Word Search in Grid Using Backtracking in DSA Python - Cheat Sheet & Quick Revision

Recall & Review

beginner

What is the main idea behind backtracking in word search problems?

Backtracking tries to build a solution step-by-step, exploring all possible paths, and if a path doesn't lead to a solution, it goes back (backtracks) to try another path.

beginner

In a word search grid, how do we decide the next cell to explore during backtracking?

We explore adjacent cells (up, down, left, right) that match the next character of the word and have not been visited yet.

beginner

Why do we mark cells as visited during backtracking in word search?

To avoid revisiting the same cell in the current path, which prevents infinite loops and incorrect matches.

beginner

What is the base case for the recursive backtracking function in word search?

When all characters of the word have been matched successfully, meaning the word is found in the grid.

intermediate

How does backtracking ensure all possible paths are checked in the grid?

By exploring one path fully, then undoing the last step (backtracking) to try other directions until all options are exhausted.

Which directions are typically checked in a word search backtracking algorithm?

A Only Left and Right

B Diagonal directions only

C Up, Down, Left, Right

D Up and Down only

Backtracking checks all four adjacent directions: up, down, left, and right.

What should you do if the current cell's character does not match the word's current character?

A Backtrack immediately

B Continue exploring neighbors

C Mark cell as visited

D Add cell to the path

If characters don't match, backtrack immediately to try other paths.

Why is it important to unmark a cell as visited when backtracking?

A To allow other paths to use that cell

B To speed up the search

C To avoid matching the word twice

D To mark the word as found

Unmarking allows other paths to explore that cell later.

What does the backtracking function return when the word is found?

A The path taken

B True

C The matched word

D False

It returns True to indicate the word was found.

If the word is not found starting from a particular cell, what happens next?

A Mark the word as found

B Stop the search

C Return the partial match

D Try starting from the next cell in the grid

The algorithm tries starting from the next cell until all cells are checked.

Explain how backtracking works to find a word in a grid.

Think about trying paths and undoing steps when stuck.

Describe the role of the visited cells tracking in the word search backtracking algorithm.

Why can't we use the same cell twice in one word path?

Practice

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

easy

A. To count the number of words in the grid

B. To sort the grid alphabetically

C. To replace characters in the grid

D. To explore all possible paths to find the word

Solution

1. Step 1: Understand backtracking in word search

   Backtracking tries all possible paths in the grid to find the target word by exploring neighbors recursively.

2. Step 2: Eliminate unrelated 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 -> Option D

4. Quick Check:

   Backtracking = Explore paths [OK]

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

Common Mistakes:

• Thinking backtracking sorts or modifies the grid
• Confusing backtracking with simple searching
• Ignoring the recursive exploration nature

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

easy

A. if 0 <= row < len(grid) and 0 <= col < len(grid[0]):

B. if row > 0 and col > 0 and row < len(grid) and col < len(grid[0]):

C. if row <= len(grid) and col <= len(grid[0]):

D. if row >= 0 or col >= 0:

Solution

1. Step 1: Understand grid boundary conditions

   Valid indices must be greater or equal to 0 and less than the number of rows or columns.

2. Step 2: Analyze each option

   if 0 <= row < len(grid) and 0 <= col < len(grid[0]): correctly uses inclusive lower bound and exclusive upper bound checks. if row > 0 and col > 0 and row < len(grid) and col < len(grid[0]): excludes 0 index, which is invalid. if row <= len(grid) and col <= len(grid[0]): allows equal to length, which is out of range. if row >= 0 or col >= 0: uses 'or' which is incorrect logic.

3. Final Answer:

   if 0 <= row < len(grid) and 0 <= col < len(grid[0]): -> Option A

4. Quick Check:

   Boundary check = if 0 <= row < len(grid) and 0 <= col < len(grid[0]): [OK]

Hint: Use 0 <= index < length for valid grid positions [OK]

Common Mistakes:

• Using > 0 instead of >= 0 excludes first row/column
• Allowing index equal to length causes errors
• Using 'or' instead of 'and' in conditions

3. What will be the output of the following code snippet?

grid = [['A','B','C','E'],
        ['S','F','C','S'],
        ['A','D','E','E']]
word = "ABCCED"

# Function to check word existence
result = word_search(grid, word)
print(result)

# Assume word_search uses backtracking to find the word

medium

A. True

B. False

C. None

D. Error

Solution

1. Step 1: Understand the word and grid

   The word "ABCCED" can be found by moving right and down in the grid starting at grid[0][0].

2. Step 2: Confirm backtracking finds the word

   Backtracking explores neighbors and matches letters sequentially, so it returns True if the word exists.

3. Final Answer:

   True -> Option A

4. Quick Check:

   Word found = True [OK]

Hint: Trace letters stepwise to confirm word presence [OK]

Common Mistakes:

• Assuming word is not found without tracing
• Confusing output with None or error
• Ignoring backtracking's path exploration

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

def backtrack(r, c, index):
    if index == len(word):
        return True
    if r < 0 or c < 0 or r >= len(grid) or c >= len(grid[0]) or grid[r][c] != word[index]:
        return False
    temp = grid[r][c]
    grid[r][c] = '#'
    found = (backtrack(r+1, c, index+1) or
             backtrack(r-1, c, index+1) or
             backtrack(r, c+1, index+1) or
             backtrack(r, c-1, index+1))
    grid[r][c] = temp
    return found

medium

A. Using '#' as marker can conflict with grid letters

B. Missing base case for index == 0

C. No bug; code is correct

D. Modifying grid in place without restoring causes errors

Solution

1. Step 1: Analyze the backtracking logic

   The code marks visited cells with '#', explores neighbors recursively, and restores the cell after exploration.

2. Step 2: Check for bugs

   Restoring the cell after recursion prevents errors. Base case for index == len(word) is correct. '#' is a common marker and safe if not in original grid.

3. Final Answer:

   No bug; code is correct -> Option C

4. Quick Check:

   Code logic = Correct [OK]

Hint: Check if visited cells are restored after recursion [OK]

Common Mistakes:

• Forgetting to restore grid cell after recursion
• Adding unnecessary base cases
• Assuming marker conflicts without checking grid content

5. Given a 4x4 grid and a list of words, which approach best finds all words using backtracking efficiently? A) Run backtracking separately for each word on the entire grid. B) Sort words alphabetically and search in order. C) Search only the first letter of each word and stop. D) Build a Trie of words and backtrack once to find all words simultaneously.

hard

A. Run backtracking separately for each word on the entire grid.

B. Build a Trie of words and backtrack once to find all words simultaneously.

C. Search only the first letter of each word and stop.

D. Sort words alphabetically and search in order.

Solution

1. Step 1: Understand the problem of multiple words

   Running backtracking separately for each word is inefficient because it repeats searches.

2. Step 2: Use Trie to optimize search

   Building a Trie of all words allows a single backtracking pass to find all words by prefix matching, improving efficiency.

3. Step 3: Eliminate other options

   Searching only first letters or sorting words does not improve search efficiency significantly.

4. Final Answer:

   Build a Trie of words and backtrack once to find all words simultaneously. -> Option B

5. Quick Check:

   Trie + backtracking = Efficient multi-word search [OK]

Hint: Use Trie to find multiple words in one backtracking pass [OK]

Common Mistakes:

• Running backtracking repeatedly wastes time
• Ignoring prefix sharing among words
• Assuming sorting helps search efficiency