222nd Commit

README.md

@@ -78,6 +78,7 @@ $ pnpm bench twoSum.bench.ts
 - [二分搜尋 (Binary Search)](./algorithms/binary-search/README.md)
 - 回溯 (Backtracking)
 - 動態規劃 (Dynamic Programming)
+  - [0-1 背包問題 (0-1 Knapsack Problem)](./algorithms/dynamic-programming/README.md)
 
 ## Basic - LeetCode 75
 
@@ -267,12 +268,14 @@ Ace Coding Interview with 75 Qs
 
 [17]: ./src/page-1/17.%20Letter%20Combinations%20of%20a%20Phone%20Number/letterCombinations.ts
 
-| DP - 1D                        |          |        |
-| ------------------------------ | -------- | ------ |
-| 1137. N-th Tribonacci Number   | Solution | Easy   |
-| 746. Min Cost Climbing Stairs  | Solution | Easy   |
-| 198. House Robber              | Solution | Medium |
-| 790. Domino and Tromino Tiling | Solution | Medium |
+| DP - 1D                        |                  |        |
+| ------------------------------ | ---------------- | ------ |
+| 1137. N-th Tribonacci Number   | [Solution][1137] | Easy   |
+| 746. Min Cost Climbing Stairs  | Solution         | Easy   |
+| 198. House Robber              | Solution         | Medium |
+| 790. Domino and Tromino Tiling | Solution         | Medium |
+
+[1137]: ./src/page-11/1137.%20N-th%20Tribonacci%20Number/tribonacci.ts
 
 | DP - Multidimensional                                     |          |        |
 | --------------------------------------------------------- | -------- | ------ |

algorithms/dynamic-programming/README.md

@@ -0,0 +1,63 @@
+# 動態規劃 (Dynamic Programming)
+
+## 0-1 背包問題 (0-1 Knapsack Problem)
+
+問題：給定 n 個物品，每個物品有一定的重量和價值，並給定一個背包的容量。你需要在不超過背包容量的情況下選擇物品，使得所選物品的總價值最大。
+
+- `Array(n)`: 物品數量
+- `weights[]`: 物品的重量
+- `values[]`: 物品的價值
+- `capacity`: 背包的容量
+
+```ts
+function knapsack(weights: number[], values: number[], capacity: number): number {
+  const n = weights.length;
+
+  // 建立一個 n+1 橫行 (Row)、capacity+1 縱列 (Column) 的二維 DP 表，初始值都設為 0
+  const dp: number[][] = Array.from({ length: n + 1 }, () => Array(capacity + 1).fill(0));
+
+  for (let i = 1; i <= n; i++) {
+    for (let c = 0; c <= capacity; c++) {
+      if (weights[i - 1] <= c) {
+        // 遞推公式
+        dp[i][c] = Math.max(
+          dp[i - 1][c], // 不選擇物品
+          dp[i - 1][c - weights[i - 1]] + values[i - 1], // 選擇物品
+        );
+      } else {
+        dp[i][c] = dp[i - 1][c];
+      }
+    }
+  }
+
+  // dp[n][capacity] 即為考慮所有物品且容量為 capacity 時的最大價值
+  return dp[n][capacity];
+}
+
+const weights = [1, 2, 3];
+const values = [5, 11, 15];
+const capacity = 4;
+console.log(knapsack(weights, values, capacity)); // 20
+```
+
+直接使用前一橫行 (Row) 的容量做優化：
+
+```ts
+function knapsack(weights: number[], values: number[], capacity: number): number {
+  const n = weights.length;
+  const dp: number[] = Array(capacity + 1).fill(0);
+
+  for (let i = 0; i < n; i++) {
+    for (let c = capacity; c >= weights[i]; c--) {
+      dp[c] = Math.max(dp[c], dp[c - weights[i]] + values[i]);
+    }
+  }
+
+  return dp[capacity];
+}
+
+const weights = [1, 2, 3];
+const values = [5, 11, 15];
+const capacity = 4;
+console.log(knapsack(weights, values, capacity)); // 20
+```

src/page-11/1137. N-th Tribonacci Number/tribonacci.ts

@@ -4,10 +4,6 @@ type Tribonacci = (n: number) => number;
  * Accepted
  */
 export const tribonacci: Tribonacci = (n) => {
-  // Base cases
-  if (n === 0) return 0;
-  if (n === 1 || n === 2) return 1;
-
   // Initialize the array with the first three values
   const T = [0, 1, 1];