Why Word Search in Grid Using Backtracking in DSA C?

The Big Idea

What if you could find any hidden word in a grid without missing a single path or getting lost?

The Scenario

Imagine you have a big crossword puzzle on paper, and you want to find if a certain word is hidden inside it by checking every letter manually.

You look at each letter, then try to see if the next letters match the word by moving up, down, left, or right.

This takes a lot of time and you can easily lose track or miss some paths.

The Problem

Manually checking every possible path for the word is slow and confusing.

You might forget which letters you already checked or accidentally reuse letters that should only be used once.

It's easy to make mistakes and miss the word even if it's there.

The Solution

Backtracking is like having a smart helper who tries every possible path step-by-step.

If the path doesn't lead to the word, the helper goes back and tries a different path.

This way, you don't miss any possibilities and avoid repeating letters incorrectly.

Before vs After

✗ Before

for each cell in grid:
  if cell matches first letter:
    check all directions manually

✓ After

bool searchWord(grid, word, row, col, index) {
  if (index == word length) return true;
  if (row < 0 || col < 0 || row >= grid rows || col >= grid cols || grid[row][col] != word[index]) return false;
  mark cell visited;
  for each direction {
    if (searchWord(next cell, index + 1)) return true;
  }
  unmark cell;
  return false;
}

What It Enables

You can efficiently find if a word exists in a grid by exploring all paths without missing or repeating letters wrongly.

Real Life Example

Finding hidden words in a word search puzzle game on your phone, where letters can be connected in multiple directions.

Key Takeaways

Manual search is slow and error-prone for word search in grids.

Backtracking tries all paths systematically and backtracks on dead ends.

This method ensures all possibilities are checked correctly and efficiently.

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