2]）上，他到离他最近的人的距离为 2 。
 如果亚历克斯坐在其它任何一个空位上，他到离他最近的人的距离为 1 。
-因此，他到离他最近的人的最大距离是 2 。
+因此，他到离他最近的人的最大距离是 2 。 
 </pre>
 
 <p><strong>示例 2：</strong></p>

solution/1300-1399/1310.XOR Queries of a Subarray/README.md

@@ -32,17 +32,17 @@ tags:
 
 <pre>
 <strong>输入：</strong>arr = [1,3,4,8], queries = [[0,1],[1,2],[0,3],[3,3]]
-<strong>输出：</strong>[2,7,14,8]
+<strong>输出：</strong>[2,7,14,8] 
 <strong>解释：</strong>
 数组中元素的二进制表示形式是：
-1 = 0001
-3 = 0011
-4 = 0100
-8 = 1000
+1 = 0001 
+3 = 0011 
+4 = 0100 
+8 = 1000 
 查询的 XOR 值为：
-[0,1] = 1 xor 3 = 2
-[1,2] = 3 xor 4 = 7
-[0,3] = 1 xor 3 xor 4 xor 8 = 14
+[0,1] = 1 xor 3 = 2 
+[1,2] = 3 xor 4 = 7 
+[0,3] = 1 xor 3 xor 4 xor 8 = 14 
 [3,3] = 8
 </pre>
 

solution/1300-1399/1310.XOR Queries of a Subarray/README_EN.md

@@ -31,17 +31,17 @@ tags:
 
 <pre>
 <strong>Input:</strong> arr = [1,3,4,8], queries = [[0,1],[1,2],[0,3],[3,3]]
-<strong>Output:</strong> [2,7,14,8]
-<strong>Explanation:</strong>
+<strong>Output:</strong> [2,7,14,8] 
+<strong>Explanation:</strong> 
 The binary representation of the elements in the array are:
-1 = 0001
-3 = 0011
-4 = 0100
-8 = 1000
+1 = 0001 
+3 = 0011 
+4 = 0100 
+8 = 1000 
 The XOR values for queries are:
-[0,1] = 1 xor 3 = 2
-[1,2] = 3 xor 4 = 7
-[0,3] = 1 xor 3 xor 4 xor 8 = 14
+[0,1] = 1 xor 3 = 2 
+[1,2] = 3 xor 4 = 7 
+[0,3] = 1 xor 3 xor 4 xor 8 = 14 
 [3,3] = 8
 </pre>
 

solution/1300-1399/1311.Get Watched Videos by Your Friends/README.md

@@ -37,7 +37,7 @@ tags:
 <p><strong><img alt="" src="https://fastly.jsdelivr.net/gh/doocs/leetcode@main/solution/1300-1399/1311.Get%20Watched%20Videos%20by%20Your%20Friends/images/leetcode_friends_1.png" style="height: 179px; width: 129px;"></strong></p>
 
 <pre><strong>输入：</strong>watchedVideos = [[&quot;A&quot;,&quot;B&quot;],[&quot;C&quot;],[&quot;B&quot;,&quot;C&quot;],[&quot;D&quot;]], friends = [[1,2],[0,3],[0,3],[1,2]], id = 0, level = 1
-<strong>输出：</strong>[&quot;B&quot;,&quot;C&quot;]
+<strong>输出：</strong>[&quot;B&quot;,&quot;C&quot;] 
 <strong>解释：</strong>
 你的 id 为 0（绿色），你的朋友包括（黄色）：
 id 为 1 -&gt; watchedVideos = [&quot;C&quot;]&nbsp;

solution/1300-1399/1311.Get Watched Videos by Your Friends/README_EN.md

@@ -33,8 +33,8 @@ tags:
 
 <pre>
 <strong>Input:</strong> watchedVideos = [[&quot;A&quot;,&quot;B&quot;],[&quot;C&quot;],[&quot;B&quot;,&quot;C&quot;],[&quot;D&quot;]], friends = [[1,2],[0,3],[0,3],[1,2]], id = 0, level = 1
-<strong>Output:</strong> [&quot;B&quot;,&quot;C&quot;]
-<strong>Explanation:</strong>
+<strong>Output:</strong> [&quot;B&quot;,&quot;C&quot;] 
+<strong>Explanation:</strong> 
 You have id = 0 (green color in the figure) and your friends are (yellow color in the figure):
 Person with id = 1 -&gt; watchedVideos = [&quot;C&quot;]&nbsp;
 Person with id = 2 -&gt; watchedVideos = [&quot;B&quot;,&quot;C&quot;]&nbsp;
@@ -50,7 +50,7 @@ C -&gt; 2
 <pre>
 <strong>Input:</strong> watchedVideos = [[&quot;A&quot;,&quot;B&quot;],[&quot;C&quot;],[&quot;B&quot;,&quot;C&quot;],[&quot;D&quot;]], friends = [[1,2],[0,3],[0,3],[1,2]], id = 0, level = 2
 <strong>Output:</strong> [&quot;D&quot;]
-<strong>Explanation:</strong>
+<strong>Explanation:</strong> 
 You have id = 0 (green color in the figure) and the only friend of your friends is the person with id = 3 (yellow color in the figure).
 </pre>
 

