In this guide, you will get 79. Word Search LeetCode Solution with the best time and space complexity. The solution to Word Search problem is provided in various programming languages like C++, Java, and Python. This will be helpful for you if you are preparing for placements, hackathons, interviews, or practice purposes. The solutions provided here are very easy to follow and include detailed explanations.
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?
Complexity Analysis
Time Complexity: O(mn4^{|\texttt{word}|})
Space Complexity: O(4^{|\texttt{word}|})
79. Word Search LeetCode Solution in C++
class Solution {
public:
bool exist(vector<vector<char>>& board, string word) {
for (int i = 0; i < board.size(); ++i)
for (int j = 0; j < board[0].size(); ++j)
if (dfs(board, word, i, j, 0))
return true;
return false;
}
private:
bool dfs(vector<vector<char>>& board, const string& word, int i, int j,
int s) {
if (i < 0 || i == board.size() || j < 0 || j == board[0].size())
return false;
if (board[i][j] != word[s] || board[i][j] == '*')
return false;
if (s == word.length() - 1)
return true;
const char cache = board[i][j];
board[i][j] = '*';
const bool isExist = dfs(board, word, i + 1, j, s + 1) ||
dfs(board, word, i - 1, j, s + 1) ||
dfs(board, word, i, j + 1, s + 1) ||
dfs(board, word, i, j - 1, s + 1);
board[i][j] = cache;
return isExist;
}
}; /* code provided by PROGIEZ */
79. Word Search LeetCode Solution in Java
class Solution {
public boolean exist(char[][] board, String word) {
for (int i = 0; i < board.length; ++i)
for (int j = 0; j < board[0].length; ++j)
if (dfs(board, word, i, j, 0))
return true;
return false;
}
private boolean dfs(char[][] board, String word, int i, int j, int s) {
if (i < 0 || i == board.length || j < 0 || j == board[0].length)
return false;
if (board[i][j] != word.charAt(s) || board[i][j] == '*')
return false;
if (s == word.length() - 1)
return true;
final char cache = board[i][j];
board[i][j] = '*';
final boolean isExist = dfs(board, word, i + 1, j, s + 1) || //
dfs(board, word, i - 1, j, s + 1) || //
dfs(board, word, i, j + 1, s + 1) || //
dfs(board, word, i, j - 1, s + 1);
board[i][j] = cache;
return isExist;
}
} // code provided by PROGIEZ
79. Word Search LeetCode Solution in Python
class Solution:
def exist(self, board: list[list[str]], word: str) -> bool:
m = len(board)
n = len(board[0])
def dfs(i: int, j: int, s: int) -> bool:
if i < 0 or i == m or j < 0 or j == n:
return False
if board[i][j] != word[s] or board[i][j] == '*':
return False
if s == len(word) - 1:
return True
cache = board[i][j]
board[i][j] = '*'
isExist = (dfs(i + 1, j, s + 1) or
dfs(i - 1, j, s + 1) or
dfs(i, j + 1, s + 1) or
dfs(i, j - 1, s + 1))
board[i][j] = cache
return isExist
return any(dfs(i, j, 0)
for i in range(m)
for j in range(n)) # code by PROGIEZ
