feat: add solutions to lc problem: No.3460 (doocs#4073)

No.3460.Longest Common Prefix After at Most One Removal

Author: yanglbme

solution/1200-1299/1289.Minimum Falling Path Sum II/README_EN.md

@@ -30,7 +30,7 @@ tags:
 <pre>
 <strong>Input:</strong> grid = [[1,2,3],[4,5,6],[7,8,9]]
 <strong>Output:</strong> 13
-<strong>Explanation:</strong>
+<strong>Explanation:</strong> 
 The possible falling paths are:
 [1,5,9], [1,5,7], [1,6,7], [1,6,8],
 [2,4,8], [2,4,9], [2,6,7], [2,6,8],

solution/3400-3499/3460.Longest Common Prefix After at Most One Removal/README.md

@@ -0,0 +1,268 @@
+---
+comments: true
+difficulty: 中等
+edit_url: https://github.com/doocs/leetcode/edit/main/solution/3400-3499/3460.Longest%20Common%20Prefix%20After%20at%20Most%20One%20Removal/README.md
+---
+
+<!-- problem:start -->
+
+# [3460. Longest Common Prefix After at Most One Removal 🔒](https://leetcode.cn/problems/longest-common-prefix-after-at-most-one-removal)
+
+[English Version](/solution/3400-3499/3460.Longest%20Common%20Prefix%20After%20at%20Most%20One%20Removal/README_EN.md)
+
+## 题目描述
+
+<!-- description:start -->
+
+<p>You are given two strings <code>s</code> and <code>t</code>.</p>
+
+<p>Return the <strong>length</strong> of the <strong>longest common <span data-keyword="string-prefix">prefix</span></strong> between <code>s</code> and <code>t</code> after removing <strong>at most</strong> one character from <code>s</code>.</p>
+
+<p><strong>Note:</strong> <code>s</code> can be left without any removal.</p>
+
+<p>&nbsp;</p>
+<p><strong class="example">Example 1:</strong></p>
+
+<div class="example-block">
+<p><strong>Input:</strong> <span class="example-io">s = &quot;madxa&quot;, t = &quot;madam&quot;</span></p>
+
+<p><strong>Output:</strong> <span class="example-io">4</span></p>
+
+<p><strong>Explanation:</strong></p>
+
+<p>Removing <code>s[3]</code> from <code>s</code> results in <code>&quot;mada&quot;</code>, which has a longest common prefix of length 4 with <code>t</code>.</p>
+</div>
+
+<p><strong class="example">Example 2:</strong></p>
+
+<div class="example-block">
+<p><strong>Input:</strong> <span class="example-io">s = &quot;leetcode&quot;, t = &quot;eetcode&quot;</span></p>
+
+<p><strong>Output:</strong> <span class="example-io">7</span></p>
+
+<p><strong>Explanation:</strong></p>
+
+<p>Removing <code>s[0]</code> from <code>s</code> results in <code>&quot;eetcode&quot;</code>, which matches <code>t</code>.</p>
+</div>
+
+<p><strong class="example">Example 3:</strong></p>
+
+<div class="example-block">
+<p><strong>Input:</strong> <span class="example-io">s = &quot;one&quot;, t = &quot;one&quot;</span></p>
+
+<p><strong>Output:</strong> <span class="example-io">3</span></p>
+
+<p><strong>Explanation:</strong></p>
+
+<p>No removal is needed.</p>
+</div>
+
+<p><strong class="example">Example 4:</strong></p>
+
+<div class="example-block">
+<p><strong>Input:</strong> <span class="example-io">s = &quot;a&quot;, t = &quot;b&quot;</span></p>
+
+<p><strong>Output:</strong> <span class="example-io">0</span></p>
+
+<p><strong>Explanation:</strong></p>
+
+<p><code>s</code> and <code>t</code> cannot have a common prefix.</p>
+</div>
+
+<p>&nbsp;</p>
+<p><strong>Constraints:</strong></p>
+
+<ul>
+	<li><code>1 &lt;= s.length &lt;= 10<sup>5</sup></code></li>
+	<li><code>1 &lt;= t.length &lt;= 10<sup>5</sup></code></li>
+	<li><code>s</code> and <code>t</code> contain only lowercase English letters.</li>
+</ul>
+
+<!-- description:end -->
+
+## 解法
+
+<!-- solution:start -->
+
+### 方法一：双指针
+
