README.md

Word Search

Problem can be found in here!

Solution: Depth-First Search

def exist(board: List[List[str]], word: str) -> bool:
    def DFS(row: int, column: int, word_index: int) -> None:
        nonlocal is_word_exist
        # To avoid further function calls as we have already found the answer
        if is_word_exist:
            return
        # This means that we have found the answer
        if word_index == len(word):
            is_word_exist = True
            return
        # Check if we exceed boundaries
        if row == -1 or row == row_number or column == -1 or column == column_number:
            return

        # Store its value for restoring purpose
        data = board[row][column]
        # Update its value to empty char to avoid using same cell twice
        board[row][column] = ""
        if data == word[word_index]:
            DFS(row+1, column, word_index+1)
            DFS(row-1, column, word_index+1)
            DFS(row, column+1, word_index+1)
            DFS(row, column-1, word_index+1)
        # If no solution is found, then updated to its initial value
        board[row][column] = data

    is_word_exist = False
    row_number, column_number = len(board), len(board[0])
    for row in range(row_number):
        for column in range(column_number):
            if board[row][column] == word[0]:
                DFS(row, column, 0)
            if is_word_exist:
                break

    return is_word_exist

Time Complexity: , Space Complexity: , where m is the number of row, n is the number of column, and w is the length of word.