Word Search in Grid Using Backtracking in DSA Java - Build from Scratch

Word Search in Grid Using Backtracking

📖 Scenario: You are building a simple word search puzzle solver. The puzzle is a grid of letters, and you want to find if a given word exists in the grid by moving horizontally or vertically to adjacent letters.This is like searching for a hidden word in a crossword puzzle.

🎯 Goal: Build a Java program that checks if a given word can be found in the letter grid by moving up, down, left, or right without revisiting the same cell.

📋 What You'll Learn

Create a 2D character array called board with the exact letters given.

Create a string variable called word with the exact value given.

Implement a backtracking method called exist that returns true if the word is found in the grid.

Use a helper method called dfs to explore the grid recursively.

Print true or false depending on whether the word exists in the grid.

💡 Why This Matters

🌍 Real World

Word search puzzles are popular games and educational tools. This algorithm helps in solving such puzzles automatically.

💼 Career

Backtracking and grid search algorithms are common in technical interviews and real-world problems like pathfinding, puzzle solving, and game development.

1

Create the letter grid

Create a 2D character array called board with these exact rows:
{'A','B','C','E'},
{'S','F','C','S'},
{'A','D','E','E'}

DSA Java

// Create the 2D character array board
// Your code here

Hint

Use curly braces to create rows inside the 2D array.

2

Add the word to search

Add a string variable called word and set it to the exact value "ABCCED" inside the main method after the board declaration.

DSA Java

public class WordSearch {
    public static void main(String[] args) {
        char[][] board = {
            {'A','B','C','E'},
            {'S','F','C','S'},
            {'A','D','E','E'}
        };
        // Create the string variable word
        // Your code here
    }
}

Hint

Use double quotes to create the string.

3

Implement the backtracking search

Add a static method exist that takes char[][] board and String word and returns boolean. Inside main, call exist(board, word) and store the result in a boolean variable found. Implement a helper method dfs to do the recursive search. Use backtracking to mark visited cells. The search moves up, down, left, and right.

DSA Java

public class WordSearch {
    public static void main(String[] args) {
        char[][] board = {
            {'A','B','C','E'},
            {'S','F','C','S'},
            {'A','D','E','E'}
        };
        String word = "ABCCED";
        // Call exist and store result in found
        // Your code here
    }

    // Add exist and dfs methods below
    // Your code here
}

Hint

Use a 2D boolean array to track visited cells. Recursively check neighbors for the next character.

4

Print the search result

Add a print statement in main to display the value of the boolean variable found.

DSA Java

public class WordSearch {
    public static void main(String[] args) {
        char[][] board = {
            {'A','B','C','E'},
            {'S','F','C','S'},
            {'A','D','E','E'}
        };
        String word = "ABCCED";
        boolean found = exist(board, word);
        // Print the value of found
        // Your code here
    }

    public static boolean exist(char[][] board, String word) {
        int rows = board.length;
        int cols = board[0].length;
        boolean[][] visited = new boolean[rows][cols];
        for (int i = 0; i = board.length || col = board[0].length) {
            return false;
        }
        if (visited[row][col] || board[row][col] != word.charAt(index)) {
            return false;
        }
        visited[row][col] = true;
        boolean found = dfs(board, word, index + 1, row - 1, col, visited) ||
                        dfs(board, word, index + 1, row + 1, col, visited) ||
                        dfs(board, word, index + 1, row, col - 1, visited) ||
                        dfs(board, word, index + 1, row, col + 1, visited);
        visited[row][col] = false;
        return found;
    }
}

Hint

Use System.out.println(found); to print the result.

Practice

1. What is the main purpose of backtracking in the Word Search problem on a grid?

easy

A. To sort the grid alphabetically

B. To explore all possible paths to find the word

C. To count the number of words in the grid

D. To replace characters in the grid

Solution

1. Step 1: Understand backtracking role

   Backtracking tries all possible paths in the grid to find the target word by moving in allowed directions.

2. Step 2: Eliminate other options

   Sorting, counting, or replacing characters are unrelated to searching paths for the word.

