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| 1 | +# 2257. Count Unguarded Cells in the Grid |
| 2 | + |
| 3 | +You are given two integers `m` and `n` representing a 0-indexed `m x n` grid. |
| 4 | +You are also given two 2D integer arrays `guards` and `walls` where `guards[i] = [row_i, col_i]` and `walls[j] = [row_j, col_j]` represent the positions of the $i^{th}$ guard and $j^{th}$ wall respectively. |
| 5 | + |
| 6 | +A guard can see every cell in the four cardinal directions (north, east, south, or west) starting from their position unless obstructed by a wall or another guard. |
| 7 | +A cell is guarded if there is at least one guard that can see it. |
| 8 | + |
| 9 | +Return the number of unoccupied cells that are not guarded. |
| 10 | + |
| 11 | +**Constraints:** |
| 12 | + |
| 13 | +- `1 <= m, n <= 10^5` |
| 14 | +- `2 <= m * n <= 10^5` |
| 15 | +- `1 <= guards.length, walls.length <= 5 * 10^4` |
| 16 | +- `2 <= guards.length + walls.length <= m * n` |
| 17 | +- `guards[i].length == walls[j].length == 2` |
| 18 | +- `0 <= row_i, row_j < m` |
| 19 | +- `0 <= col_i, col_j < n` |
| 20 | +- All the positions in `guards` and `walls` are unique. |
| 21 | + |
| 22 | +## 基礎思路 |
| 23 | + |
| 24 | +本題要求在一個由牆壁 (`walls`) 與警衛 (`guards`) 組成的 $m \times n$ 網格中,計算「沒有被警衛看守、也沒有被佔據」的格子數量。警衛能沿著四個基本方向(上、下、左、右)直線觀察,直到被牆或另一名警衛阻擋。 |
| 25 | + |
| 26 | +在分析問題時,我們需要考慮以下幾個重點: |
| 27 | + |
| 28 | +- **遮蔽條件**:警衛的視線會被「牆壁」或「其他警衛」阻擋。 |
| 29 | +- **觀察方向**:警衛可以同時往四個方向延伸視線。 |
| 30 | +- **可見格子標記**:若某格被至少一名警衛看到,就視為被看守。 |
| 31 | +- **效率要求**:$m \times n \le 10^5$,必須避免二維 BFS 或全表掃描(會超時),需設計線性時間掃描法。 |
| 32 | + |
| 33 | +為此,我們採用以下策略: |
| 34 | + |
| 35 | +- **一維壓平儲存結構**:將二維座標 $(r, c)$ 轉為一維索引 $r \times n + c$,使用 TypedArray (`Uint8Array`) 儲存每格狀態,節省記憶體。 |
| 36 | +- **分方向線性掃描(Sweep Line)**: |
| 37 | + - 先以**橫向掃描**處理左右兩方向的可視格; |
| 38 | + - 再以**縱向掃描**處理上下兩方向的可視格。 |
| 39 | + 每一行與每一列都僅被掃描兩次(正向與反向),確保 $O(mn)$ 時間內完成。 |
| 40 | +- **狀態標記**: |
| 41 | + - `0`:空格(可被看守); |
| 42 | + - `1`:牆壁(阻斷視線); |
| 43 | + - `2`:警衛; |
| 44 | + - `3`:被看守的空格。 |
| 45 | +- **統計未看守格數**:起初總空格為 $m \times n -$(牆數 + 警衛數),最後扣除被看守格即可。 |
| 46 | + |
| 47 | +此設計可在不使用 BFS 的情況下高效地模擬警衛的直線視線,達成題目要求。 |
| 48 | + |
| 49 | +## 解題步驟 |
| 50 | + |
| 51 | +### Step 1:初始化常數與狀態陣列 |
| 52 | + |
| 53 | +建立四種狀態常數,並配置 `Uint8Array` 儲存每格狀態。 |
| 54 | + |
| 55 | +```typescript |
| 56 | +// 定義狀態常數:0=空格、1=牆、2=警衛、3=被看守 |
| 57 | +const STATE_EMPTY = 0; |
| 58 | +const STATE_WALL = 1; |
| 59 | +const STATE_GUARD = 2; |
| 60 | +const STATE_GUARDED = 3; |
| 61 | + |
| 62 | +// 以一維陣列表示 m × n 網格,節省記憶體開銷 |
| 63 | +const totalCells = m * n; |
| 64 | +const state = new Uint8Array(totalCells); |
