Word Search in Grid Using Backtracking in DSA C++ - 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 and word sizes increase.

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.

bool backtrack(vector<vector<char>>& board, string& word, int i, int j, int index) {
    if (index == word.size()) return true;
    if (i < 0 || j < 0 || i >= board.size() || j >= board[0].size() || board[i][j] != word[index]) return false;
    char temp = board[i][j];
    board[i][j] = '#';
    bool 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;
}

bool exist(vector<vector<char>>& board, string word) {
    for (int i = 0; i < board.size(); i++) {
        for (int j = 0; j < board[0].size(); j++) {
            if (backtrack(board, word, i, j, 0)) return true;
        }
    }
    return false;
}
    

This code tries to find a word in a 2D grid by exploring all possible paths using backtracking.

Identify Repeating Operations

Identify the loops, recursion, array traversals that repeat.

• Primary operation: Recursive backtracking calls exploring up to 4 directions for each letter in the word.
• How many times: For each cell in the grid (m x n), it may start a search that explores up to 4 choices per letter in the word (length k).

How Execution Grows With Input

As the grid size (m x n) and word length (k) grow, the search tries many paths.

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: The operations grow very fast because for each letter, up to 4 directions are explored, leading to exponential growth with word length.

Final Time Complexity

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

This means the time grows quickly with the word length, as each letter can lead to 4 new paths, and we try starting from every cell.

Common Mistake

[X] Wrong: "The search only takes time proportional to the grid size (m * n)."

[OK] Correct: Because for each cell, the search explores many possible paths for the word letters, causing exponential growth with word length, not just grid size.

Interview Connect

Understanding this complexity helps you explain why backtracking can be slow and when it is suitable, showing your grasp of algorithm behavior in real problems.

Self-Check

"What if we limited the search to only move right and down instead of all 4 directions? How would the time complexity change?"

Practice

1. What is the main purpose of backtracking in the Word Search in Grid problem?

easy

A. To explore all possible paths to find the word in the grid

B. To sort the grid alphabetically before searching

C. To count the number of words in the grid

D. To replace characters in the grid with special symbols

Solution

1. Step 1: Understand backtracking role

   Backtracking tries all possible paths in the grid to find the word by moving in allowed directions and undoing moves if they don't lead to a solution.

2. Step 2: Eliminate other options

   Sorting, counting words, or replacing characters are not related to backtracking's purpose in this problem.

3. Final Answer:

   To explore all possible paths to find the word in the grid -> Option A

4. Quick Check:

   Backtracking = Explore paths [OK]

Hint: Backtracking means trying all paths and undoing wrong steps [OK]

Common Mistakes:

• Thinking backtracking sorts or modifies the grid
• Confusing backtracking with simple searching
• Ignoring the undo step in backtracking

2. Which of the following is the correct way to check if a cell (row, col) is inside the grid boundaries in C++?

easy

A. if (row == 0 && col == 0)

B. if (row > 0 && row <= rows && col > 0 && col <= cols)

C. if (row >= 0 && row < rows && col >= 0 && col < cols)

D. if (row >= 0 || row < rows || col >= 0 || col < cols)

Solution

1. Step 1: Understand grid indexing

   Grid rows and columns start at 0, so valid indices are from 0 to rows-1 and 0 to cols-1.

2. Step 2: Check boundary condition

   if (row >= 0 && row < rows && col >= 0 && col < cols) correctly checks if row and col are within these ranges using && (and) operator.

3. Final Answer:

   if (row >= 0 && row < rows && col >= 0 && col < cols) -> Option C

4. Quick Check:

   Boundary check = if (row >= 0 && row < rows && col >= 0 && col < cols) [OK]

Hint: Use >= 0 and < size with && to check boundaries [OK]

Common Mistakes:

• Using > 0 instead of >= 0 excludes first row/column
• Using || instead of && allows invalid indices
• Checking only (0,0) cell

3. What will be the output of the following C++ code snippet for word search?

char grid[3][3] = {{'A','B','C'},{'D','E','F'},{'G','H','I'}};
string word = "BE";
// Assume function exist: bool exist(char grid[][], string word)
cout << (exist(grid, word) ? "Found" : "Not Found") << endl;

Assuming exist uses backtracking correctly.

medium

A. Found

B. Not Found

C. Compilation Error

D. Runtime Error

Solution

1. Step 1: Analyze grid and word

   The grid is:
   A B C
   D E F
   G H I
   The word is "BE".

