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WordSearch.java
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WordSearch.java
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package by.andd3dfx.search.wordsearch;
/**
* <pre>
* <a href="https://leetcode.com/problems/word-search/">Task description</a>
*
* 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
* </pre>
*
* @see <a href="https://youtu.be/FsKU04anMtE">Video solution</a>
*/
public class WordSearch {
public boolean exist(char[][] board, String word) {
var m = board.length;
var n = board[0].length;
for (int i = 0; i < m; i++) {
for (int j = 0; j < n; j++) {
if (checkIsCharacterSuitable(board, i, j, word, 0)) {
boolean[][] visited = new boolean[m][n];
visited[i][j] = true;
if (findNextCharacter(board, i, j, word, 1, visited)) {
return true;
}
}
}
}
return false;
}
private boolean findNextCharacter(char[][] board, int i, int j, String word, int pos, boolean[][] visited) {
if (pos >= word.length()) {
return true;
}
if (checkNewPos(board, i - 1, j, word, pos, visited)) {
return true;
}
if (checkNewPos(board, i, j - 1, word, pos, visited)) {
return true;
}
if (checkNewPos(board, i + 1, j, word, pos, visited)) {
return true;
}
if (checkNewPos(board, i, j + 1, word, pos, visited)) {
return true;
}
return false;
}
private boolean checkNewPos(char[][] board, int i, int j, String word, int pos, boolean[][] visited) {
var m = board.length;
var n = board[0].length;
if (i < 0 || j < 0 || i >= m || j >= n) {
return false;
}
if (!visited[i][j] && checkIsCharacterSuitable(board, i, j, word, pos)) {
visited[i][j] = true;
if (findNextCharacter(board, i, j, word, pos + 1, visited)) {
return true;
}
visited[i][j] = false;
}
return false;
}
private boolean checkIsCharacterSuitable(char[][] board, int i, int j, String word, int pos) {
return board[i][j] == word.charAt(pos);
}
}