Add a solution to Minesweeper

soapyigu · soapyigu · commit 618b3ce8bb5f · 2022-10-06T15:31:35.000-07:00
diff --git a/DFS/Minesweeper.swift b/DFS/Minesweeper.swift
@@ -0,0 +1,64 @@
+/**
+ * Question Link: https://leetcode.com/problems/minesweeper/
+ * Primary idea: Classic Depth-first Search. Check current node and dfs all directions if mine count is 0, update the board accordingly.
+ * 
+ * Time Complexity: O(mn), Space Complexity: O(mn)
+ *
+ */
+
+class Minesweeper {
+    private let dirs = [(0, 1), (0, -1), (1, 0), (-1, 0), (1, 1), (-1, 1), (1, -1), (-1, -1)]
+
+    func updateBoard(_ board: [[Character]], _ click: [Int]) -> [[Character]] {
+        let i = click[0], j = click[1]
+        var board = board
+
+        if board[i][j] == "M" {
+            board[i][j] = "X".first!
+            return board
+        }
+
+        let m = board.count, n = board[0].count
+        var isVisited = Array(repeating: Array(repeating: false, count: n), count: m)
+
+        dfs(&board, i, j, &isVisited)
+
+        return board
+    }
+
+    private func dfs(_ board: inout [[Character]], _ i: Int, _ j: Int, _ isVisited: inout [[Bool]]) {
+        guard isValid(i, j, board) && !isVisited[i][j] else {
+            return
+        }
+
+        isVisited[i][j] = true
+
+        if board[i][j] == "E" {
+            var count = 0
+            for dir in dirs {
+                let x = i + dir.0, y = j + dir.1
+
+                if isValid(x, y, board) {
+                    if board[x][y] == "X" || board[x][y] == "M" {
+                        count += 1
+                    }
+                }
+            }
+
+            if count == 0 {
+                board[i][j] = "B".first!
+
+                for dir in dirs {
+                    let x = i + dir.0, y = j + dir.1
+                    dfs(&board, x, y, &isVisited)
+                }
+            } else {
+                board[i][j] = String(count).first!
+            }
+        }
+    }
+
+    private func isValid(_ i: Int, _ j: Int, _ board: [[Character]]) -> Bool {
+        return i >= 0 && i < board.count && j >= 0 && j < board[0].count
+    }
+}
