Enhance class & function documentation in WordSearch.java (#5578)

Author: Hardvan

src/main/java/com/thealgorithms/backtracking/WordSearch.java

@@ -1,51 +1,72 @@
 package com.thealgorithms.backtracking;
 
-/*
-Word Search Problem (https://en.wikipedia.org/wiki/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 cell, where "adjacent" cells are
-those horizontally or vertically neighboring. The same letter cell may not be used more than once.
-
-For example,
-Given board =
-
-[
- ['A','B','C','E'],
- ['S','F','C','S'],
- ['A','D','E','E']
-]
-word = "ABCCED", -> returns true,
-word = "SEE", -> returns true,
-word = "ABCB", -> returns false.
-*/
-
-/*
-   Solution
-   Depth First Search in matrix (as multiple sources possible) with backtracking
-   like finding cycle in a directed graph. Maintain a record of path
-
-   Tx = O(m * n * 3^L): for each cell, we look at 3 options (not 4 as that one will be visited), we
-   do it L times Sx = O(L) : stack size is max L
-*/
-
+/**
+ * Word Search Problem
+ *
+ * This class solves the word search problem where given an m x n grid of characters (board)
+ * and a target word, the task is to check if the word exists in the grid.
+ * The word can be constructed from sequentially adjacent cells (horizontally or vertically),
+ * and the same cell may not be used more than once in constructing the word.
+ *
+ * Example:
+ * - For board =
+ *     [
+ *       ['A','B','C','E'],
+ *       ['S','F','C','S'],
+ *       ['A','D','E','E']
+ *     ]
+ *   and word = "ABCCED", -> returns true
+ *   and word = "SEE",    -> returns true
+ *   and word = "ABCB",   -> returns false
+ *
+ * Solution:
+ * - Depth First Search (DFS) with backtracking is used to explore possible paths from any cell
+ *   matching the first letter of the word. DFS ensures that we search all valid paths, while
+ *   backtracking helps in reverting decisions when a path fails to lead to a solution.
+ *
+ * Time Complexity: O(m * n * 3^L)
+ *  - m = number of rows in the board
+ *  - n = number of columns in the board
+ *  - L = length of the word
+ *  - For each cell, we look at 3 possible directions (since we exclude the previously visited direction),
+ *    and we do this for L letters.
+ *
+ * Space Complexity: O(L)
+ *  - Stack space for the recursive DFS function, where L is the maximum depth of recursion (length of the word).
+ */
 public class WordSearch {
     private final int[] dx = {0, 0, 1, -1};
     private final int[] dy = {1, -1, 0, 0};
     private boolean[][] visited;
     private char[][] board;
     private String word;
 
+    /**
+     * Checks if the given (x, y) coordinates are valid positions in the board.
+     *
+     * @param x The row index.
+     * @param y The column index.
+     * @return True if the coordinates are within the bounds of the board; false otherwise.
+     */
     private boolean isValid(int x, int y) {
         return x >= 0 && x < board.length && y >= 0 && y < board[0].length;
     }
 
+    /**
+     * Performs Depth First Search (DFS) from the cell (x, y)
+     * to search for the next character in the word.
+     *
+     * @param x       The current row index.
+     * @param y       The current column index.
+     * @param nextIdx The index of the next character in the word to be matched.
+     * @return True if a valid path is found to match the remaining characters of the word; false otherwise.
+     */
     private boolean doDFS(int x, int y, int nextIdx) {
         visited[x][y] = true;
         if (nextIdx == word.length()) {
             return true;
         }
+
         for (int i = 0; i < 4; ++i) {
             int xi = x + dx[i];
             int yi = y + dy[i];
@@ -56,10 +77,19 @@ private boolean doDFS(int x, int y, int nextIdx) {
                 }
             }
         }
-        visited[x][y] = false;
+
+        visited[x][y] = false; // Backtrack
         return false;
     }
 
+    /**
+     * Main function to check if the word exists in the board. It initiates DFS from any
+     * cell that matches the first character of the word.
+     *
+     * @param board The 2D grid of characters (the board).
+     * @param word  The target word to search for in the board.
+     * @return True if the word exists in the board; false otherwise.
+     */
     public boolean exist(char[][] board, String word) {
         this.board = board;
         this.word = word;