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Solution/1143. Longest Common Subsequence/readme.md

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+---
+comments: true
+difficulty: Medium
+edit_url: https://github.com/doocs/leetcode/edit/main/solution/1100-1199/1143.Longest%20Common%20Subsequence/README_EN.md
+tags:
+    - String
+    - Dynamic Programming
+---
+
+<!-- problem:start -->
+
+# [1143. Longest Common Subsequence](https://leetcode.com/problems/longest-common-subsequence)
+
+[中文文档](/solution/1100-1199/1143.Longest%20Common%20Subsequence/README.md)
+
+## Description
+
+<!-- description:start -->
+
+<p>Given two strings <code>text1</code> and <code>text2</code>, return <em>the length of their longest <strong>common subsequence</strong>. </em>If there is no <strong>common subsequence</strong>, return <code>0</code>.</p>
+
+<p>A <strong>subsequence</strong> of a string is a new string generated from the original string with some characters (can be none) deleted without changing the relative order of the remaining characters.</p>
+
+<ul>
+	<li>For example, <code>&quot;ace&quot;</code> is a subsequence of <code>&quot;abcde&quot;</code>.</li>
+</ul>
+
+<p>A <strong>common subsequence</strong> of two strings is a subsequence that is common to both strings.</p>
+
+<p>&nbsp;</p>
+<p><strong class="example">Example 1:</strong></p>
+
+<pre>
+<strong>Input:</strong> text1 = &quot;abcde&quot;, text2 = &quot;ace&quot; 
+<strong>Output:</strong> 3  
+<strong>Explanation:</strong> The longest common subsequence is &quot;ace&quot; and its length is 3.
+</pre>
+
+<p><strong class="example">Example 2:</strong></p>
+
+<pre>
+<strong>Input:</strong> text1 = &quot;abc&quot;, text2 = &quot;abc&quot;
+<strong>Output:</strong> 3
+<strong>Explanation:</strong> The longest common subsequence is &quot;abc&quot; and its length is 3.
+</pre>
+
+<p><strong class="example">Example 3:</strong></p>
+
+<pre>
+<strong>Input:</strong> text1 = &quot;abc&quot;, text2 = &quot;def&quot;
+<strong>Output:</strong> 0
+<strong>Explanation:</strong> There is no such common subsequence, so the result is 0.
+</pre>
+
+<p>&nbsp;</p>
+<p><strong>Constraints:</strong></p>
+
+<ul>
+	<li><code>1 &lt;= text1.length, text2.length &lt;= 1000</code></li>
+	<li><code>text1</code> and <code>text2</code> consist of only lowercase English characters.</li>
+</ul>
+
+<!-- description:end -->
+
+## Solutions
+
+<!-- solution:start -->
+
+### Solution 1: Dynamic Programming
+
+We define $f[i][j]$ as the length of the longest common subsequence of the first $i$ characters of $text1$ and the first $j$ characters of $text2$. Therefore, the answer is $f[m][n]$, where $m$ and $n$ are the lengths of $text1$ and $text2$, respectively.
+
+If the $i$th character of $text1$ and the $j$th character of $text2$ are the same, then $f[i][j] = f[i - 1][j - 1] + 1$; if the $i$th character of $text1$ and the $j$th character of $text2$ are different, then $f[i][j] = max(f[i - 1][j], f[i][j - 1])$. The state transition equation is:
+
+$$
+f[i][j] =
+\begin{cases}
+f[i - 1][j - 1] + 1, & \textit{if } text1[i - 1] = text2[j - 1] \\
+\max(f[i - 1][j], f[i][j - 1]), & \textit{if } text1[i - 1] \neq text2[j - 1]
+\end{cases}
+$$
+
+The time complexity is $O(m \times n)$, and the space complexity is $O(m \times n)$. Here, $m$ and $n$ are the lengths of $text1$ and $text2$, respectively.
