Add solution for Word Search problem using backtracking

Author: jamiebase

word-search/gcount85.py

@@ -0,0 +1,42 @@
+"""
+# Intuition
+백트래킹으로 풀었습니다.
+
+# Complexity
+- Time complexity: O(M*N*3^L) => M*N 보드 한 칸을 모두 방문하며,
+word 의 길이(L) 횟수 단계 만큼 탐색해야 한다. 매 단계 탐색할 때는 매번 3갈래씩 길을 찾는다.
+
+- Space complexity: 재귀 스택 O(L), visited 크기 O(L)
+"""
+
+
+class Solution:
+    def exist(self, board: list[list[str]], word: str) -> bool:
+        m, n = len(board[0]), len(board)
+        direction = [[0, 1], [0, -1], [1, 0], [-1, 0]]
+
+        def dfs(x, y, cur_step, visited):
+            if cur_step == len(word) - 1:
+                return True
+
+            for dx, dy in direction:
+                nx, ny = x + dx, y + dy
+                if nx < 0 or ny < 0 or nx >= n or ny >= m:
+                    continue
+                if board[nx][ny] != word[cur_step + 1]:
+                    continue
+                if (nx, ny) in visited:
+                    continue
+                visited.add((nx, ny))
+                if dfs(nx, ny, cur_step + 1, visited):
+                    return True
+                visited.discard((nx, ny))
+            return False
+
+        for x, row in enumerate(board):
+            for y, start_word in enumerate(row):
+                if word[0] != start_word:
+                    continue
+                if dfs(x, y, 0, {(x, y)}):
+                    return True
+        return False