Word Search

Solution:

Scan matrix and pass board, coords and word to dfs recursive search for 'word' by letter
Pretty much identical to "Number of Islands" only difference is implementation of temp variable to hold the preserved board
As letters are found they are updated to '#' to prevent double counting a letter but if the recurvise search finds some but not all letters in a word we still need to be able to restore the board to continue the scan of the matrix once the 'dfs' function completely finishes

Output: True
class Solution:
    def exist(self, board, word):
        if not board:
            return False

        for i in range(len(board)):
            for j in range(len(board(i))):
                if self.dfs(board, i, j, word):
                    return True

    def dfs(self, board, i, j, word):
        if len(word) == 0
            return True

        if i < 0 or j < 0 or i >= len(board) or j >= len(board[i]) or word[0] != board[i][j]
            return False

        temp = board[i][j]
        boardp[i][j] = "#"

        result = self.dfs(board, i+1, j, word[1:]) or self.dfs(board, i-1, j, word[1:]) or self.dfs(board, i, j+1, word[1:]) or self.dfs(board, i, j-1, word[1:])
        board[i][j] = temp
        return result

board = [
    ["A","B","C","E"],["S","F","C","S"],["A","D","E","E"]
]
word = "ABCCED"

p1 = Solution()
print(p1.exist(board, word))