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Feb 18, 2025
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feat: add solutions to lc problem: No.3460 #4073

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2 changes: 1 addition & 1 deletion solution/1200-1299/1289.Minimum Falling Path Sum II/README_EN.md
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Expand Up @@ -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],
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268 changes: 268 additions & 0 deletions solution/3400-3499/3460.Longest Common Prefix After at Most One Removal/README.md
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---
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 -->
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