| 65 | +``` |
| 66 | + |
| 67 | +### Step 2:初始化未佔據格數 |
| 68 | + |
| 69 | +先計算所有非牆與非警衛格子的初始數量,方便最終扣除。 |
| 70 | + |
| 71 | +```typescript |
| 72 | +// 初始未佔據格數 = 總格數 - (牆 + 警衛) |
| 73 | +const wallsLength = walls.length; |
| 74 | +const guardsLength = guards.length; |
| 75 | +let unoccupiedCells = totalCells - wallsLength - guardsLength; |
| 76 | +``` |
| 77 | + |
| 78 | +### Step 3:標記牆壁位置 |
| 79 | + |
| 80 | +遍歷 `walls` 陣列,將對應格狀態設為 `STATE_WALL`。 |
| 81 | + |
| 82 | +```typescript |
| 83 | +// 標記牆壁位置(視線阻斷) |
| 84 | +for (let i = 0; i < wallsLength; i += 1) { |
| 85 | + const rowIndex = walls[i][0]; |
| 86 | + const columnIndex = walls[i][1]; |
| 87 | + const index = rowIndex * n + columnIndex; |
| 88 | + state[index] = STATE_WALL; |
| 89 | +} |
| 90 | +``` |
| 91 | + |
| 92 | +### Step 4:標記警衛位置 |
| 93 | + |
| 94 | +遍歷 `guards` 陣列,將對應格狀態設為 `STATE_GUARD`。 |
| 95 | + |
| 96 | +```typescript |
| 97 | +// 標記警衛位置(視線起點) |
| 98 | +for (let i = 0; i < guardsLength; i += 1) { |
| 99 | + const rowIndex = guards[i][0]; |
| 100 | + const columnIndex = guards[i][1]; |
| 101 | + const index = rowIndex * n + columnIndex; |
| 102 | + state[index] = STATE_GUARD; |
| 103 | +} |
| 104 | +``` |
| 105 | + |
| 106 | +### Step 5:橫向掃描(每行左→右與右→左) |
| 107 | + |
| 108 | +對每一行進行兩次掃描: |
| 109 | + |
| 110 | +- 第一次由左至右(模擬往右視線) |
| 111 | +- 第二次由右至左(模擬往左視線) |
| 112 | + |
| 113 | +若視線活躍(前方有警衛未被牆阻斷),且格為空則標記為被看守。 |
| 114 | + |
| 115 | +```typescript |
| 116 | +// 被看守空格計數 |
| 117 | +let guardedEmptyCount = 0; |
| 118 | + |
| 119 | +// 逐行橫向掃描 |
| 120 | +for (let rowIndex = 0; rowIndex < m; rowIndex += 1) { |
| 121 | + // ←→方向視線模擬 |
| 122 | + let hasActiveGuard = false; |
| 123 | + let index = rowIndex * n; |
| 124 | + |
| 125 | + // 左→右掃描 |
| 126 | + for (let columnIndex = 0; columnIndex < n; columnIndex += 1) { |
| 127 | + const cell = state[index]; |
| 128 | + |
| 129 | + if (cell === STATE_WALL) { |
| 130 | + hasActiveGuard = false; // 牆阻擋視線 |
| 131 | + } else if (cell === STATE_GUARD) { |
| 132 | + hasActiveGuard = true; // 新警衛開始發出視線 |
| 133 | + } else if (hasActiveGuard && cell === STATE_EMPTY) { |
| 134 | + // 標記被看守格 |
| 135 | + state[index] = STATE_GUARDED; |
| 136 | + guardedEmptyCount += 1; |
| 137 | + } |
| 138 | + |
| 139 | + index += 1; |
| 140 | + } |
| 141 | + |
| 142 | + // 右→左掃描 |
| 143 | + hasActiveGuard = false; |
| 144 | + index = rowIndex * n + (n - 1); |
| 145 | + |
| 146 | + for (let columnIndex = n - 1; columnIndex >= 0; columnIndex -= 1) { |
| 147 | + const cell = state[index]; |
| 148 | + |
| 149 | + if (cell === STATE_WALL) { |
| 150 | + hasActiveGuard = false; // 牆阻斷反向視線 |
