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79. Word Search

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:

e1

Input: board = [["A","B","C","E"],["S","F","C","S"],["A","D","E","E"]], word = "ABCCED" Output: true

Example 2:

e2

Input: board = [["A","B","C","E"],["S","F","C","S"],["A","D","E","E"]], word = "SEE" Output: true

Example 3:

e3

Input: board = [["A","B","C","E"],["S","F","C","S"],["A","D","E","E"]], word = "ABCB" Output: false

Constraints:

  • m == board.length
  • n = board[i].length
  • 1 <= m, n <= 6
  • 1 <= word.length <= 15
  • board and word consists of only lowercase and uppercase English letters.

Follow up: Could you use search pruning to make your solution faster with a larger board?

Solution

class Solution:
    def exist(self, board: List[List[str]], word: str) -> bool:
        def dfs(row, col, index):
            if index == len(word):
                return True
            if row < 0 or col < 0 or row >= len(board) or col >= len(board[0]) or board[row][col] != word[index]:
                return False
            temp = board[row][col]
            board[row][col] = '#'  # Mark as visited
            found = (dfs(row + 1, col, index + 1) or dfs(row - 1, col, index + 1) or
                     dfs(row, col + 1, index + 1) or dfs(row, col - 1, index + 1))
            board[row][col] = temp  # Backtrack
            return found

        for row in range(len(board)):
            for col in range(len(board[0])):
                if dfs(row, col, 0):
                    return True
        return False

Thoughts

Time Complexity

The time complexity of this solution is O(M X N X 4^L), where M and N are the dimensions of the board, and L is the length of the word. This is because, in the worst case, we might need to explore all 4 directions for each character in the word.

Space Complexity

The space complexity is O(L) for the recursion stack, where L is the length of the word. In the worst case, the word could be as long as the number of cells in the board.