Word Search in Grid Using Backtracking in DSA Python - Time & Space Complexity

Time Complexity: Word Search in Grid Using Backtracking

O(M * N * 4^L)

Understanding Time Complexity

We want to understand how the time needed to find a word in a grid changes as the grid or word size grows.

How does the search effort increase when the grid or word gets bigger?

Scenario Under Consideration

Analyze the time complexity of the following code snippet.

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

    def backtrack(r, c, index):
        if index == len(word):
            return True
        if r < 0 or c < 0 or r >= rows or c >= cols or board[r][c] != word[index]:
            return False

        temp = board[r][c]
        board[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))

        board[r][c] = temp
        return found

    for i in range(rows):
        for j in range(cols):
            if backtrack(i, j, 0):
                return True
    return False

This code tries to find if a word exists in the grid by exploring all possible paths using backtracking.

Identify Repeating Operations

Identify the loops, recursion, array traversals that repeat.

• Primary operation: The backtracking function explores up to 4 directions recursively for each letter in the word.
• How many times: For each cell in the grid (rows * cols), backtracking can explore up to 4 recursive calls per letter in the word.

How Execution Grows With Input

As the grid size or word length grows, the number of paths to explore grows very fast because each letter can lead to 4 new directions.

Input Size (n = word length)	Approx. Operations
3	Up to 4^3 = 64 recursive calls per starting cell
5	Up to 4^5 = 1024 recursive calls per starting cell
10	Up to 4^10 = 1,048,576 recursive calls per starting cell

Pattern observation: The number of operations grows exponentially with the word length, making the search much slower for longer words.

Final Time Complexity

Time Complexity: O(M * N * 4^L)

This means the time grows very fast with the word length L, and also depends on the grid size M rows and N columns.

Common Mistake

[X] Wrong: "The search only checks each cell once, so it is O(M*N)."

[OK] Correct: Because the search explores many paths recursively for each letter, it can revisit cells in different paths, causing exponential growth, not just a simple grid scan.

Interview Connect

Understanding this complexity helps you explain why backtracking can be slow and when it is suitable, showing your grasp of algorithm efficiency in real problems.

Self-Check

"What if we limited the search to only move right and down instead of all four directions? How would the time complexity change?"

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