--- /dev/null	Thu Jan 01 00:00:00 1970 +0000
+++ b/async_multi_threads/alphabet_search.py	Thu Dec 25 20:07:46 2025 -0800
@@ -0,0 +1,56 @@
+# Given an m x n grid of characters board and a string word, return true if word exists in the grid.
+# 
+# The word can be constructed from letters of sequentially adjacent cells, where adjacent cells are horizontally or vertically neighboring. The same letter cell may not be used more than once.
+# 
+#  
+# 
+# Example 1:
+# 
+# 
+# Input: board = [
+#   ["A","B","C","E"],
+#   ["S","F","C","S"],
+#   ["A","D","E","E"]
+# ], word = "ABCCED"
+# Output: true
+
+def main(board, word):
+    ROWS = len(board)
+    COLS = len(board[0])
+
+    pos = 0
+
+    directions = [
+        (0, 1),
+        (1, 0),
+        (0, -1),
+        (-1, 0),
+    ]
+    def dfs(row, col, pos):
+        # import pdb; pdb.set_trace()
+        if pos == len(word):
+            return True
+
+        if (row < 0 or row >= ROWS) or (col < 0 or col >= COLS) or board[row][col] == "#":
+            return False
+
+        letter = board[row][col]
+        if letter == word[pos]:
+            pos += 1
+            board[row][col] = "#"
+            for direction in directions:
+                if dfs(row + direction[0], col + direction[1], pos):
+                    return True
+            board[row][col] = letter
+
+    for r in range(ROWS):
+        for c in range(COLS):
+            if dfs(r, c, 0):
+                return True
+    return False
+
+
+board = [["b"],["a"],["b"],["b"],["a"]]
+word = "baa"
+
+print(main(board, word))