Add: Add 2025/1/22

Author: whats2000

1765-Map of Highest Peak/Note.md

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+# 1765. Map of Highest Peak
+
+You are given an integer matrix `isWater` of size `m x n` that represents a map of land and water cells.
+
+* If `isWater[i][j] == 0`, cell `(i, j)` is a land cell.
+* If `isWater[i][j] == 1`, cell `(i, j)` is a water cell.
+
+You must assign each cell a height in a way that follows these rules:
+
+* The height of each cell must be non-negative.
+* If the cell is a water cell, its height must be `0`.
+* Any two adjacent cells must have an absolute height difference of at most `1`. 
+* A cell is adjacent to another cell if the former is directly north, east, south, or west of the latter (i.e., their sides are touching).
+
+Find an assignment of heights such that the maximum height in the matrix is maximized.
+
+Return an integer matrix height of size `m x n` where `height[i][j]` is cell `(i, j)`'s height. 
+If there are multiple solutions, return any of them.
+
+## 基礎思路
+這可以視為一個 Multi-Source BFS 問題，由水源開始，逐漸往空白地區擴散，直到所有地區都被填滿。
+
+> Tips 這題很容易超時，或炸記憶體，這邊引入一些小技巧
+> 1. 用簡化後 Sliding Window 方向陣列，可以從 4 x 2 大小簡化到 5 x 1 (移動一次取兩位，一樣是4個方向)
+> 2. 用 nextQueue 來存放下一次要處理的點，減少使用 `unshift` 所帶來的效能損耗
+
+## 解題步驟
+
+### Step 1: 紀錄地圖大小
+
+```typescript
+const m = isWater.length;    // 行數
+const n = isWater[0].length; // 列數
+```
+
+### Step 2: 初始化地圖與待處理佇列
+
+```typescript
+// 初始化地圖，設置所有點的高度為 -1，代表未訪問過
+const heights = Array.from({ length: m }, () => Array(n).fill(-1));
+let queue: [number, number, number][] = [];
+```
+
+### Step 3: 初始化水源
+
+```typescript
+// 初始化水源，將所有水源點的高度設為 0，並加入待處理佇列
+for (let row = 0; row < m; row++) {
+  for (let col = 0; col < n; col++) {
+    // 如果是水源，高度設為 0，並加入待處理佇列
+    if (isWater[row][col] === 1) {
+      heights[row][col] = 0;
+      queue.push([row, col, 0]);
+    }
+  }
+}
+```
+
+### Step 4: 開始 BFS
+
+```typescript
+while (queue.length > 0) {
+  // 用來標記下個 BFS level 的點
+  const nextQueue: [number, number, number][] = [];
+  
+  // 遍歷當前 BFS level 的所有點
+  for (const [currentRow, currentCol, currentHeight] of queue) {
+    for (let i = 0; i < 4; i++) {
+      // 利用 Sliding Window 方向陣列，取得下一個點的位置
+      const nextRow = currentRow + HIGHEST_PEEK_DIRECTIONS[i];
+      const nextCol = currentCol + HIGHEST_PEEK_DIRECTIONS[i + 1];
+
+      // 如果下一個點超出地圖範圍，或已經訪問過，則跳過
+      if (
+        nextRow < 0 ||
+        nextRow >= m ||
+        nextCol < 0 ||
+        nextCol >= n ||
+        heights[nextRow][nextCol] !== -1
+      ) {
+        continue;
+      }
+
+      // 設置旁邊的點的高度，並加入下一次處理的佇列
+      heights[nextRow][nextCol] = currentHeight + 1;
+      queue.push([nextRow, nextCol, currentHeight + 1]);
+    }
+  }
+  // 移動到下一個 BFS level
+  queue = nextQueue;
+}
+```
+
+## 時間複雜度
+
+- 標記水源：$O(m \times n)$
+- BFS 會遍歷所有點，時間複雜度為 $O(m \times n)$
+- 總時間複雜度為 $O(m \times n)$
+
+> $O(m \times n)$
+
+## 空間複雜度
+- 額外使用了一個二維陣列 `heights` 來記錄每個點的高度，空間複雜度為 $O(m \times n)$
+- `queue` 與 `nextQueue` 佇列的空間複雜度為 $O(m \times n)$
+- 總空間複雜度為 $O(m \times n)$
+
+> $O(m \times n)$

1765-Map of Highest Peak/answer.ts

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+// Direction array for moving north, east, south, and west
+// This using sliding window technique, for less memory usage
+const HIGHEST_PEEK_DIRECTIONS: number[] = [0, 1, 0, -1, 0];
+
+function highestPeak(isWater: number[][]): number[][] {
+  // Get the number of rows and columns
+  const m = isWater.length;
+  const n = isWater[0].length;
+
+  // Initialize the result matrix
+  const heights = Array.from({ length: m }, () => Array(n).fill(-1));
+
+  // Queue for BFS, starting with all water cells
+  let queue: [number, number, number][] = [];
+  for (let row = 0; row < m; row++) {
+    for (let col = 0; col < n; col++) {
+      // If the cell is water, set the height to 0 and add to the queue
+      if (isWater[row][col] === 1) {
+        heights[row][col] = 0;
+        queue.push([row, col, 0]);
+      }
+    }
+  }
+
+  // Perform BFS to calculate heights for land cells
+  while (queue.length > 0) {
+    const nextQueue: [number, number, number][] = [];
+    for (const [currentRow, currentCol, currentHeight] of queue) {
+      for (let i = 0; i < 4; i++) {
+        const nextRow = currentRow + HIGHEST_PEEK_DIRECTIONS[i];
+        const nextCol = currentCol + HIGHEST_PEEK_DIRECTIONS[i + 1];
+
+        // Check bounds and if the cell is unvisited
+        if (
+          nextRow < 0 ||
+          nextRow >= m ||
+          nextCol < 0 ||
+          nextCol >= n ||
+          heights[nextRow][nextCol] !== -1
+        ) {
+          continue;
+        }
+
+        heights[nextRow][nextCol] = currentHeight + 1;
+        queue.push([nextRow, nextCol, currentHeight + 1]);
+      }
+    }
+    // Move to the next BFS level
+    queue = nextQueue;
+  }
+
+  return heights;
+}