Snake in Matrix

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There is a snake in an n x n matrix grid and can move in four possible directions. Each cell in the grid is identified by the position: grid[i][j] = (i * n) + j.

The snake starts at cell 0 and follows a sequence of commands.

You are given an integer n representing the size of the grid and an array of strings commands where each command[i] is either "UP", "RIGHT", "DOWN", and "LEFT". It's guaranteed that the snake will remain within the grid boundaries throughout its movement.

Return the position of the final cell where the snake ends up after executing commands.

Example 1:

Input: n = 2, commands = ["RIGHT","DOWN"]

Output: 3

Explanation:

0	1
2	3

0	1
2	3

0	1
2	3

Example 2:

Input: n = 3, commands = ["DOWN","RIGHT","UP"]

Output: 1

Explanation:

0	1	2
3	4	5
6	7	8

0	1	2
3	4	5
6	7	8

0	1	2
3	4	5
6	7	8

0	1	2
3	4	5
6	7	8

Constraints:

2 <= n <= 10
1 <= commands.length <= 100
commands consists only of "UP", "RIGHT", "DOWN", and "LEFT".
The input is generated such the snake will not move outside of the boundaries.

Try coding it in LeetCode

Solution

Clarifying Questions

When you get asked this question in a real-life environment, it will often be ambiguous (especially at FAANG). Make sure to ask these questions in that case:

What are the dimensions of the matrix (number of rows and columns)? Are there any constraints on these dimensions?
What type of data will the matrix contain (e.g., integers, characters)? Are there any restrictions on the values within the matrix (e.g., positive only, a specific range)?
What defines a valid "snake"? Does the snake have to move only horizontally and vertically, or are diagonal movements allowed? How long does it have to be?
If there are multiple possible snake paths in the matrix, should I return any one of them, or is there a specific path I should prioritize (e.g., the longest, the first one found starting from the top-left)?
If no snake path exists in the matrix according to the defined rules, what should the function return (e.g., null, an empty list, a specific error code)?

Brute Force Solution

Approach

The goal is to find the longest 'snake' path in a grid where each step increases by exactly one. The brute force way checks every possible path, one by one, to find the longest valid snake.

Here's how the algorithm would work step-by-step:

Start at any cell in the grid. Think of this as the beginning of a possible snake.
From that starting point, explore all the neighboring cells (up, down, left, right).
For each neighbor, check if its value is exactly one greater than the current cell's value. If it is, that neighbor can be the next step in our snake.
If a neighbor is a valid step, continue the snake by exploring its neighbors in the same way.
Keep doing this, building longer and longer snake paths. When you reach a dead end (no valid neighbors), record the length of that snake.
Go back to the beginning and start at a different cell in the grid. Repeat the whole process to find more snakes.
Compare the lengths of all the snakes you found, and the longest one is your answer.

Code Implementation

def longest_snake_path_brute_force(matrix):
    rows = len(matrix)
    cols = len(matrix[0])
    longest_path = 0

    def explore_path(row, col, current_length):
        nonlocal longest_path
        longest_path = max(longest_path, current_length)

        # Explore all 4 neighbors
        for delta_row, delta_col in [(0, 1), (0, -1), (1, 0), (-1, 0)]:
            new_row = row + delta_row
            new_col = col + delta_col

            if 0 <= new_row < rows and 0 <= new_col < cols:
                # Check if neighbor is a valid next step in the snake
                if matrix[new_row][new_col] == matrix[row][col] + 1:

                    explore_path(new_row, new_col, current_length + 1)

    # Iterate through each cell as a possible starting point.
    for start_row in range(rows):
        for start_col in range(cols):

            explore_path(start_row, start_col, 1)

    return longest_path

Big(O) Analysis

Time Complexity

O(4^n) – The brute force approach explores all possible paths starting from each cell in the matrix. In the worst case, from each cell, we could potentially explore up to 4 neighboring cells recursively. This recursive exploration can continue for a depth related to the dimensions of the matrix, denoted as 'n' representing the maximum possible length of a snake (which is bounded by the number of cells). Therefore, at each step, we have 4 possible directions to explore, resulting in a time complexity of approximately O(4^n), where n is related to the number of cells and effectively represents the maximum snake length.

Space Complexity

O(N) – The algorithm, as described, explores paths using recursion. In the worst-case scenario, where each cell leads to another valid cell, the recursion depth can reach up to N, where N is the total number of cells in the grid (rows * cols). Each recursive call adds a frame to the call stack, thus consuming memory. Therefore, the auxiliary space complexity is proportional to the maximum depth of the recursion, resulting in O(N) space complexity due to the recursion stack.

Optimal Solution

Approach

The goal is to find the longest path within a grid that forms a snake-like sequence (increasing by one at each step). Instead of checking every possible path, we'll use a method to remember the longest snake found so far originating from each spot. This way, we avoid repeating work and quickly find the best snake.

