Written by Ron on 010111 April 000112011. Posted in Blog

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

from typing import List
from collections import Counter

class Solution:
    def exist(self, board: List[List[str]], word: str) -> bool:
        
        R = len(board)
        C = len(board[0])
        
        # if word > board
        if len(word) > R*C: return False
        
        # count chars board < chars word
        count = Counter(sum(board, []))
        for c, countWord in Counter(word).items():
            if count[c] < countWord: return False
                        
        seen = set()
        
        def dfs(r, c, i):
            if i == len(word): return True
            
            print(r, c, i, seen)
            
            if r < 0  \
            or c < 0  \
            or r >= R \
            or c >= C \
            or word[i] != board[r][c] \
            or (r,c) in seen:
                return False
                
            print(r, c, i, seen, "find")
            
            seen.add((r,c))
            res = (
                dfs(r+1,c,i+1) or 
                dfs(r-1,c,i+1) or
                dfs(r,c+1,i+1) or
                dfs(r,c-1,i+1) 
            )
            print("res", res)
            seen.remove((r,c))
            print(seen)

            return res
        
        for r in range(R):
            for c in range(C):
                print("---")
                if dfs(r,c,0): return True
                
        return False
    
       
s = Solution()       

print(s.exist([
["C","A","A"],
["A","A","A"],
["B","C","D"]], "AAB"))#true

print(s.exist([["a","a"]], "a"))#true
print(s.exist([
["A","B","C","E"],
["S","F","C","S"],
["A","D","E","E"]], "ABCCED"))#true
print(s.exist([
["A","B","C","E"],
["S","F","C","S"],
["A","D","E","E"]], "SEE"))#true
print(s.exist([
["A","B","C","E"],
["S","F","C","S"],
["A","D","E","E"]], "ABCB"))#false