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

Practice - 5 Tasks

1fill in blank

easy

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

DSA Python

for i in range(len(board)):
    for j in range(len(board[0])):
        if board[i][j] == word[[1]]:
            if backtrack(i, j, 0):
                return True

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

A1

Blen(word)

C-1

D0

Attempts:

3 left

💡 Hint

Common Mistakes

Using 1 instead of 0 for the first character index.

Using len(word) which is out of range.

The search starts by matching the first character of the word, which is at index 0.

2fill in blank

medium

Complete the code to check if the current position is out of grid bounds.

DSA Python

if i < 0 or i >= len(board) or j < 0 or j >= [1]:
    return False

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

Alen(word)

Blen(board)

Clen(board[0])

Dlen(board) - 1

Attempts:

3 left

💡 Hint

Common Mistakes

Using len(board) which is the number of rows.

Using len(word) which is unrelated.

The number of columns is len(board[0]), so j must be less than that to be inside the grid.

3fill in blank

hard

Fix the error in marking the current cell as visited.

DSA Python

temp = board[i][j]
board[i][j] = [1]

# backtracking calls here

board[i][j] = temp

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

A'#'

B''

CNone

D'*'

Attempts:

3 left

💡 Hint

Common Mistakes

Using None which causes type errors.

Using empty string which may cause confusion.

Marking the cell with '#' is a common way to indicate it is visited temporarily.

4fill in blank

hard

Fill both blanks to check the next character and explore neighbors.

DSA Python

if word[[1]] != board[i][j]:
    return False

return (backtrack(i+1, j, [2]) or backtrack(i-1, j, [2]) or backtrack(i, j+1, [2]) or backtrack(i, j-1, [2]))

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

Aindex

Bindex + 1

Cindex - 1

D0

Attempts:

3 left

💡 Hint

Common Mistakes

Using index - 1 which goes backward in the word.

Using 0 which restarts the search incorrectly.

We check if the current board cell matches word[index], then explore neighbors with index + 1 for the next character.

5fill in blank

hard

Fill all three blanks to complete the backtracking function signature and base case.

DSA Python

def backtrack(i, j, [1]):
    if [2] == len(word):
        return True
    if i < 0 or i >= len(board) or j < 0 or j >= len(board[0]) or board[i][j] != word[[3]]:
        return False

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

Aindex

Bindex == 0

Dword

Attempts:

3 left

💡 Hint

Common Mistakes

Using 'word' as parameter instead of index.

Using 'index == 0' as parameter which is a boolean, not a variable.

The function uses 'index' to track the current character position. The base case checks if index equals word length, and the character check uses word[index].

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