solution/2000-2099/2029.Stone Game IX/README.md

@@ -42,7 +42,7 @@ tags:
 <strong>输出：</strong>true
 <strong>解释：</strong>游戏进行如下：
 - 回合 1：Alice 可以移除任意一个石子。
-- 回合 2：Bob 移除剩下的石子。
+- 回合 2：Bob 移除剩下的石子。 
 已移除的石子的值总和为 1 + 2 = 3 且可以被 3 整除。因此，Bob 输，Alice 获胜。
 </pre>
 
@@ -51,7 +51,7 @@ tags:
 <pre>
 <strong>输入：</strong>stones = [2]
 <strong>输出：</strong>false
-<strong>解释：</strong>Alice 会移除唯一一个石子，已移除石子的值总和为 2 。
+<strong>解释：</strong>Alice 会移除唯一一个石子，已移除石子的值总和为 2 。 
 由于所有石子都已移除，且值总和无法被 3 整除，Bob 获胜。
 </pre>
 

solution/2000-2099/2029.Stone Game IX/README_EN.md

@@ -36,7 +36,7 @@ tags:
 <strong>Output:</strong> true
 <strong>Explanation:</strong>&nbsp;The game will be played as follows:
 - Turn 1: Alice can remove either stone.
-- Turn 2: Bob removes the remaining stone.
+- Turn 2: Bob removes the remaining stone. 
 The sum of the removed stones is 1 + 2 = 3 and is divisible by 3. Therefore, Bob loses and Alice wins the game.
 </pre>
 
@@ -45,7 +45,7 @@ The sum of the removed stones is 1 + 2 = 3 and is divisible by 3. Therefore, Bob
 <pre>
 <strong>Input:</strong> stones = [2]
 <strong>Output:</strong> false
-<strong>Explanation:</strong>&nbsp;Alice will remove the only stone, and the sum of the values on the removed stones is 2.
+<strong>Explanation:</strong>&nbsp;Alice will remove the only stone, and the sum of the values on the removed stones is 2. 
 Since all the stones are removed and the sum of values is not divisible by 3, Bob wins the game.
 </pre>
 

