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

Time Complexity: Word Search in Grid Using Backtracking

O(n * 4^k)

Understanding Time Complexity

We want to know how long it takes to find a word in a grid by trying all possible paths.

How does the search time grow as the grid or word gets bigger?

Scenario Under Consideration

Analyze the time complexity of the following code snippet.

bool dfs(char** board, int rows, int cols, int r, int c, char* word, int index) {
    if (index == strlen(word)) return true;
    if (r < 0 || c < 0 || r >= rows || c >= cols || board[r][c] != word[index]) return false;
    char temp = board[r][c];
    board[r][c] = '#';
    bool found = dfs(board, rows, cols, r+1, c, word, index+1) ||
                 dfs(board, rows, cols, r-1, c, word, index+1) ||
                 dfs(board, rows, cols, r, c+1, word, index+1) ||
                 dfs(board, rows, cols, r, c-1, word, index+1);
    board[r][c] = temp;
    return found;
}

bool exist(char** board, int rows, int cols, char* word) {
    for (int i = 0; i < rows; i++) {
        for (int j = 0; j < cols; j++) {
            if (dfs(board, rows, cols, i, j, word, 0)) return true;
        }
    }
    return false;
}
    

This code tries to find the word in the grid by searching from each cell and exploring all directions recursively.

Identify Repeating Operations

Identify the loops, recursion, array traversals that repeat.

• Primary operation: Recursive depth-first search (DFS) exploring up to 4 directions from each cell.
• How many times: For each cell in the grid (rows * cols), DFS explores paths up to the length of the word (k), branching up to 4 directions at each step.

How Execution Grows With Input

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

Input Size (n = rows*cols, k = word length)	Approx. Operations
n=10, k=3	About 10 * 4^3 = 640
n=100, k=5	About 100 * 4^5 = 102,400
n=1000, k=6	About 1000 * 4^6 = 4,096,000

Pattern observation: The operations grow exponentially with the word length and linearly with the grid size.

Final Time Complexity

Time Complexity: O(n * 4^k)

This means the search time grows linearly with the number of cells and exponentially with the word length.

Common Mistake

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

[OK] Correct: The search explores many paths from each cell, branching multiple times, so it repeats work exponentially with word length.

Interview Connect

Understanding this complexity helps you explain how backtracking explores many possibilities and why pruning or constraints matter in real problems.

Self-Check

"What if we limited the search to only move right and down? How would the time complexity change?"

Practice

1. What is the main purpose of using backtracking in a word search grid problem?

easy

A. To sort the grid alphabetically before searching

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

C. To count the number of words in the grid

D. To replace characters in the grid with a placeholder

Solution

1. Step 1: Understand backtracking in grid search

   Backtracking tries all possible paths from a starting point to find the target word.

2. Step 2: Compare options with backtracking purpose

   Only To explore all possible paths to find the word in the grid describes exploring all paths, which matches backtracking's goal.

3. Final Answer:

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

4. Quick Check:

   Backtracking = explore all paths [OK]

Hint: Backtracking means trying all paths step-by-step [OK]

Common Mistakes:

• Thinking backtracking sorts or counts words
• Confusing backtracking with simple search
• Assuming backtracking modifies the grid permanently

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 >= 1 && row < rows && col >= 1 && col < cols)

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 zero-based indexing in C arrays

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

2. Step 2: Check each option for correct boundary conditions

   if (row >= 0 && row < rows && col >= 0 && col < cols) correctly uses >= 0 and < rows/cols. Others use > 0 or <= rows which are incorrect for zero-based arrays.

