Word Search in Python
Suppose we have a 2D board and a word, we have to find if the word is present in the grid or not. The words can be made from letters of sequentially adjacent cell, "adjacent" cells are those horizontally or vertically neighboring cells. We should not use the same letter cell more than once. So if the matrix is like −
Given words are say “ABCCED”, the answer will be true, for word “SEE”, it will be true, but for “ABCB” if will be false.
Let us see the steps −
We will solve this using recursive approach. So if the recursive method name is called find(), this takes matrix mat, word, row, col and index i. Initially the index i = 0
if i = length of words, then return True
if row >= row count of mat or row < 0 or col >= col count of mat or col < 0 or word[i] is not same as mat[row, col], then return false
mat[row, col] := “*”
res := find(mat, word, row + 1, col, i + 1) or find(mat, word, row - 1, col, i + 1) or find(mat, word, row, col + 1, i + 1) or find(mat, word, row, col - 1, i + 1)
mat[row, col] := word[i]
return res
The main task will be performed like −
n := row count and m := column count
for i in range 0 to n
for j in range 0 to m
if word[0] = mat[i, j]
if find(mat, word, i, j) is not false, then return true
Example
Let us see the following implementation to get better understanding −
class Solution(object):
def exist(self, board, word):
"""
:type board: List[List[str]]
:type word: str
:rtype: bool
"""
n =len(board)
m = len(board[0])
for i in range(n):
for j in range(m):
if word[0] == board[i][j]:
if self.find(board,word,i,j):
return True
return False
def find(self, board,word,row,col,i=0):
#print(word[:i+1])
#print(board)
if i== len(word):
return True
if row>= len(board) or row <0 or col >=len(board[0]) or col<0 or word[i]!=board[row][col]:
return False
board[row][col] = '*'
res = self.find(board,word,row+1,col,i+1) or self.find(board,word,row-1,col,i+1) or self.find(board,word,row,col+1,i+1) or self.find(board,word,row,col-1,i+1)
board[row][col] = word[i]
return res
Input
[["A","B","C","E"],["S","F","C","S"],["A","D","E","E"]]
"SEE"
Output
true
Published on 03-Feb-2020 10:10:36
