diff --git a/Solution/72. Edit Distance/72. Edit Distance.py b/Solution/72. Edit Distance/72. Edit Distance.py new file mode 100644 index 0000000..d0ce251 --- /dev/null +++ b/Solution/72. Edit Distance/72. Edit Distance.py @@ -0,0 +1,14 @@ +class Solution: + def minDistance(self, word1: str, word2: str) -> int: + m, n = len(word1), len(word2) + f = [[0] * (n + 1) for _ in range(m + 1)] + for j in range(1, n + 1): + f[0][j] = j + for i, a in enumerate(word1, 1): + f[i][0] = i + for j, b in enumerate(word2, 1): + if a == b: + f[i][j] = f[i - 1][j - 1] + else: + f[i][j] = min(f[i - 1][j], f[i][j - 1], f[i - 1][j - 1]) + 1 + return f[m][n] \ No newline at end of file diff --git a/Solution/72. Edit Distance/readme.md b/Solution/72. Edit Distance/readme.md new file mode 100644 index 0000000..955318b --- /dev/null +++ b/Solution/72. Edit Distance/readme.md @@ -0,0 +1,250 @@ + + + +# [72. Edit Distance](https://leetcode.com/problems/edit-distance) + +--- +- **comments**: true +- **difficulty**: Medium +- **tags**: + - String + - Dynamic Programming +--- + +## Description + + + +

Given two strings word1 and word2, return the minimum number of operations required to convert word1 to word2.

+ +

You have the following three operations permitted on a word:

+ + + +

 

+

Example 1:

+ +
+Input: word1 = "horse", word2 = "ros"
+Output: 3
+Explanation: 
+horse -> rorse (replace 'h' with 'r')
+rorse -> rose (remove 'r')
+rose -> ros (remove 'e')
+
+ +

Example 2:

+ +
+Input: word1 = "intention", word2 = "execution"
+Output: 5
+Explanation: 
+intention -> inention (remove 't')
+inention -> enention (replace 'i' with 'e')
+enention -> exention (replace 'n' with 'x')
+exention -> exection (replace 'n' with 'c')
+exection -> execution (insert 'u')
+
+ +

 

+

Constraints:

+ + + + + +## Solutions + + + +### Solution 1: Dynamic Programming + +We define $f[i][j]$ as the minimum number of operations to convert $word1$ of length $i$ to $word2$ of length $j$. $f[i][0] = i$, $f[0][j] = j$, $i \in [1, m], j \in [0, n]$. + +We consider $f[i][j]$: + +- If $word1[i - 1] = word2[j - 1]$, then we only need to consider the minimum number of operations to convert $word1$ of length $i - 1$ to $word2$ of length $j - 1$, so $f[i][j] = f[i - 1][j - 1]$; +- Otherwise, we can consider insert, delete, and replace operations, then $f[i][j] = \min(f[i - 1][j], f[i][j - 1], f[i - 1][j - 1]) + 1$. + +Finally, we can get the state transition equation: + +$$ +f[i][j] = \begin{cases} +i, & \textit{if } j = 0 \\ +j, & \textit{if } i = 0 \\ +f[i - 1][j - 1], & \textit{if } word1[i - 1] = word2[j - 1] \\ +\min(f[i - 1][j], f[i][j - 1], f[i - 1][j - 1]) + 1, & \textit{otherwise} +\end{cases} +$$ + +Finally, we return $f[m][n]$. + +The time complexity is $O(m \times n)$, and the space complexity is $O(m \times n)$. $m$ and $n$ are the lengths of $word1$ and $word2$ respectively. + + + +#### Python3 + +```python +class Solution: + def minDistance(self, word1: str, word2: str) -> int: + m, n = len(word1), len(word2) + f = [[0] * (n + 1) for _ in range(m + 1)] + for j in range(1, n + 1): + f[0][j] = j + for i, a in enumerate(word1, 1): + f[i][0] = i + for j, b in enumerate(word2, 1): + if a == b: + f[i][j] = f[i - 1][j - 1] + else: + f[i][j] = min(f[i - 1][j], f[i][j - 1], f[i - 1][j - 1]) + 1 + return f[m][n] +``` + +#### Java + +```java +class Solution { + public int minDistance(String word1, String word2) { + int m = word1.length(), n = word2.length(); + int[][] f = new int[m + 1][n + 1]; + for (int j = 1; j <= n; ++j) { + f[0][j] = j; + } + for (int i = 1; i <= m; ++i) { + f[i][0] = i; + for (int j = 1; j <= n; ++j) { + if (word1.charAt(i - 1) == word2.charAt(j - 1)) { + f[i][j] = f[i - 1][j - 1]; + } else { + f[i][j] = Math.min(f[i - 1][j], Math.min(f[i][j - 1], f[i - 1][j - 1])) + 1; + } + } + } + return f[m][n]; + } +} +``` + +#### C++ + +```cpp +class Solution { +public: + int minDistance(string word1, string word2) { + int m = word1.size(), n = word2.size(); + int f[m + 1][n + 1]; + for (int j = 0; j <= n; ++j) { + f[0][j] = j; + } + for (int i = 1; i <= m; ++i) { + f[i][0] = i; + for (int j = 1; j <= n; ++j) { + if (word1[i - 1] == word2[j - 1]) { + f[i][j] = f[i - 1][j - 1]; + } else { + f[i][j] = min({f[i - 1][j], f[i][j - 1], f[i - 1][j - 1]}) + 1; + } + } + } + return f[m][n]; + } +}; +``` + +#### Go + +```go +func minDistance(word1 string, word2 string) int { + m, n := len(word1), len(word2) + f := make([][]int, m+1) + for i := range f { + f[i] = make([]int, n+1) + } + for j := 1; j <= n; j++ { + f[0][j] = j + } + for i := 1; i <= m; i++ { + f[i][0] = i + for j := 1; j <= n; j++ { + if word1[i-1] == word2[j-1] { + f[i][j] = f[i-1][j-1] + } else { + f[i][j] = min(f[i-1][j], min(f[i][j-1], f[i-1][j-1])) + 1 + } + } + } + return f[m][n] +} +``` + +#### TypeScript + +```ts +function minDistance(word1: string, word2: string): number { + const m = word1.length; + const n = word2.length; + const f: number[][] = Array(m + 1) + .fill(0) + .map(() => Array(n + 1).fill(0)); + for (let j = 1; j <= n; ++j) { + f[0][j] = j; + } + for (let i = 1; i <= m; ++i) { + f[i][0] = i; + for (let j = 1; j <= n; ++j) { + if (word1[i - 1] === word2[j - 1]) { + f[i][j] = f[i - 1][j - 1]; + } else { + f[i][j] = Math.min(f[i - 1][j], f[i][j - 1], f[i - 1][j - 1]) + 1; + } + } + } + return f[m][n]; +} +``` + +#### JavaScript + +```js +/** + * @param {string} word1 + * @param {string} word2 + * @return {number} + */ +var minDistance = function (word1, word2) { + const m = word1.length; + const n = word2.length; + const f = Array(m + 1) + .fill(0) + .map(() => Array(n + 1).fill(0)); + for (let j = 1; j <= n; ++j) { + f[0][j] = j; + } + for (let i = 1; i <= m; ++i) { + f[i][0] = i; + for (let j = 1; j <= n; ++j) { + if (word1[i - 1] === word2[j - 1]) { + f[i][j] = f[i - 1][j - 1]; + } else { + f[i][j] = Math.min(f[i - 1][j], f[i][j - 1], f[i - 1][j - 1]) + 1; + } + } + } + return f[m][n]; +}; +``` + + + + + + \ No newline at end of file