3. Final Answer:

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

4. Quick Check:

   Check boundaries with >=0 and < size [OK]

Hint: Use zero-based index: start at 0, less than size [OK]

Common Mistakes:

• Using > 0 instead of >= 0 excludes first row/column
• Using <= rows includes out-of-bound index
• Mixing 1-based and 0-based indexing

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

char grid[3][3] = {{'A','B','C'},{'D','E','F'},{'G','H','I'}};
int found = searchWord(grid, 3, 3, "BEF");
printf("%d", found);

Assuming searchWord correctly implements backtracking to find the word in the grid.

medium

A. 1

B. 0

C. Compilation error

D. Runtime error

Solution

1. Step 1: Analyze the grid and the word "BEF"

   The grid rows are: ABC, DEF, GHI. The word "BEF" requires letters B, E, F in sequence adjacent horizontally or vertically.

2. Step 2: Check if "BEF" exists in adjacent cells

   B is at (0,1), E is at (1,1), F is at (1,2). B to E is vertical down, E to F is horizontal right, so path exists.

3. Step 3: Confirm if backtracking finds the word

   Backtracking can follow B(0,1) -> E(1,1) -> F(1,2), so word is found.

4. Final Answer:

   1 -> Option A

5. Quick Check:

   Word found = 1 [OK]

Hint: Trace letters stepwise in grid adjacency [OK]

Common Mistakes:

• Assuming letters must be in same row only
• Ignoring vertical or diagonal adjacency rules
• Confusing output 0 and 1 meaning

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

bool dfs(char grid[][], int row, int col, int index, char* word) {
  if (index == strlen(word)) return true;
  if (row < 0 || row >= rows || col < 0 || col >= cols) return false;
  if (grid[row][col] != word[index]) return false;
  char temp = grid[row][col];
  grid[row][col] = '#';
  bool found = dfs(grid, row+1, col, index+1, word) ||
               dfs(grid, row-1, col, index+1, word) ||
               dfs(grid, row, col+1, index+1, word) ||
               dfs(grid, row, col-1, index+1, word);
  grid[row][col] = temp;
  return found;
}

medium

A. The base case should check index > strlen(word) instead of ==

B. The grid cell is not restored after recursion

C. The grid parameter lacks column size, causing compilation error

D. The function does not check diagonal neighbors

Solution

1. Step 1: Check function parameter declaration

   In C, 2D arrays must have fixed column size in parameters, e.g., char grid[][cols]. Missing column size causes compilation error.

2. Step 2: Verify other parts of the code

   Base case with index == strlen(word) is correct. Grid cell is restored after recursion. Diagonal check is optional, not a bug.

3. Final Answer:

   The grid parameter lacks column size, causing compilation error -> Option C

4. Quick Check:

   2D array params need column size [OK]

Hint: Always specify column size in 2D array parameters [OK]

Common Mistakes:

• Ignoring 2D array parameter syntax in C
• Thinking base case condition is wrong
• Assuming diagonal moves are mandatory

5. Given a 4x4 grid:

{ {'C','A','T','F'},
  {'B','G','E','S'},
  {'I','T','A','E'},
  {'S','O','N','G'} }

Which of the following words can be found using backtracking with adjacent horizontal or vertical moves only?

hard

A. "SONG"

B. "BATS"

C. "CATS"

D. "FATE"

Solution

1. Step 1: Check each word's presence in the grid with adjacency rules

   "SONG": S(3,0), O(3,1), N(3,2), G(3,3) - all adjacent horizontally, so possible.
   "BATS": B(1,0) has no adjacent A (neighbors: C(0,0), I(2,0), G(1,1)), so no.
   "CATS": C(0,0), A(0,1), T(0,2) adjacent, but T(0,2) has no adjacent S (neighbors: A(0,1), F(0,3), E(1,2), A(2,2)), so no.
   "FATE": F(0,3) has no adjacent A (neighbors: T(0,2), S(1,3)), so no.

2. Step 2: Identify which word is fully connected by adjacent moves

   Only "SONG" has all letters adjacent in sequence.

3. Final Answer:

   "SONG" -> Option A

4. Quick Check:

   Check adjacency path for each word [OK]

Hint: Trace letters stepwise horizontally or vertically [OK]

Common Mistakes:

• Assuming diagonal moves allowed
• Ignoring adjacency between letters
• Choosing words with missing adjacent letters