| 151 | + } else if (cell === STATE_GUARD) { |
| 152 | + hasActiveGuard = true; // 警衛向左發出視線 |
| 153 | + } else if (hasActiveGuard && cell === STATE_EMPTY) { |
| 154 | + state[index] = STATE_GUARDED; |
| 155 | + guardedEmptyCount += 1; |
| 156 | + } |
| 157 | + |
| 158 | + index -= 1; |
| 159 | + } |
| 160 | +} |
| 161 | +``` |
| 162 | + |
| 163 | +### Step 6:縱向掃描(每列上→下與下→上) |
| 164 | + |
| 165 | +同理,再對每一列進行兩次掃描: |
| 166 | + |
| 167 | +- 第一次由上至下(模擬往下視線) |
| 168 | +- 第二次由下至上(模擬往上視線) |
| 169 | + |
| 170 | +```typescript |
| 171 | +// 逐列縱向掃描 |
| 172 | +for (let columnIndex = 0; columnIndex < n; columnIndex += 1) { |
| 173 | + // 上→下掃描 |
| 174 | + let hasActiveGuard = false; |
| 175 | + let index = columnIndex; |
| 176 | + |
| 177 | + for (let rowIndex = 0; rowIndex < m; rowIndex += 1) { |
| 178 | + const cell = state[index]; |
| 179 | + |
| 180 | + if (cell === STATE_WALL) { |
| 181 | + hasActiveGuard = false; // 牆阻斷下視線 |
| 182 | + } else if (cell === STATE_GUARD) { |
| 183 | + hasActiveGuard = true; // 警衛向下發出視線 |
| 184 | + } else if (hasActiveGuard && cell === STATE_EMPTY) { |
| 185 | + state[index] = STATE_GUARDED; |
| 186 | + guardedEmptyCount += 1; |
| 187 | + } |
| 188 | + |
| 189 | + index += n; |
| 190 | + } |
| 191 | + |
| 192 | + // 下→上掃描 |
| 193 | + hasActiveGuard = false; |
| 194 | + index = (m - 1) * n + columnIndex; |
| 195 | + |
| 196 | + for (let rowIndex = m - 1; rowIndex >= 0; rowIndex -= 1) { |
| 197 | + const cell = state[index]; |
| 198 | + |
| 199 | + if (cell === STATE_WALL) { |
| 200 | + hasActiveGuard = false; // 牆阻斷上視線 |
| 201 | + } else if (cell === STATE_GUARD) { |
| 202 | + hasActiveGuard = true; // 警衛向上發出視線 |
| 203 | + } else if (hasActiveGuard && cell === STATE_EMPTY) { |
| 204 | + state[index] = STATE_GUARDED; |
| 205 | + guardedEmptyCount += 1; |
| 206 | + } |
| 207 | + |
| 208 | + index -= n; |
| 209 | + } |
| 210 | +} |
| 211 | +``` |
| 212 | + |
| 213 | +### Step 7:計算結果並回傳 |
| 214 | + |
| 215 | +最後,用未佔據總格扣掉被看守格數,得到「未被看守的空格數」。 |
| 216 | + |
| 217 | +```typescript |
| 218 | +// 最終結果 = 總未佔據格數 - 被看守格數 |
| 219 | +return unoccupiedCells - guardedEmptyCount; |
| 220 | +``` |
| 221 | + |
| 222 | +## 時間複雜度 |
| 223 | + |
| 224 | +- 初始化牆與警衛標記:$O(w + g)$,其中 $w$、$g$ 分別為牆與警衛數。 |
| 225 | +- 四次方向掃描(兩行兩列):每格僅訪問常數次,為 $O(m \times n)$。 |
| 226 | +- 總時間複雜度為 $O(m \times n + w + g)$。 |
| 227 | + |
| 228 | +> $O(m \times n + w + g)$ |
| 229 | +
|
| 230 | +## 空間複雜度 |
| 231 | + |
| 232 | +- 使用一個大小為 $m \times n$ 的 `Uint8Array` 儲存狀態。 |
| 233 | +- 額外僅有常數級變數,無遞迴或額外結構。 |
| 234 | +- 總空間複雜度為 $O(m \times n)$。 |
| 235 | + |
| 236 | +> $O(m \times n)$ |
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