2. Step 2: Check if "BE" can be formed

   Starting at 'B' (0,1), next letter 'E' must be adjacent (up, down, left, right). Adjacent cells to 'B' are 'A'(0,0), 'C'(0,2), 'E'(1,1). 'E' is adjacent, so path B->E exists.

3. Step 3: Confirm backtracking finds the word

   Backtracking will find "BE" successfully.

4. Final Answer:

   Found -> Option A

5. Quick Check:

   Word "BE" exists in grid = Found [OK]

Hint: Check adjacency of letters in grid for word presence [OK]

Common Mistakes:

• Ignoring adjacency rules
• Assuming diagonal moves allowed (usually not)
• Confusing output with error

4. Identify the bug in this backtracking snippet for word search:

bool backtrack(int r, int c, int index) {
  if (index == word.length()) return true;
  if (r < 0 || r >= rows || c < 0 || c >= cols) return false;
  if (grid[r][c] != word[index]) return false;
  grid[r][c] = '#';
  bool found = backtrack(r+1, c, index+1) || backtrack(r-1, c, index+1) || backtrack(r, c+1, index+1) || backtrack(r, c-1, index+1);
  return found;
}

medium

A. Incorrect boundary checks

B. Wrong base case condition

C. Using '||' instead of '&&' in recursive calls

D. Missing restoring grid[r][c] after recursion

Solution

1. Step 1: Understand backtracking marking

   The code marks visited cells by setting grid[r][c] = '#'.

2. Step 2: Check if grid is restored after recursion

   After recursive calls, the original character should be restored to allow other paths. This code does not restore it.

3. Step 3: Identify the bug

   Missing restoration causes wrong results or blocks other paths.

4. Final Answer:

   Missing restoring grid[r][c] after recursion -> Option D

5. Quick Check:

   Restore grid after recursion = Fix bug [OK]

Hint: Always undo changes after recursive calls in backtracking [OK]

Common Mistakes:

• Not restoring visited cells
• Wrong base case logic
• Incorrect boundary conditions

5. Given a 4x4 grid:

char grid[4][4] = {
  {'C','A','T','F'},
  {'B','G','E','S'},
  {'I','T','A','E'},
  {'S','E','R','T'}
};

Which of the following words can be found using backtracking moving only up, down, left, or right?
Select the correct option.

hard

A. "CATS"

B. "TEA"

C. "FEST"

D. "BAT"

Solution

1. Step 1: Check "CATS"

   "CATS" starts at (0,0) 'C', next 'A' at (0,1), next 'T' at (0,2), but no adjacent 'S' to 'T' at (0,2). Adjacent: 'A'(0,1), 'F'(0,3), 'E'(1,2). No path exists.

2. Step 2: Check "TEA"

   "TEA" letters: T, E, A.
   Possible path: (2,1) 'T' -> (3,1) 'E' -> (2,2) 'A' ? From 'E'(3,1) to 'A'(2,2) is diagonal, invalid.
   Try other 'T's: (0,2) 'T' -> (1,2) 'E' down -> (2,2) 'A' down. Valid path.

3. Step 3: Check "FEST" and "BAT"

   "FEST" letters not found in order with adjacency.
   "BAT" letters: B(1,0), A(2,2), T(0,2) no continuous path.

4. Final Answer:

   "TEA" -> Option B

5. Quick Check:

   Check adjacency for each word = "TEA" found [OK]

Hint: Trace letter adjacency step-by-step for each word [OK]

Common Mistakes:

• Assuming diagonal moves allowed
• Not checking letter order carefully
• Ignoring multiple occurrences of letters