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.

solution

• this is a pretty classic dfs problem.
• we go through each character in the grid, and when we see a valid starting character, dfs through the entire grid to try to complete the target word.
• keep a visited set that we add and remove from (when we backtrack) between dfs calls.

def exist(self, board: List[List[str]], word: str) -> bool:
	dirs = [[0,1],[1,0],[0,-1],[-1,0]]
	visited = set()
 
	def dfs(i, j, word_idx):
		if word_idx == len(word)-1:
			return True
 
		res = False
		for d in dirs:
			ni, nj = i + d[0], j + d[1]
			if ni >= 0 and ni < len(board) and nj >= 0 and nj < len(board[0]):
					if (ni,nj) not in visited and board[ni][nj] == word[word_idx+1]:
						visited.add((ni,nj))
						res = res or dfs(ni, nj, word_idx+1)
						visited.remove((ni,nj))
		return res
 
	for i in range(len(board)):
		for j in range(len(board[0])):
			if board[i][j] == word[0]:
				visited.add((i,j))
				found = dfs(i, j, 0)
				if found:
					return True
				visited.remove((i,j))
	return False