Add: Add 2025/1/23

Author: whats2000

1267-Count Servers that Communicate/Note.md

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+# 1267. Count Servers that Communicate
+
+You are given a map of a server center, represented as a `m * n` integer matrix `grid`, 
+where 1 means that on that cell there is a server and 0 means that it is no server. 
+Two servers are said to communicate if they are on the same row or on the same column.
+
+Return the number of servers that communicate with any other server.
+
+## 基礎思路
+可以溝通的條件是在同一行或同一列，那麼換句話說，就是他所處的行或列上的伺服器數量大於 1，那麼這個伺服器就可以溝通。
+那麼問題就分解成了兩個部分，一個是計算每一行與每一列的伺服器數量，然後逐一檢查這個伺服器是否可以溝通。
+
+## 解題步驟
+
+### Step 1: 紀錄每一行與每一列的伺服器數量
+
+```typescript
+const m = grid.length;
+const n = grid[0].length;
+```
+
+### Step 2: 計算每一行與每一列的伺服器數量
+
+```typescript
+// 紀錄每一行與每一列的伺服器數量
+const serverCountInRow = new Array(m).fill(0);
+const serverCountInCol = new Array(n).fill(0);
+
+// 遍歷 grid，計算每一行與每一列的伺服器數量
+for (let i = 0; i < m; i++) {
+  for (let j = 0; j < n; j++) {
+    if (grid[i][j] === 1) {
+      serverCountInRow[i]++;
+      serverCountInCol[j]++;
+    }
+  }
+}
+```
+
+### Step 3: 計算可以溝通的伺服器數量
+
+```typescript
+let count = 0;
+for (let i = 0; i < m; i++) {
+  for (let j = 0; j < n; j++) {
+    // 我們會在以下情況跳過
+    // 1. 這個位置沒有伺服器
+    // 2. 這個伺服器所在的行或列只有一個伺服器
+    if (grid[i][j] === 0 || serverCountInRow[i] === 1 && serverCountInCol[j] === 1) {
+      continue;
+    }
+
+    count++;
+  }
+}
+```
+
+## 時間複雜度
+- 計算每一行與每一列的伺服器數量需要 $O(m * n)$ 的時間
+- 計算可以溝通的伺服器數量需要 $O(m * n)$ 的時間
+- 因此總時間複雜度為 $O(m * n)$
+
+> $O(m * n)$
+
+## 空間複雜度
+- 我們需要兩個陣列來記錄每一行與每一列的伺服器數量，因此空間複雜度為 $O(m + n)$
+- 紀錄其他變數的空間複雜度為 $O(1)$
+- 因此總空間複雜度為 $O(m + n)$
+
+> $O(m + n)$

1267-Count Servers that Communicate/answer.ts

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+function countServers(grid: number[][]): number {
+  // Get the number of rows and columns
+  const m = grid.length;
+  const n = grid[0].length;
+
+  // Storage for the number of servers in each row and column
+  const serverCountInRow = new Array(m).fill(0);
+  const serverCountInCol = new Array(n).fill(0);
+
+  // Count the number of servers in each row and column
+  for (let i = 0; i < m; i++) {
+    for (let j = 0; j < n; j++) {
+      if (grid[i][j] === 1) {
+        serverCountInRow[i]++;
+        serverCountInCol[j]++;
+      }
+    }
+  }
+
+  // Count the number of servers that can communicate
+  let count = 0;
+  for (let i = 0; i < m; i++) {
+    for (let j = 0; j < n; j++) {
+      // When the cell is empty
+      // Or the server is the only server in the row and column
+      // We skip this cell
+      if (grid[i][j] === 0 || serverCountInRow[i] === 1 && serverCountInCol[j] === 1) {
+        continue;
+      }
+
+      count++;
+    }
+  }
+
+  return count;
+}