Word Search in Grid Using Backtracking in DSA Python - Build from Scratch

Word Search in Grid Using Backtracking

📖 Scenario: You are creating a simple word search puzzle solver. The puzzle is a grid of letters, and you want to find if a given word exists in the grid by moving horizontally or vertically to adjacent letters.This is like searching for a word hidden in a crossword puzzle.

🎯 Goal: Build a program that checks if a given word can be found in the letter grid by moving up, down, left, or right without revisiting the same cell.

📋 What You'll Learn

Create a 2D list called board with the exact letters given.

Create a string variable called word with the exact value given.

Write a function called exist that uses backtracking to find the word in the board.

Print True if the word is found, otherwise False.

💡 Why This Matters

🌍 Real World

Word search puzzles are common in newspapers and apps. This technique helps solve or create such puzzles automatically.

💼 Career

Backtracking is a key algorithm technique used in many problems like puzzles, games, and constraint satisfaction tasks in software development.

1

Create the letter grid

Create a 2D list called board with these exact rows:
['A', 'B', 'C', 'E'],
['S', 'F', 'C', 'S'],
['A', 'D', 'E', 'E']

DSA Python

# Create the 2D list called board with the exact rows
# Your code here

Hint

Use square brackets to create a list of lists. Each inner list is a row of letters.

2

Set the word to search

Create a string variable called word and set it to 'ABCCED'

DSA Python

board = [
    ['A', 'B', 'C', 'E'],
    ['S', 'F', 'C', 'S'],
    ['A', 'D', 'E', 'E']
]
# Create the string variable word with value 'ABCCED'
# Your code here

Hint

Use single or double quotes to create the string.

3

Write the backtracking function

Write a function called exist that takes board and word as parameters and returns True if the word exists in the board using backtracking. Use helper function dfs inside exist to explore neighbors. Use variables row, col, and index to track position and progress.

DSA Python

board = [
    ['A', 'B', 'C', 'E'],
    ['S', 'F', 'C', 'S'],
    ['A', 'D', 'E', 'E']
]
word = 'ABCCED'
# Write the exist function with backtracking logic
# Your code here

Hint

Use a helper function dfs inside exist to check each letter recursively. Mark visited cells temporarily.

4

Print the result

Write a print statement to display the result of calling exist(board, word)

DSA Python

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

def exist(board, word):
    rows = len(board)
    cols = len(board[0])

    def dfs(row, col, index):
        if index == len(word):
            return True
        if row = rows or col = cols:
            return False
        if board[row][col] != word[index]:
            return False

        temp = board[row][col]
        board[row][col] = '#'

        found = (dfs(row + 1, col, index + 1) or
                 dfs(row - 1, col, index + 1) or
                 dfs(row, col + 1, index + 1) or
                 dfs(row, col - 1, index + 1))

        board[row][col] = temp
        return found

    for r in range(rows):
        for c in range(cols):
            if dfs(r, c, 0):
                return True
    return False

# Print the result of exist(board, word)
# Your code here

Hint

Call the exist function with board and word and print the result.

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