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

Practice - 5 Tasks

1fill in blank

easy

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

DSA C++

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 (board[i][j] == word[0] && dfs(board, word, i, j, 0)) {
                return [1];
            }
        }
    }
    return false;
}

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

Atrue

Bfalse

Cboard[i][j]

Dword.size()

Attempts:

3 left

💡 Hint

Common Mistakes

Returning false immediately instead of true when word is found.

Returning the character instead of a boolean.

If the first character matches and dfs returns true, the word exists, so return true.

2fill in blank

medium

Complete the code to mark the current cell as visited during DFS.

DSA C++

bool dfs(vector<vector<char>>& board, string& word, int i, int j, int index) {
    if (index == word.size()) return true;
    if (i < 0 || i >= board.size() || j < 0 || j >= board[0].size() || board[i][j] != word[index]) return false;
    char temp = board[i][j];
    board[i][j] = [1];
    bool found = dfs(board, word, i + 1, j, index + 1) ||
                 dfs(board, word, i - 1, j, index + 1) ||
                 dfs(board, word, i, j + 1, index + 1) ||
                 dfs(board, word, i, j - 1, index + 1);
    board[i][j] = temp;
    return found;
}

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

A' '

B'0'

C'*'

D'#'

Attempts:

3 left

💡 Hint

Common Mistakes

Not marking the cell as visited, causing infinite loops.

Using a character that might appear in the board.

Marking the cell with '#' indicates it is visited to avoid revisiting during this path.

3fill in blank

hard

Fix the error in the base case that checks if the entire word is found.

DSA C++

bool dfs(vector<vector<char>>& board, string& word, int i, int j, int index) {
    if (index [1] word.size()) return true;
    // rest of code omitted for brevity
}

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

A<

B==

C>

D<=

Attempts:

3 left

💡 Hint

Common Mistakes

Using less than or greater than instead of equality.

Off-by-one errors in base case.

The base case is when index equals the word length, meaning all characters matched.

4fill in blank

hard

Fill both blanks to complete the dictionary comprehension that maps words to their lengths if length > 3.

DSA C++

auto wordLengths = std::unordered_map<std::string, int>{};
for (const auto& word : words) {
    if (word.size() [1] 3) {
        wordLengths[word] = [2];
    }
}

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

A>

Bword.size()

C<

Dword.length()

Attempts:

3 left

💡 Hint

Common Mistakes

Using less than instead of greater than.

Using word.length() inconsistently.

We check if word size is greater than 3, then store the size for that word.

5fill in blank

hard

Fill all three blanks to complete the recursive DFS call and backtracking steps.

DSA C++

bool dfs(vector<vector<char>>& board, string& word, int i, int j, int index) {
    if (index == word.size()) return true;
    if (i < 0 || i >= board.size() || j < 0 || j >= board[0].size() || board[i][j] != word[index]) return false;
    char temp = board[i][j];
    board[i][j] = [1];
    bool found = dfs(board, word, i + [2], j, index + 1) ||
                 dfs(board, word, i - [2], j, index + 1) ||
                 dfs(board, word, i, j + [3], index + 1) ||
                 dfs(board, word, i, j - [3], index + 1);
    board[i][j] = temp;
    return found;
}

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

A'#'

B1

D0

Attempts:

3 left

💡 Hint

Common Mistakes

Not marking visited cell.

Using 0 instead of 1 for movement steps.

Mark visited with '#', move one step in each direction (up/down/left/right) by adding or subtracting 1.

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