add Word Search

Author: HoonDongKang

word-search/HoonDongKang.ts

@@ -0,0 +1,83 @@
+/**
+ * [Problem]: [79] Word Search
+ *
+ * (https://leetcode.com/problems/word-search/description/)
+ */
+
+function exist(board: string[][], word: string): boolean {
+    //시간복잡도 O(m * n * 4^w)
+    //공간복잡도 O(w)
+    function dfsChangingBoard(board: string[][], word: string): boolean {
+        const rows = board.length;
+        const cols = board[0].length;
+
+        for (let i = 0; i < rows; i++) {
+            for (let j = 0; j < cols; j++) {
+                if (dfs(i, j, 0)) return true;
+            }
+        }
+
+        return false;
+
+        function dfs(i: number, j: number, k: number): boolean {
+            if (k === word.length) return true;
+
+            if (i < 0 || j < 0) return false;
+            if (i >= rows || j >= cols) return false;
+            if (board[i][j] !== word[k]) return false;
+
+            const temp = board[i][j];
+            board[i][j] = "";
+
+            const result =
+                dfs(i + 1, j, k + 1) ||
+                dfs(i - 1, j, k + 1) ||
+                dfs(i, j + 1, k + 1) ||
+                dfs(i, j - 1, k + 1);
+
+            board[i][j] = temp;
+            return result;
+        }
+    }
+
+    //시간복잡도 O(m * n * 4^w)
+    //공간복잡도 O(w)
+    function dfsMaintainingBoard(board: string[][], word: string): boolean {
+        const rows = board.length;
+        const cols = board[0].length;
+
+        // array보다 Set<string>이 더 효율적
+        // Set.has: O(1)  / Array.includes: O(n)
+        const passed = new Set<string>();
+
+        for (let i = 0; i < rows; i++) {
+            for (let j = 0; j < cols; j++) {
+                if (dfs(i, j, 0)) return true;
+            }
+        }
+
+        return false;
+
+        function dfs(i: number, j: number, k: number): boolean {
+            if (k === word.length) return true;
+
+            if (i < 0 || j < 0) return false;
+            if (i >= rows || j >= cols) return false;
+            if (board[i][j] !== word[k]) return false;
+            if (passed.has(`${i},${j}`)) return false;
+
+            passed.add(`${i},${j}`);
+
+            const result =
+                dfs(i + 1, j, k + 1) ||
+                dfs(i - 1, j, k + 1) ||
+                dfs(i, j + 1, k + 1) ||
+                dfs(i, j - 1, k + 1);
+
+            passed.delete(`${i},${j}`);
+            return result;
+        }
+    }
+
+    return dfsMaintainingBoard(board, word);
+}