259th Commit

src/page-1/72. Edit Distance/README.md

@@ -0,0 +1,112 @@
+# 72. 編輯距離 (Edit Distance)
+
+給定兩個字串 word1 和 word2，返回將 word1 轉換為 word2 所使用的最少操作數。
+
+你可以對字詞進行以下三種操作：
+
+- 插入一個字元
+- 刪除一個字元
+- 替換一個字元
+
+範例 1：
+
+```coffee
+輸入: word1 = "horse", word2 = "ros"
+輸出: 3
+說明:
+horse -> rorse (將 'h' 替換為 'r')
+rorse -> rose (刪除 'r')
+rose -> ros (刪除 'e')
+```
+
+範例 2：
+
+```coffee
+輸入: word1 = "intention", word2 = "execution"
+輸出: 5
+說明:
+intention -> inention (刪除 't')
+inention -> enention (將 'i' 替換為 'e')
+enention -> exention (將 'n' 替換為 'x')
+exention -> exection (將 'n' 替換為 'c')
+exection -> execution (插入 'u')
+```
+
+## 解題
+
+- 插入操作：`dp[i][j-1] + 1`
+
+```coffee
+word1: "abc" -> "abcd"
+word2: "abcd"
+```
+
+|     | ""  | a   | b   | c   | d   |
+| --- | --- | --- | --- | --- | --- |
+| ""  | 0   | 1   | 2   | 3   | 4   |
+| a   | 1   | 0   | 1   | 2   | 3   |
+| ab  | 2   | 1   | 0   | 1   | 2   |
+| abc | 3   | 2   | 1   | 0   | 1   |
+
+- 刪除操作：`dp[i-1][j] + 1`
+
+```coffee
+word1: "abc" -> "ab"
+word2: "ab"
+```
+
+|     | ""  | a   | b   | c   |
+| --- | --- | --- | --- | --- |
+| ""  | 0   | 1   | 2   | 3   |
+| a   | 1   | 0   | 1   | 2   |
+| ab  | 2   | 1   | 0   | 1   |
+| abc | 3   | 2   | 1   | 1   |
+
+- 替換操作：`dp[i-1][j-1] + 1`
+
+```coffee
+word1: "abc" -> "abd"
+word2: "abd"
+```
+
+|     | ""  | a   | b   | c   |
+| --- | --- | --- | --- | --- |
+| ""  | 0   | 1   | 2   | 3   |
+| a   | 1   | 0   | 1   | 2   |
+| ab  | 2   | 1   | 0   | 1   |
+| abc | 3   | 2   | 1   | 1   |
+| adc | 3   | 2   | 1   | 1   |
+
+```ts
+export const minDistance: MinDistance = (word1, word2) => {
+  const m = word1.length;
+  const n = word2.length;
+
+  // 建立一個二維 DP 表
+  const dp: number[][] = Array.from({ length: m + 1 }, () => Array(n + 1).fill(0));
+
+  // 空字串情況處理
+  for (let i = 0; i <= m; i++) dp[i][0] = i; // 刪除 word1 中的所有字元
+  for (let j = 0; j <= n; j++) dp[0][j] = j; // 插入所有字元以匹配 word2
+
+  // 填充 DP 表
+  for (let i = 1; i <= m; i++) {
+    for (let j = 1; j <= n; j++) {
+      if (word1[i - 1] === word2[j - 1]) {
+        // 字元匹配，無需操作
+        dp[i][j] = dp[i - 1][j - 1];
+      } else {
+        // 選擇最少的插入、刪除或替換操作
+        dp[i][j] = Math.min(
+          dp[i][j - 1] + 1, // 插入
+          dp[i - 1][j] + 1, // 刪除
+          dp[i - 1][j - 1] + 1, // 替換
+        );
+      }
+    }
+  }
+
+  // 結果位於右下角格子中
+  return dp[m][n];
+};
+```

src/page-1/72. Edit Distance/minDistance.test.ts

@@ -0,0 +1,8 @@
+import { minDistance } from './minDistance';
+
+describe('72. Edit Distance', () => {
+  test('minDistance', () => {
+    expect(minDistance('horse', 'ros')).toBe(3);
+    expect(minDistance('intention', 'execution')).toBe(5);
+  });
+});

src/page-1/72. Edit Distance/minDistance.ts

@@ -0,0 +1,36 @@
+type MinDistance = (word1: string, word2: string) => number;
+
+/**
+ * Accepted
+ */
+export const minDistance: MinDistance = (word1, word2) => {
+  const m = word1.length;
+  const n = word2.length;
+
+  // Create a 2D DP array
+  const dp: number[][] = Array.from({ length: m + 1 }, () => Array(n + 1).fill(0));
+
+  // If one of the strings is empty, we need to delete or insert all characters of the other string
+  for (let i = 0; i <= m; i++) dp[i][0] = i; // Deleting all characters from word1
+  for (let j = 0; j <= n; j++) dp[0][j] = j; // Inserting all characters to match word2
+
+  // Fill the DP table
+  for (let i = 1; i <= m; i++) {
+    for (let j = 1; j <= n; j++) {
+      if (word1[i - 1] === word2[j - 1]) {
+        // Characters match, no operation needed
+        dp[i][j] = dp[i - 1][j - 1];
+      } else {
+        // Choose the minimum of insert, delete, or replace operation
+        dp[i][j] = Math.min(
+          dp[i][j - 1] + 1, // Insert
+          dp[i - 1][j] + 1, // Delete
+          dp[i - 1][j - 1] + 1, // Replace
+        );
+      }
+    }
+  }
+
+  // The result is in the bottom-right cell
+  return dp[m][n];
+};

src/page-17/1768. Merge Strings Alternately/README.md

@@ -6,7 +6,7 @@
 
 範例 1：
 
-```ts
+```coffee
 輸入: word1 = "abc", word2 = "pqr"
 輸出: "apbqcr"
 說明: 字串合併情況如下所示：
@@ -17,7 +17,7 @@ word2:   p   q   r
 
 範例 2：
 
-```ts
+```coffee
 輸入: word1 = "ab", word2 = "pqrs"
 輸出: "apbqrs"
 說明: 注意，word2 比 word1 長，"rs" 需要追加到合併後字串的結尾。
@@ -28,7 +28,7 @@ word2:   p   q   r   s
 
 範例 3：
 
-```ts
+```coffee
 輸入: word1 = "abcd", word2 = "pq"
 輸出: "apbqcd"
 說明: 注意，word1 比 word2 長，"cd" 需要追加到合併後字串的結尾。

src/page-7/714. Best Time to Buy and Sell Stock with Transaction Fee/README.md

@@ -9,6 +9,26 @@
 - 你不得同時進行多項交易 (即，您必須在再次購買之前出售股票)。
 - 每次買賣股票只收取一次手續費。
 
+範例 1：
+
+```coffee
+輸入: prices = [1,3,2,8,4,9], fee = 2
+輸出: 8
+說明: 可以透過以下方式實現最大利潤：
+- 以 prices[0] = 1 買入
+- 以 prices[3] = 8 賣出
+- 以 prices[4] = 4 買入
+- 以 prices[5] = 9 賣出
+總利潤為 ((8 - 1) - 2) + ((9 - 4) - 2) = 8。
+```
+
+範例 2：
+
+```coffee
+輸入: prices = [1,3,7,5,10,3], fee = 3
+輸出: 6
+```
+
 ## 解題
 
 使用動態規劃 (Dynamic Programming)