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

Practice - 5 Tasks

1fill in blank

easy

Complete the code to check if the current cell matches the current character of the word.

DSA C

if (board[row][col] == [1]) {
    // proceed with backtracking
}

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

A'*'

Bword[0]

C'\0'

Dword[index]

Attempts:

3 left

💡 Hint

Common Mistakes

Comparing with word[0] always checks only the first character.

Using '*' or '\0' does not check the correct character.

We compare the board cell with the current character of the word at position index.

2fill in blank

medium

Complete the code to mark the current cell as visited by replacing it with a special character.

DSA C

char temp = board[row][col];
board[row][col] = [1];

// backtracking calls

board[row][col] = temp;

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

A'*'

B'#'

C'\0'

D' '

Attempts:

3 left

💡 Hint

Common Mistakes

Using '\0' can cause string termination issues.

Using ' ' (space) might be confusing if spaces are in the grid.

We mark the cell as visited by replacing it with '*', a character not in the word.

3fill in blank

hard

Fix the error in the recursive call to explore the next character in the word.

DSA C

if (backtrack(board, word, row + 1, col, [1], rows, cols) ||
    backtrack(board, word, row - 1, col, [1], rows, cols) ||
    backtrack(board, word, row, col + 1, [1], rows, cols) ||
    backtrack(board, word, row, col - 1, [1], rows, cols)) {
    return 1;
}

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

Aindex - 1

Bindex

Cindex + 1

D0

Attempts:

3 left

💡 Hint

Common Mistakes

Using index without increment causes infinite loops.

Decrementing index goes backward in the word.

We move to the next character in the word by increasing index by 1.

4fill in blank

hard

Fill both blanks to check if the current position is out of bounds or the cell is already visited.

DSA C

if (row [1] 0 || row [2] rows || col < 0 || col >= cols || board[row][col] == '*') {
    return 0;
}

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

A<

B>=

C<=

D>

Attempts:

3 left

💡 Hint

Common Mistakes

Using '<=' for lower bound allows -1 which is invalid.

Using '>' for upper bound misses the last valid index.

We check if row is less than 0 or greater or equal to rows to stay inside the grid.

5fill in blank

hard

Fill all three blanks to complete the main function that starts the search for the word in the grid.

DSA C

int exist(char** board, int boardSize, int* boardColSize, char* word) {
    for (int i = 0; i < boardSize; i++) {
        for (int j = 0; j < [1]; j++) {
            if (backtrack(board, word, i, j, 0, boardSize, [2])) {
                return [3];
            }
        }
    }
    return 0;
}

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

A*boardColSize

BboardColSize[0]

C1

DboardSize

Attempts:

3 left

💡 Hint

Common Mistakes

Using boardSize for columns is incorrect.

Returning 0 inside the loop misses found words.

We use boardColSize[0] to get the number of columns, and return 1 if word is found.

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