3. Final Answer:

   To explore all possible paths to find the word -> Option B

4. Quick Check:

   Backtracking = Explore paths [OK]

Hint: Backtracking means trying all paths until word found [OK]

Common Mistakes:

• Thinking backtracking sorts or modifies the grid
• Confusing backtracking with simple search
• Assuming backtracking counts words instead of searching

2. Which of the following Java method signatures correctly represents a backtracking helper for word search in a grid?

easy

A. boolean backtrack(char[][] board, String word, int index, int row, int col)

B. void backtrack(char[] board, String word, int index)

C. int backtrack(char[][] board, String word, int row, int col)

D. String backtrack(char[][] board, String word, int index, int row, int col)

Solution

1. Step 1: Check method parameters

   The method needs the board (2D char array), the word, current index in word, and current position (row, col) to explore.

2. Step 2: Check return type

   Backtracking returns boolean to indicate if word is found along this path.

3. Final Answer:

   boolean backtrack(char[][] board, String word, int index, int row, int col) -> Option A

4. Quick Check:

   Correct signature = boolean backtrack(char[][] board, String word, int index, int row, int col) [OK]

Hint: Backtracking returns boolean and needs position + index [OK]

Common Mistakes:

• Using 1D array instead of 2D for board
• Wrong return type like int or String
• Missing index or position parameters

3. Given the grid:

[['A','B','C','E'],
 ['S','F','C','S'],
 ['A','D','E','E']]

and the word "ABCCED", what will the backtracking function return?

medium

A. throws Exception

B. false

C. null

D. true

Solution

1. Step 1: Trace the word in the grid

   The word "ABCCED" can be found by path: A(0,0) -> B(0,1) -> C(0,2) -> C(1,2) -> E(2,2) -> D(2,1).

2. Step 2: Backtracking confirms path exists

   Since the path exists without revisiting cells, backtracking returns true.

3. Final Answer:

   true -> Option D

4. Quick Check:

   Word found = true [OK]

Hint: Trace path step-by-step to confirm word presence [OK]

Common Mistakes:

• Assuming word is not found without checking all directions
• Confusing similar letters in different positions
• Ignoring backtracking's ability to explore multiple paths

4. Consider this backtracking snippet for word search:

if (row < 0 || col < 0 || row >= board.length || col >= board[0].length || board[row][col] != word.charAt(index)) {
    return false;
}

What is the main issue if we forget to mark visited cells during recursion?

medium

A. The function may revisit the same cell causing infinite loops

B. The function will always return true

C. The function will throw an ArrayIndexOutOfBoundsException

D. The function will skip valid paths

Solution

1. Step 1: Understand visited marking role

   Marking visited cells prevents revisiting the same cell in the current path, avoiding cycles.

2. Step 2: Consequence of missing visited marking

   Without marking, recursion may revisit cells endlessly, causing infinite loops or stack overflow.

3. Final Answer:

   The function may revisit the same cell causing infinite loops -> Option A

4. Quick Check:

   Missing visited mark = infinite loops [OK]

Hint: Always mark visited cells to avoid loops [OK]

Common Mistakes:

• Assuming no visited marking is safe
• Confusing out-of-bound checks with visited logic
• Thinking missing visited causes exceptions instead of loops

5. You want to find if the word "SEE" exists in the grid:

[['A','B','C','E'],
 ['S','F','E','S'],
 ['A','D','E','E']]

Using backtracking, which approach correctly handles multiple possible paths and avoids revisiting cells?

hard

A. Only search in one direction (right) to reduce complexity

B. Modify the board characters to '*' when visited and never revert them

C. Use a boolean visited[][] array to mark cells during recursion and backtrack by unmarking after recursion returns

D. Use a global counter to limit recursion depth

Solution

1. Step 1: Understand visited marking with backtracking

   Using a boolean visited array allows marking cells as visited during recursion and unmarking them after exploring all paths from that cell.

2. Step 2: Evaluate other options

   Modifying board permanently breaks original data, searching only right misses paths, and limiting recursion depth does not solve revisiting.

3. Final Answer:

   Use a boolean visited[][] array to mark cells during recursion and backtrack by unmarking after recursion returns -> Option C

4. Quick Check:

   Visited array + backtrack unmark = correct approach [OK]

Hint: Mark visited with boolean array and unmark after recursion [OK]

Common Mistakes:

• Permanently changing board characters
• Ignoring directions other than right
• Using recursion limits instead of visited marking