Here's how the algorithm would work step-by-step:

Start by imagining you're at one of the numbers in the grid.
Check all the spots around it to see if any of them have a number that's exactly one bigger.
If you find a bigger number, that extends the 'snake' you're building. Remember how long the snake is at that new spot (including the spot you started from).
Now, imagine you're at *that* new spot. Repeat the process: check its neighbors for a number that's one bigger, and update the snake length if you find one.
The trick is: if you ever come back to a spot you've already been to, and you already know how long the snake can be from that spot, don't bother recalculating it. Just use the length you already know. This prevents a lot of wasted effort.
Keep doing this for every spot in the grid. The longest snake length you find during this process is your answer.

Code Implementation

def find_longest_snake_path(matrix):
    if not matrix or not matrix[0]:
        return 0

    rows = len(matrix)
    cols = len(matrix[0])
    longest_path_from_cell = [[0] * cols for _ in range(rows)]
    max_snake_length = 0

    def get_longest_path_from(row_index, column_index):
        #If we already know the longest path, reuse it.
        if longest_path_from_cell[row_index][column_index] != 0:
            return longest_path_from_cell[row_index][column_index]

        current_snake_length = 1
        #Check adjacent cells for a value one greater.
        for row_offset, col_offset in [(0, 1), (0, -1), (1, 0), (-1, 0)]:
            neighbor_row = row_index + row_offset
            neighbor_col = column_index + col_offset

            if 0 <= neighbor_row < rows and 0 <= neighbor_col < cols and \
               matrix[neighbor_row][neighbor_col] == matrix[row_index][column_index] + 1:
                
                current_snake_length = max(current_snake_length, 1 + get_longest_path_from(neighbor_row, neighbor_col))

        longest_path_from_cell[row_index][column_index] = current_snake_length
        return current_snake_length

    #Iterate through the matrix and calculate each path.
    for row_index in range(rows):
        for column_index in range(cols):
            current_path_length = get_longest_path_from(row_index, column_index)
            #Keep track of the maximum snake length.
            max_snake_length = max(max_snake_length, current_path_length)

    return max_snake_length

Big(O) Analysis

Time Complexity

O(m*n) – The algorithm iterates through each cell of the m x n matrix. For each cell, it potentially explores its neighbors to extend the snake. Due to memoization (remembering the longest snake from each spot), each cell is visited and processed at most once. The neighbor exploration takes constant time. Therefore, the time complexity is proportional to the number of cells in the matrix, resulting in O(m*n).

Space Complexity

O(N*M) – The algorithm uses a memoization technique to store the longest snake length found so far originating from each cell in the grid. This is implemented using a data structure, likely a 2D array or matrix, of the same dimensions as the input grid to store the lengths. If the input grid has dimensions N rows and M columns, then the auxiliary space required for memoization is proportional to N*M. Thus, the space complexity is O(N*M).

Edge Cases

Case	How to Handle
Null or empty matrix	Return an empty list or null, indicating no snake traversal is possible.
Matrix with only one row or one column	Handle traversal as a simple linear sequence rather than a 2D 'snake'.
Non-rectangular matrix (rows with differing lengths)	Return an error, throw an exception, or pad the matrix to make it rectangular.
Matrix with negative numbers or other non-integer values	The algorithm should specify intended behavior and appropriately handle number conversion or type-checking errors.
Extremely large matrix exceeding memory limitations	Consider using iterative approach or generator functions to avoid loading the entire matrix into memory at once.
Matrix contains only identical values	Ensure traversal direction alternates as intended even with constant values.
Integer overflow in matrix dimensions	Use appropriate data types (e.g., long) to store indices and calculated values.
No valid snake path exists based on problem constraints	Return an empty list or a specific indicator (e.g., null) if a valid path cannot be found.

Null or empty matrix

How to Handle:

Return an empty list or null, indicating no snake traversal is possible.

Matrix with only one row or one column

How to Handle:

Handle traversal as a simple linear sequence rather than a 2D 'snake'.

Non-rectangular matrix (rows with differing lengths)

How to Handle:

Return an error, throw an exception, or pad the matrix to make it rectangular.

Matrix with negative numbers or other non-integer values

How to Handle:

The algorithm should specify intended behavior and appropriately handle number conversion or type-checking errors.

Extremely large matrix exceeding memory limitations

How to Handle:

Consider using iterative approach or generator functions to avoid loading the entire matrix into memory at once.

Matrix contains only identical values

How to Handle:

Ensure traversal direction alternates as intended even with constant values.

Integer overflow in matrix dimensions

How to Handle:

Use appropriate data types (e.g., long) to store indices and calculated values.

No valid snake path exists based on problem constraints

How to Handle:

Return an empty list or a specific indicator (e.g., null) if a valid path cannot be found.