Word Search – LeetCode Solution Java , Python 3, Python 2 , C , C++, Best and Optimal Solutions , All you need.
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.lengthn = board[i].length1 <= m, n <= 61 <= word.length <= 15boardandwordconsists of only lowercase and uppercase English letters.
C++ Word Search LeetCode Solution
bool exist(vector<vector<char>>& board, string word) {
for (unsigned int i = 0; i < board.size(); i++)
for (unsigned int j = 0; j < board[0].size(); j++)
if (dfs(board, i, j, word))
return true;
return false;
}
bool dfs(vector<vector<char>>& board, int i, int j, string& word) {
if (!word.size())
return true;
if (i<0 || i>=board.size() || j<0 || j>=board[0].size() || board[i][j] != word[0])
return false;
char c = board[i][j];
board[i][j] = '*';
string s = word.substr(1);
bool ret = dfs(board, i-1, j, s) || dfs(board, i+1, j, s) || dfs(board, i, j-1, s) || dfs(board, i, j+1, s);
board[i][j] = c;
return ret;
}
Java Word Search LeetCode Solution
public class Solution {
static boolean[][] visited;
public boolean exist(char[][] board, String word) {
visited = new boolean[board.length][board[0].length];
for(int i = 0; i < board.length; i++){
for(int j = 0; j < board[i].length; j++){
if((word.charAt(0) == board[i][j]) && search(board, word, i, j, 0)){
return true;
}
}
}
return false;
}
private boolean search(char[][]board, String word, int i, int j, int index){
if(index == word.length()){
return true;
}
if(i >= board.length || i < 0 || j >= board[i].length || j < 0 || board[i][j] != word.charAt(index) || visited[i][j]){
return false;
}
visited[i][j] = true;
if(search(board, word, i-1, j, index+1) ||
search(board, word, i+1, j, index+1) ||
search(board, word, i, j-1, index+1) ||
search(board, word, i, j+1, index+1)){
return true;
}
visited[i][j] = false;
return false;
}
}
Python 3 Word Search LeetCode Solution
def exist(self, board, word):
if not board:
return False
for i in xrange(len(board)):
for j in xrange(len(board[0])):
if self.dfs(board, i, j, word):
return True
return False
# check whether can find word, start at (i,j) position
def dfs(self, board, i, j, word):
if len(word) == 0: # all the characters are checked
return True
if i<0 or i>=len(board) or j<0 or j>=len(board[0]) or word[0]!=board[i][j]:
return False
tmp = board[i][j] # first character is found, check the remaining part
board[i][j] = "#" # avoid visit agian
# check whether can find "word" along one direction
res = self.dfs(board, i+1, j, word[1:]) or self.dfs(board, i-1, j, word[1:]) \
or self.dfs(board, i, j+1, word[1:]) or self.dfs(board, i, j-1, word[1:])
board[i][j] = tmp
return res
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