+我们记录字符串 $s$ 和 $t$ 的长度分别为 $n$ 和 $m$，然后用两个指针 $i$ 和 $j$ 分别指向字符串 $s$ 和 $t$ 的开头，用一个布尔变量 $\textit{rem}$ 记录是否已经删除过字符。
+
+接下来，我们开始遍历字符串 $s$ 和 $t$，如果 $s[i]$ 不等于 $t[j]$，我们就判断是否已经删除过字符，如果已经删除过字符，我们就退出循环，否则我们标记已经删除过字符，然后跳过 $s[i]$；否则，我们跳过 $s[i]$ 和 $t[j]$。继续遍历，直到 $i \geq n$ 或 $j \geq m$。
+
+最后返回 $j$ 即可。
+
+时间复杂度 $O(n+m)$，其中 $n$ 和 $m$ 分别是字符串 $s$ 和 $t$ 的长度。
+
+<!-- tabs:start -->
+
+#### Python3
+
+```python
+class Solution:
+    def longestCommonPrefix(self, s: str, t: str) -> int:
+        n, m = len(s), len(t)
+        i = j = 0
+        rem = False
+        while i < n and j < m:
+            if s[i] != t[j]:
+                if rem:
+                    break
+                rem = True
+            else:
+                j += 1
+            i += 1
+        return j
+```
+
+#### Java
+
+```java
+class Solution {
+    public int longestCommonPrefix(String s, String t) {
+        int n = s.length(), m = t.length();
+        int i = 0, j = 0;
+        boolean rem = false;
+        while (i < n && j < m) {
+            if (s.charAt(i) != t.charAt(j)) {
+                if (rem) {
+                    break;
+                }
+                rem = true;
+            } else {
+                ++j;
+            }
+            ++i;
+        }
+        return j;
+    }
+}
+```
+
+#### C++
+
+```cpp
+class Solution {
+public:
+    int longestCommonPrefix(string s, string t) {
+        int n = s.length(), m = t.length();
+        int i = 0, j = 0;
+        bool rem = false;
+        while (i < n && j < m) {
+            if (s[i] != t[j]) {
+                if (rem) {
+                    break;
+                }
+                rem = true;
+            } else {
+                ++j;
+            }
+            ++i;
+        }
+        return j;
+    }
+};
+```
+
+#### Go
+
+```go
+func longestCommonPrefix(s string, t string) int {
+	n, m := len(s), len(t)
+	i, j := 0, 0
+	rem := false
+	for i < n && j < m {
+		if s[i] != t[j] {
+			if rem {
+				break
+			}
+			rem = true
+		} else {
+			j++
+		}
+		i++
+	}
+	return j
+}
+```
+
+#### TypeScript
+
+```ts
+function longestCommonPrefix(s: string, t: string): number {
+    const [n, m] = [s.length, t.length];
+    let [i, j] = [0, 0];
+    let rem: boolean = false;
+    while (i < n && j < m) {
+        if (s[i] !== t[j]) {
+            if (rem) {
+                break;
+            }
+            rem = true;
+        } else {
+            ++j;
+        }
+        ++i;
+    }
+    return j;
+}
+```
+
+#### Rust
+
+```rust
+impl Solution {
+    pub fn longest_common_prefix(s: String, t: String) -> i32 {
+        let (n, m) = (s.len(), t.len());
+        let (mut i, mut j) = (0, 0);
+        let mut rem = false;
+
+        while i < n && j < m {
+            if s.as_bytes()[i] != t.as_bytes()[j] {
+                if rem {
+                    break;
+                }
+                rem = true;
+            } else {
+                j += 1;
+            }
+            i += 1;
+        }
+
+        j as i32
+    }
+}
+```
+
+#### JavaScript
+
+```js
+/**
+ * @param {string} s
+ * @param {string} t
+ * @return {number}
+ */
+var longestCommonPrefix = function (s, t) {
+    const [n, m] = [s.length, t.length];
+    let [i, j] = [0, 0];
+    let rem = false;
+    while (i < n && j < m) {
+        if (s[i] !== t[j]) {
+            if (rem) {
+                break;
+            }
+            rem = true;
+        } else {
+            ++j;
+        }
+        ++i;
+    }
+    return j;
+};
+```
+
+<!-- tabs:end -->
+
+<!-- solution:end -->
+
+<!-- problem:end -->