feat: [Week 04-3] solve word-search

Author: Chaedie

word-search/Chaedie.py

@@ -0,0 +1,126 @@
+"""
+Solution: 
+    1) board의 상하좌우를 탐색하되 아래 조건을 base case로 걸러준다.
+    1.1) index 가 word의 길이이면 결과값 판단
+    1.2) out of bounds 판단
+    1.3) index를 통해 현재 글자와 board의 글자의 일치 판단
+    1.4) 방문 여부 판단
+    2) board를 돌면서 backtrack 이 True 인 케이스가 있으면 return True
+
+m = row_len
+n = col_len
+L = 단어 길이
+Time: O(m n 4^L)
+Space: O(mn + L^2) = visit set O(mn) + 호출 스택 및 cur_word O(L^2)
+"""
+
+
+class Solution:
+    def exist(self, board: List[List[str]], word: str) -> bool:
+        ROWS, COLS = len(board), len(board[0])
+        visit = set()
+
+        def backtrack(r, c, index, cur_word):
+            if index == len(word):
+                return word == cur_word
+            if r < 0 or c < 0 or r == ROWS or c == COLS:
+                return False
+            if word[index] != board[r][c]:
+                return False
+            if (r, c) in visit:
+                return False
+
+            visit.add((r, c))
+            condition = (
+                backtrack(r + 1, c, index + 1, cur_word + board[r][c])
+                or backtrack(r - 1, c, index + 1, cur_word + board[r][c])
+                or backtrack(r, c + 1, index + 1, cur_word + board[r][c])
+                or backtrack(r, c - 1, index + 1, cur_word + board[r][c])
+            )
+            visit.remove((r, c))
+            return condition
+
+        for i in range(ROWS):
+            for j in range(COLS):
+                if backtrack(i, j, 0, ""):
+                    return True
+        return False
+
+
+"""
+Solution: 
+    공간 복잡도 낭비를 줄이기 위해 cur_word 제거
+Time: O(m n 4^L)
+Space: O(mn + L)
+"""
+
+
+class Solution:
+    def exist(self, board: List[List[str]], word: str) -> bool:
+        ROWS, COLS = len(board), len(board[0])
+        visit = set()
+
+        def backtrack(r, c, index):
+            if index == len(word):
+                return True
+            if r < 0 or c < 0 or r == ROWS or c == COLS:
+                return False
+            if word[index] != board[r][c]:
+                return False
+            if (r, c) in visit:
+                return False
+
+            visit.add((r, c))
+            condition = (
+                backtrack(r + 1, c, index + 1)
+                or backtrack(r - 1, c, index + 1)
+                or backtrack(r, c + 1, index + 1)
+                or backtrack(r, c - 1, index + 1)
+            )
+            visit.remove((r, c))
+            return condition
+
+        for i in range(ROWS):
+            for j in range(COLS):
+                if backtrack(i, j, 0):
+                    return True
+        return False
+
+
+"""
+Solution: 
+    공간 복잡도를 줄이기 위해 visit set 제거 
+    -> board[r][c]에 빈문자열을 잠깐 추가하는것으로 대체
+Time: O(m n 4^L)
+Space: O(L)
+"""
+
+
+class Solution:
+    def exist(self, board: List[List[str]], word: str) -> bool:
+        ROWS, COLS = len(board), len(board[0])
+
+        def backtrack(r, c, index):
+            if index == len(word):
+                return True
+            if r < 0 or c < 0 or r == ROWS or c == COLS:
+                return False
+            if word[index] != board[r][c]:
+                return False
+
+            temp = board[r][c]
+            board[r][c] = ""
+            condition = (
+                backtrack(r + 1, c, index + 1)
+                or backtrack(r - 1, c, index + 1)
+                or backtrack(r, c + 1, index + 1)
+                or backtrack(r, c - 1, index + 1)
+            )
+            board[r][c] = temp
+            return condition
+
+        for i in range(ROWS):
+            for j in range(COLS):
+                if backtrack(i, j, 0):
+                    return True
+        return False