LC 79 - Word Search

Question

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

Example 2:

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

Example 3:

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?

Links

Question here and solution here

Solution

concept

We use backtracking on every cells. Each backtracking is essentially a DFS on 4 directions.

code

class Solution:
    def exist(self, board: List[List[str]], word: str) -> bool:
        ROW, COL = len(board), len(board[0])
        path = set()

        def dfs(r, c, idx):
            if idx == len(word):
                return True
            if (r < 0 or c < 0 or
                r >= ROW or c >= COL or
                board[r][c] != word[idx] or
                (r,c) in path):
                return False

            path.add((r,c))
            ans = (dfs(r + 1, c, idx + 1) or
                    dfs(r - 1, c, idx + 1) or
                    dfs(r, c + 1, idx + 1) or
                    dfs(r , c - 1, idx + 1))
            path.remove((r,c))
            return ans

        for r in range(ROW):
            for c in range(COL):
                exist = dfs(r, c, 0)
                if exist:
                    return True
        return False

Complexity

time: $O(m*4^n)$, where $m$ is the number of cells in the board and $n$ is the length of the word.
space: $O(n)$