Word Search in Grid Using Backtracking in DSA Python - Practice Problems & Challenges

Challenge - 5 Problems

❓ Predict Output

intermediate

Output of Word Search Result

What is the output of the following code that checks if the word "CAT" exists in the grid?

DSA Python

def exist(board, word):
    rows, cols = len(board), len(board[0])
    def backtrack(r, c, i):
        if i == len(word):
            return True
        if r < 0 or c < 0 or r >= rows or c >= cols or board[r][c] != word[i]:
            return False
        temp = board[r][c]
        board[r][c] = '#'
        found = (backtrack(r+1, c, i+1) or backtrack(r-1, c, i+1) or
                 backtrack(r, c+1, i+1) or backtrack(r, c-1, i+1))
        board[r][c] = temp
        return found
    for row in range(rows):
        for col in range(cols):
            if backtrack(row, col, 0):
                return True
    return False

board = [
    ['C','A','T'],
    ['X','Z','Y'],
    ['B','N','M']
]
print(exist(board, "CAT"))

A false

B False

C True

D true

💡 Hint

Check if the word "CAT" can be found by moving horizontally or vertically in the grid.

The function returns True because the word "CAT" is found in the first row of the grid moving left to right.

❓ Predict Output

intermediate

Output of Word Search for Non-Existing Word

What will be the output of this code when searching for the word "DOG" in the given grid?

DSA Python

def exist(board, word):
    rows, cols = len(board), len(board[0])
    def backtrack(r, c, i):
        if i == len(word):
            return True
        if r < 0 or c < 0 or r >= rows or c >= cols or board[r][c] != word[i]:
            return False
        temp = board[r][c]
        board[r][c] = '#'
        found = (backtrack(r+1, c, i+1) or backtrack(r-1, c, i+1) or
                 backtrack(r, c+1, i+1) or backtrack(r, c-1, i+1))
        board[r][c] = temp
        return found
    for row in range(rows):
        for col in range(cols):
            if backtrack(row, col, 0):
                return True
    return False

board = [
    ['D','O','G'],
    ['A','B','C'],
    ['X','Y','Z']
]
print(exist(board, "CAT"))

A False

B true

C True

D false

💡 Hint

The word "CAT" is not present in the grid.

The function returns False because the word "CAT" cannot be found in the grid.

🔧 Debug

advanced

Identify the Error in Backtracking Code

What error will this code produce when trying to find the word "HELLO" in the grid?

DSA Python

def exist(board, word):
    rows, cols = len(board), len(board[0])
    def backtrack(r, c, i):
        if i == len(word):
            return True
        if r < 0 or c < 0 or r >= rows or c >= cols or board[r][c] != word[i]:
            return False
        board[r][c] = '#'
        found = (backtrack(r+1, c, i+1) or backtrack(r-1, c, i+1) or
                 backtrack(r, c+1, i+1) or backtrack(r, c-1, i+1))
        board[r][c] = word[i]
        return found
    for row in range(rows):
        for col in range(cols):
            if backtrack(row, col, 0):
                return True
    return False

board = [
    ['H','E','L','L','O'],
    ['A','B','C','D','E'],
    ['F','G','H','I','J']
]
print(exist(board, "HELLO"))

A TypeError: unsupported operand type(s)

B No error, output is True

C ValueError: assignment to immutable element

D IndexError: list index out of range

💡 Hint

Consider the mutability of the board elements.

The board is a list of lists of strings, which are immutable. Assigning board[r][c] = '#' causes a ValueError because strings are immutable in Python.

🚀 Application

advanced

Number of Words Found in Grid

Given a list of words and a grid, how many words can be found in the grid using the backtracking approach?

DSA Python

words = ["CAT", "DOG", "BIRD"]
board = [
    ['C','A','T'],
    ['D','O','G'],
    ['B','I','R']
]

# Assume exist function is defined as before
count = 0
for w in words:
    if exist(board, w):
        count += 1
print(count)

A 2

B 3

C 1

D 0

💡 Hint

Check which words can be found by moving horizontally or vertically.

"CAT" and "DOG" can be found in the grid, but "BIRD" cannot be found completely.

🧠 Conceptual

expert

Why Backtracking is Suitable for Word Search in Grid

Why is backtracking a good approach to solve the word search problem in a grid?

A Because it sorts the grid to find the word faster.

B Because it converts the grid into a graph and uses breadth-first search only.

C Because it uses hashing to store all words in the grid.

D Because it tries all possible paths and abandons paths that do not lead to a solution, saving time.

💡 Hint

Think about how backtracking explores paths and prunes wrong ones.

Backtracking explores all possible paths for the word and abandons paths early if they don't match, making it efficient for this problem.

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