solution/2800-2899/2800.Shortest String That Contains Three Strings/README.md

@@ -288,4 +288,230 @@ impl Solution {
 
 <!-- solution:end -->
 
+<!-- solution:start -->
+
+### 方法二：枚举 + KMP
+
+我们可以使用 KMP 算法来优化字符串的合并过程。
+
+时间复杂度 $O(n)$，空间复杂度 $O(n)$。其中 $n$ 是三个字符串的长度之和。
+
+<!-- tabs:start -->
+
+#### Python3
+
+```python
+class Solution:
+    def minimumString(self, a: str, b: str, c: str) -> str:
+        def f(s: str, t: str) -> str:
+            if s in t:
+                return t
+            if t in s:
+                return s
+            p = t + "#" + s + "$"
+            n = len(p)
+            next = [0] * n
+            next[0] = -1
+            i, j = 2, 0
+            while i < n:
+                if p[i - 1] == p[j]:
+                    j += 1
+                    next[i] = j
+                    i += 1
+                elif j:
+                    j = next[j]
+                else:
+                    next[i] = 0
+                    i += 1
+            return s + t[next[-1] :]
+
+        ans = ""
+        for a, b, c in permutations((a, b, c)):
+            s = f(f(a, b), c)
+            if ans == "" or len(s) < len(ans) or (len(s) == len(ans) and s < ans):
+                ans = s
+        return ans
+```
+
+#### Java
+
+```java
+class Solution {
+    public String minimumString(String a, String b, String c) {
+        String[] s = {a, b, c};
+        int[][] perm = {{0, 1, 2}, {0, 2, 1}, {1, 0, 2}, {1, 2, 0}, {2, 1, 0}, {2, 0, 1}};
+        String ans = "";
+        for (var p : perm) {
+            int i = p[0], j = p[1], k = p[2];
+            String t = f(f(s[i], s[j]), s[k]);
+            if ("".equals(ans) || t.length() < ans.length()
+                || (t.length() == ans.length() && t.compareTo(ans) < 0)) {
+                ans = t;
+            }
+        }
+        return ans;
+    }
+
+    private String f(String s, String t) {
+        if (s.contains(t)) {
+            return s;
+        }
+        if (t.contains(s)) {
+            return t;
+        }
+        char[] p = (t + "#" + s + "$").toCharArray();
+        int n = p.length;
+        int[] next = new int[n];
+        next[0] = -1;
+        for (int i = 2, j = 0; i < n;) {
+            if (p[i - 1] == p[j]) {
+                next[i++] = ++j;
+            } else if (j > 0) {
+                j = next[j];
+            } else {
+                next[i++] = 0;
+            }
+        }
+        return s + t.substring(next[n - 1]);
+    }
+}
+```
+
+#### C++
+
+```cpp
+class Solution {
+public:
+    string minimumString(string a, string b, string c) {
+        vector<string> s = {a, b, c};
+        vector<vector<int>> perm = {{0, 1, 2}, {0, 2, 1}, {1, 0, 2}, {1, 2, 0}, {2, 1, 0}, {2, 0, 1}};
+        string ans = "";
+        for (auto& p : perm) {
+            int i = p[0], j = p[1], k = p[2];
+            string t = f(f(s[i], s[j]), s[k]);
+            if (ans == "" || t.size() < ans.size() || (t.size() == ans.size() && t < ans)) {
+                ans = t;
+            }
+        }
+        return ans;
+    }
+
+    string f(string s, string t) {
+        if (s.find(t) != string::npos) {
+            return s;
+        }
+        if (t.find(s) != string::npos) {
+            return t;
+        }
+        string p = t + "#" + s + "$";
+        int n = p.size();
+        int next[n];
+        next[0] = -1;
+        next[1] = 0;
+        for (int i = 2, j = 0; i < n;) {
+            if (p[i - 1] == p[j]) {
+                next[i++] = ++j;
+            } else if (j > 0) {
+                j = next[j];
+            } else {
+                next[i++] = 0;
+            }
+        }
+        return s + t.substr(next[n - 1]);
+    };
+};
+```
+
+#### Go
+
+```go
+func minimumString(a string, b string, c string) string {
+	f := func(s, t string) string {
+		if strings.Contains(s, t) {
+			return s
+		}
+		if strings.Contains(t, s) {
+			return t
+		}
+		p := t + "#" + s + "$"
+		n := len(p)
+		next := make([]int, n)
+		next[0] = -1
+		for i, j := 2, 0; i < n; {
+			if p[i-1] == p[j] {
+				j++
+				next[i] = j
+				i++
+			} else if j > 0 {
+				j = next[j]
+			} else {
+				next[i] = 0
+				i++
+			}
+		}
+		return s + t[next[n-1]:]
+	}
+	s := [3]string{a, b, c}
+	ans := ""
+	for _, p := range [][]int{{0, 1, 2}, {0, 2, 1}, {1, 0, 2}, {1, 2, 0}, {2, 0, 1}, {2, 1, 0}} {
+		i, j, k := p[0], p[1], p[2]
+		t := f(f(s[i], s[j]), s[k])
+		if ans == "" || len(t) < len(ans) || (len(t) == len(ans) && t < ans) {
+			ans = t
+		}
+	}
+	return ans
+}
+```
+
+#### TypeScript
+
+```ts
+function minimumString(a: string, b: string, c: string): string {
+    const f = (s: string, t: string): string => {
+        if (s.includes(t)) {
+            return s;
+        }
+        if (t.includes(s)) {
+            return t;
+        }
+        const p = t + '#' + s + '$';
+        const n = p.length;
+        const next: number[] = Array(n).fill(0);
+        next[0] = -1;
+        for (let i = 2, j = 0; i < n; ) {
+            if (p[i - 1] === p[j]) {
+                next[i++] = ++j;
+            } else if (j) {
+                j = next[j];
+            } else {
+                next[i++] = 0;
+            }
+        }
+        return s + t.slice(next[n - 1]);
+    };
+    const s: string[] = [a, b, c];
+    const perm: number[][] = [
+        [0, 1, 2],
+        [0, 2, 1],
+        [1, 0, 2],
+        [1, 2, 0],
+        [2, 0, 1],
+        [2, 1, 0],
+    ];
+    let ans = '';
+    for (const [i, j, k] of perm) {
+        const t = f(f(s[i], s[j]), s[k]);
+        if (ans === '' || t.length < ans.length || (t.length === ans.length && t < ans)) {
+            ans = t;
+        }
+    }
+    return ans;
+}
+```
+
+<!-- tabs:end -->
+
+<!-- solution:end -->
+
 <!-- problem:end -->