Word Search in Grid Using Backtracking in DSA Java - Time & Space Complexity

Time Complexity: Word Search in Grid Using Backtracking

O(m * n * 4^k)

Understanding Time Complexity

We want to understand how the time needed to find a word in a grid grows as the grid or word size increases.

How does the search effort change when the grid or word gets bigger?

Scenario Under Consideration

Analyze the time complexity of the following code snippet.

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 (backtrack(board, word, i, j, 0)) return true;
        }
    }
    return false;
}

private boolean backtrack(char[][] board, String word, int i, int j, int index) {
    if (index == word.length()) return true;
    if (i < 0 || j < 0 || i >= board.length || j >= board[0].length || board[i][j] != word.charAt(index)) return false;
    char temp = board[i][j];
    board[i][j] = '#';
    boolean found = backtrack(board, word, i + 1, j, index + 1) ||
                    backtrack(board, word, i - 1, j, index + 1) ||
                    backtrack(board, word, i, j + 1, index + 1) ||
                    backtrack(board, word, i, j - 1, index + 1);
    board[i][j] = temp;
    return found;
}

This code tries to find a word in a 2D grid by checking each cell and exploring neighbors recursively.

Identify Repeating Operations

Identify the loops, recursion, array traversals that repeat.

• Primary operation: Recursive backtracking exploring up to 4 directions per character.
• How many times: For each cell in the grid (m x n), it may explore up to 4 recursive calls per character in the word (length k).

How Execution Grows With Input

As the grid size or word length grows, the search tries many paths, growing very fast.

Input Size (m x n, k)	Approx. Operations
3 x 3 grid, word length 3	Up to 9 * 4^3 = 576
5 x 5 grid, word length 5	Up to 25 * 4^5 = 25 * 1024 = 25600
10 x 10 grid, word length 6	Up to 100 * 4^6 = 100 * 4096 = 409600

Pattern observation: Operations grow exponentially with word length and linearly with grid size.

Final Time Complexity

Time Complexity: O(m * n * 4^k)

This means the time grows quickly with the word length and also depends on the grid size.

Common Mistake

[X] Wrong: "The search only checks each cell once, so it is O(m * n)."

[OK] Correct: The search explores many paths recursively for each character, causing exponential growth with word length, not just linear with grid size.

Interview Connect

Understanding this complexity helps you explain how backtracking explores many possibilities and why it can be slow for long words or big grids.

Self-Check

"What if we limited the search to only move right and down instead of all four directions? How would the time complexity change?"

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