Word Search in Grid Using Backtracking in DSA Java - Interactive Practice

Practice - 5 Tasks

1fill in blank

easy

Complete the code to start the search from the first cell in the grid.

DSA Java

for (int i = 0; i < board.length; i++) {
    for (int j = 0; j < board[0].length; j++) {
        if (exist(board, word, i, j, 0)) {
            return [1];
        }
    }
}
return false;

Drag options to blanks, or click blank then click option'

A0

Bfalse

Ctrue

Dnull

Attempts:

3 left

💡 Hint

Common Mistakes

Returning false inside the loop instead of true.

Returning 0 or null which are invalid for boolean return type.

If the word is found starting from any cell, we return true to indicate success.

2fill in blank

medium

Complete the base case to check if the entire word has been matched.

DSA Java

if (index == word.length()) {
    return [1];
}

Drag options to blanks, or click blank then click option'

Atrue

Bword.length()

Cindex

Dfalse

Attempts:

3 left

💡 Hint

Common Mistakes

Returning false in the base case.

Returning the index or word length instead of a boolean.

When index equals the word length, it means all characters matched, so return true.

3fill in blank

hard

Fix the error in boundary check to avoid ArrayIndexOutOfBoundsException.

DSA Java

if (row < 0 || row >= board.length || col < 0 || col >= [1] || board[row][col] != word.charAt(index)) {
    return false;
}

Drag options to blanks, or click blank then click option'

Aword.length()

Bboard[0].length

Cboard.length

Dindex

Attempts:

3 left

💡 Hint

Common Mistakes

Using board.length for columns which is the row count.

Using word length or index which are unrelated.

The number of columns is board[0].length, so col must be less than that to be valid.

4fill in blank

hard

Fill both blanks to mark the cell as visited and then restore it after backtracking.

DSA Java

char temp = board[row][col];
board[row][col] = [1];
boolean found = exist(board, word, row + 1, col, index + 1) ||
                exist(board, word, row - 1, col, index + 1) ||
                exist(board, word, row, col + 1, index + 1) ||
                exist(board, word, row, col - 1, index + 1);
board[row][col] = [2];
return found;

Drag options to blanks, or click blank then click option'

A'#'

Btemp

C'*'

D' '

Attempts:

3 left

💡 Hint

Common Mistakes

Not restoring the original character after recursion.

Using the wrong character to mark visited cells.

We mark the cell with '#' to indicate visited, then restore original char stored in temp after recursion.

5fill in blank

hard

Fill all three blanks to complete the recursive calls for all four directions.

DSA Java

return exist(board, word, row + [1], col + [2], index + 1) ||
       exist(board, word, row + [3], col, index + 1) ||
       exist(board, word, row, col + 1, index + 1) ||
       exist(board, word, row, col - 1, index + 1);

Drag options to blanks, or click blank then click option'

A1

B-1

C0

D2

Attempts:

3 left

💡 Hint

Common Mistakes

Using wrong increments causing out of bounds or skipping cells.

Not covering all four directions.

We move down (row+1,col), up (row-1,col), right (row,col+1), and left (row,col-1).

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