+
+<!-- tabs:start -->
+
+#### Python3
+
+```python
+class Solution:
+    def longestCommonSubsequence(self, text1: str, text2: str) -> int:
+        m, n = len(text1), len(text2)
+        f = [[0] * (n + 1) for _ in range(m + 1)]
+        for i in range(1, m + 1):
+            for j in range(1, n + 1):
+                if text1[i - 1] == text2[j - 1]:
+                    f[i][j] = f[i - 1][j - 1] + 1
+                else:
+                    f[i][j] = max(f[i - 1][j], f[i][j - 1])
+        return f[m][n]
+```
+
+#### Java
+
+```java
+class Solution {
+    public int longestCommonSubsequence(String text1, String text2) {
+        int m = text1.length(), n = text2.length();
+        int[][] f = new int[m + 1][n + 1];
+        for (int i = 1; i <= m; ++i) {
+            for (int j = 1; j <= n; ++j) {
+                if (text1.charAt(i - 1) == text2.charAt(j - 1)) {
+                    f[i][j] = f[i - 1][j - 1] + 1;
+                } else {
+                    f[i][j] = Math.max(f[i - 1][j], f[i][j - 1]);
+                }
+            }
+        }
+        return f[m][n];
+    }
+}
+```
+
+#### C++
+
+```cpp
+class Solution {
+public:
+    int longestCommonSubsequence(string text1, string text2) {
+        int m = text1.size(), n = text2.size();
+        int f[m + 1][n + 1];
+        memset(f, 0, sizeof f);
+        for (int i = 1; i <= m; ++i) {
+            for (int j = 1; j <= n; ++j) {
+                if (text1[i - 1] == text2[j - 1]) {
+                    f[i][j] = f[i - 1][j - 1] + 1;
+                } else {
+                    f[i][j] = max(f[i - 1][j], f[i][j - 1]);
+                }
+            }
+        }
+        return f[m][n];
+    }
+};
+```
+
+#### Go
+
+```go
+func longestCommonSubsequence(text1 string, text2 string) int {
+	m, n := len(text1), len(text2)
+	f := make([][]int, m+1)
+	for i := range f {
+		f[i] = make([]int, n+1)
+	}
+	for i := 1; i <= m; i++ {
+		for j := 1; j <= n; j++ {
+			if text1[i-1] == text2[j-1] {
+				f[i][j] = f[i-1][j-1] + 1
+			} else {
+				f[i][j] = max(f[i-1][j], f[i][j-1])
+			}
+		}
+	}
+	return f[m][n]
+}
+```
+
+#### TypeScript
+
+```ts
+function longestCommonSubsequence(text1: string, text2: string): number {
+    const m = text1.length;
+    const n = text2.length;
+    const f = Array.from({ length: m + 1 }, () => Array(n + 1).fill(0));
+    for (let i = 1; i <= m; i++) {
+        for (let j = 1; j <= n; j++) {
+            if (text1[i - 1] === text2[j - 1]) {
+                f[i][j] = f[i - 1][j - 1] + 1;
+            } else {
+                f[i][j] = Math.max(f[i - 1][j], f[i][j - 1]);
+            }
+        }
+    }
+    return f[m][n];
+}
+```
+
+#### Rust
+
+```rust
+impl Solution {
+    pub fn longest_common_subsequence(text1: String, text2: String) -> i32 {
+        let (m, n) = (text1.len(), text2.len());
+        let (text1, text2) = (text1.as_bytes(), text2.as_bytes());
+        let mut f = vec![vec![0; n + 1]; m + 1];
+        for i in 1..=m {
+            for j in 1..=n {
+                f[i][j] = if text1[i - 1] == text2[j - 1] {
+                    f[i - 1][j - 1] + 1
+                } else {
+                    f[i - 1][j].max(f[i][j - 1])
+                };
+            }
+        }
+        f[m][n]
+    }
+}
+```
+
+#### JavaScript
+
+```js
+/**
+ * @param {string} text1
+ * @param {string} text2
+ * @return {number}
+ */
+var longestCommonSubsequence = function (text1, text2) {
+    const m = text1.length;
+    const n = text2.length;
+    const f = Array.from({ length: m + 1 }, () => Array(n + 1).fill(0));
+    for (let i = 1; i <= m; ++i) {
+        for (let j = 1; j <= n; ++j) {
+            if (text1[i - 1] == text2[j - 1]) {
+                f[i][j] = f[i - 1][j - 1] + 1;
+            } else {
+                f[i][j] = Math.max(f[i - 1][j], f[i][j - 1]);
+            }
+        }
+    }
+    return f[m][n];
+};
+```
+
+#### C#
+
+```cs
+public class Solution {
+    public int LongestCommonSubsequence(string text1, string text2) {
+        int m = text1.Length, n = text2.Length;
+        int[,] f = new int[m + 1, n + 1];
+        for (int i = 1; i <= m; ++i) {
+            for (int j = 1; j <= n; ++j) {
+                if (text1[i - 1] == text2[j - 1]) {
+                    f[i, j] = f[i - 1, j - 1] + 1;
+                } else {
+                    f[i, j] = Math.Max(f[i - 1, j], f[i, j - 1]);
+                }
+            }
+        }
+        return f[m, n];
+    }
+}
+```
+
+#### Kotlin
+
+```kotlin
+class Solution {
+    fun longestCommonSubsequence(text1: String, text2: String): Int {
+        val m = text1.length
+        val n = text2.length
+        val f = Array(m + 1) { IntArray(n + 1) }
+        for (i in 1..m) {
+            for (j in 1..n) {
+                if (text1[i - 1] == text2[j - 1]) {
+                    f[i][j] = f[i - 1][j - 1] + 1
+                } else {
+                    f[i][j] = Math.max(f[i - 1][j], f[i][j - 1])
+                }
+            }
+        }
+        return f[m][n]
+    }
+}
+```
+
+<!-- tabs:end -->
+
+<!-- solution:end -->
+
+<!-- problem:end -->