Merge pull request #359 from soapyigu/DFS

Add a solution to Check if Word Can Be Placed In Crossword

DFS/CheckWordCrossword.swift

@@ -0,0 +1,54 @@
+/**
+ * Question Link: https://leetcode.com/problems/check-if-word-can-be-placed-in-crossword/
+ * Primary idea: Classic Depth-first Search, check possible position and go through four directions
+ * 
+ * Time Complexity: O(mnl), Space Complexity: O(1)
+ *
+ */
+
+class CheckWordCrossword {
+    func placeWordInCrossword(_ board: [[Character]], _ word: String) -> Bool {
+        let dirs = [(0, 1), (0, -1), (1, 0), (-1, 0)], m = board.count, n = board[0].count
+
+        for i in 0..<m {
+            for j in 0..<n {
+                for dir in dirs {
+                    let prev = (i - dir.0, j - dir.1)
+
+                    guard !isValid(prev.0, prev.1, board) else {
+                        continue
+                    }
+
+                    if dfs(0, Array(word), board, i, j, dir) {
+                        return true
+                    }
+                }
+
+            }
+        }
+
+        return false
+    }
+
+    private func isValid(_ i: Int, _ j: Int, _ board: [[Character]]) -> Bool {
+        let m = board.count, n = board[0].count
+
+        return i >= 0 && i < m && j >= 0 && j < n && board[i][j] != "#"
+    }
+
+    private func dfs(_ idx: Int, _ word: [Character], _ board: [[Character]], _ i: Int, _ j: Int, _ dir: (Int, Int)) -> Bool {
+        if idx == word.count {
+            return !isValid(i, j, board)
+        }
+
+        guard isValid(i, j, board) else {
+            return false
+        }
+
+        guard board[i][j] == " " || board[i][j] == word[idx] else {
+            return false
+        }
+
+        return dfs(idx + 1, word, board, i + dir.0, j + dir.1, dir)
+    }
+}