Leetcode 75

Solution/72. Edit Distance/72. Edit Distance.py

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+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]

Solution/72. Edit Distance/readme.md

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+
+<!-- problem:start -->
+
+# [72. Edit Distance](https://leetcode.com/problems/edit-distance)
+
+---
+- **comments**: true
+- **difficulty**: Medium
+- **tags**:
+    - String
+    - Dynamic Programming
+---
+
+## Description
+
+<!-- description:start -->
+
+<p>Given two strings <code>word1</code> and <code>word2</code>, return <em>the minimum number of operations required to convert <code>word1</code> to <code>word2</code></em>.</p>
+
+<p>You have the following three operations permitted on a word:</p>
+
+<ul>
+	<li>Insert a character</li>
+	<li>Delete a character</li>
+	<li>Replace a character</li>
+</ul>
+
+<p>&nbsp;</p>
+<p><strong class="example">Example 1:</strong></p>
+
+<pre>
+<strong>Input:</strong> word1 = &quot;horse&quot;, word2 = &quot;ros&quot;
+<strong>Output:</strong> 3
+<strong>Explanation:</strong> 
+horse -&gt; rorse (replace &#39;h&#39; with &#39;r&#39;)
+rorse -&gt; rose (remove &#39;r&#39;)
+rose -&gt; ros (remove &#39;e&#39;)
+</pre>
+
+<p><strong class="example">Example 2:</strong></p>
+
+<pre>
+<strong>Input:</strong> word1 = &quot;intention&quot;, word2 = &quot;execution&quot;
+<strong>Output:</strong> 5
+<strong>Explanation:</strong> 
+intention -&gt; inention (remove &#39;t&#39;)
+inention -&gt; enention (replace &#39;i&#39; with &#39;e&#39;)
+enention -&gt; exention (replace &#39;n&#39; with &#39;x&#39;)
+exention -&gt; exection (replace &#39;n&#39; with &#39;c&#39;)
+exection -&gt; execution (insert &#39;u&#39;)
+</pre>
+
+<p>&nbsp;</p>
+<p><strong>Constraints:</strong></p>
+
+<ul>
+	<li><code>0 &lt;= word1.length, word2.length &lt;= 500</code></li>
+	<li><code>word1</code> and <code>word2</code> consist of lowercase English letters.</li>
+</ul>
+
+<!-- description:end -->
+
+## Solutions
+
+<!-- solution:start -->
+
+### 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.
+
+<!-- tabs:start -->
+
+#### 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];
+};
+```
+
+<!-- tabs:end -->
+
+<!-- solution:end -->
+
+<!-- problem:end -->