Fill a Special Grid

#541 Most AskedMedium

Fill a Special Grid #541 Most Asked

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ArraysRecursion

You are given a non-negative integer n representing a 2ⁿ x 2ⁿ grid. You must fill the grid with integers from 0 to 2²ⁿ - 1 to make it special. A grid is special if it satisfies all the following conditions:

All numbers in the top-right quadrant are smaller than those in the bottom-right quadrant.
All numbers in the bottom-right quadrant are smaller than those in the bottom-left quadrant.
All numbers in the bottom-left quadrant are smaller than those in the top-left quadrant.
Each of its quadrants is also a special grid.

Return the special 2ⁿ x 2ⁿ grid.

Note: Any 1x1 grid is special.

Example 1:

Input: n = 0

Output: [[0]]

Explanation:

The only number that can be placed is 0, and there is only one possible position in the grid.

Example 2:

Input: n = 1

Output: [[3,0],[2,1]]

Explanation:

The numbers in each quadrant are:

Top-right: 0
Bottom-right: 1
Bottom-left: 2
Top-left: 3

Since 0 < 1 < 2 < 3, this satisfies the given constraints.

Example 3:

Input: n = 2

Output: [[15,12,3,0],[14,13,2,1],[11,8,7,4],[10,9,6,5]]

Explanation:

The numbers in each quadrant are:

Top-right: 3, 0, 2, 1
Bottom-right: 7, 4, 6, 5
Bottom-left: 11, 8, 10, 9
Top-left: 15, 12, 14, 13
max(3, 0, 2, 1) < min(7, 4, 6, 5)
max(7, 4, 6, 5) < min(11, 8, 10, 9)
max(11, 8, 10, 9) < min(15, 12, 14, 13)

This satisfies the first three requirements. Additionally, each quadrant is also a special grid. Thus, this is a special grid.

Constraints:

0 <= n <= 10

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 grid (number of rows and columns), and what is the maximum size of each dimension?
What data type will be used to represent the grid cells (e.g., integers, characters)? What is the range of possible values for the cells?
Are there any constraints on the pattern or the special property that the filled grid needs to satisfy? Is there a specific objective function to optimize?
What should be returned if it's impossible to fill the grid according to the specified constraints or pattern?
Are all cells initially empty, or can some cells have predefined values that must be respected when filling the grid?

Brute Force Solution

Approach

The brute force approach to filling a special grid involves trying every possible combination of numbers to see if any fit the rules. It's like trying to solve a puzzle by randomly placing pieces until you find a working solution. We methodically go through each possibility until we find one that meets all the grid's requirements.

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

Start by trying to fill the first empty spot in the grid with every possible valid number.
For each of those numbers, move on to the next empty spot and again try filling it with every possible valid number.
Continue this process until the entire grid is filled with numbers.
Once the grid is filled, check if the numbers meet all the special rules of the grid.
If the rules are met, we've found a valid solution and we can store it.
If the rules are not met, we discard this attempt and go back to the previous empty spot to try a different number.
Repeat this entire process, exploring every possible number combination, until we've exhausted all options.
Finally, compare all the valid solutions we've found, and pick the best one based on the specific problem criteria.

Code Implementation

def fill_special_grid_brute_force(grid, rules):
    rows = len(grid)
    cols = len(grid[0])
    solutions = []

    def is_valid(current_grid):
        return rules(current_grid)

    def find_empty_cell(current_grid):
        for row_index in range(rows):
            for col_index in range(cols):
                if current_grid[row_index][col_index] is None:
                    return row_index, col_index
        return None, None

    def solve_grid(current_grid):
        row_index, col_index = find_empty_cell(current_grid)

        # If no empty cells, the grid is full
        if row_index is None:
            if is_valid(current_grid):
                solutions.append([row[:] for row in current_grid])
            return

        # Try every possible number in the empty spot
        for possible_number in range(1, 10):  # Assuming numbers 1-9 are valid
            current_grid[row_index][col_index] = possible_number
            solve_grid([row[:] for row in current_grid])
            current_grid[row_index][col_index] = None # Reset for backtracking

    solve_grid(grid)

    # Pick the best solution based on specific criteria (example: smallest sum)
    if solutions:
        best_solution = solutions[0]
        min_sum = sum(sum(row) for row in solutions[0])
        for solution in solutions[1:]:
            current_sum = sum(sum(row) for row in solution)
            # Find optimal solution out of valid ones
            if current_sum < min_sum:
                min_sum = current_sum
                best_solution = solution
        return best_solution
    else:
        return None

Big(O) Analysis

Time Complexity

O(m^n) – The provided brute force approach explores every possible combination of numbers to fill the grid. Let's say the grid has n empty spots to fill, and each spot can be filled with up to m possible numbers (where m depends on the problem constraints). For each empty spot, we have m choices, and we repeat this for n spots. This leads to a total of m * m * ... * m (n times) possible combinations. Therefore, the time complexity is O(m^n), where m is the number of choices for each cell, and n is the number of empty cells to fill.

Space Complexity

O(N) – The brute force approach uses recursion to fill the grid. In the worst-case scenario, where the initial grid is mostly empty and each spot needs to be explored deeply, the depth of the recursion could reach N, where N is the number of cells in the grid. Each recursive call adds a frame to the call stack, consuming memory to store local variables and the return address. Therefore, the auxiliary space used by the recursion stack is proportional to N, resulting in O(N) space complexity.

Optimal Solution

Approach

The best way to fill this grid is to think step-by-step, placing values to maximize the remaining space. By prioritizing filling from the edges inward and handling each row and column systematically, we can ensure the grid is filled optimally without trying every single possibility.

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

Start by focusing on the outer edges of the grid (top, bottom, left, and right).
Fill the longest possible continuous sections along these edges with the given values.
Notice that the inner parts of the grid are constrained by what you've already placed along the outer edges.
Continue filling the largest possible remaining sections, moving towards the center of the grid.
Always make sure you're filling the section that lets you put down the longest continuous block of the given values next.
If you find yourself stuck, or at a point where you can go in multiple directions, go to one of those directions.
If you cannot place any more values, the grid is full and you're done.

Code Implementation

def fill_special_grid(grid_rows, grid_columns, values_to_place):
    grid = [[' ' for _ in range(grid_columns)] for _ in range(grid_rows)]

    def can_place(row_start, column_start, row_end, column_end, value):
        if row_start > row_end or column_start > column_end:
            return False
        for row_index in range(row_start, row_end + 1):
            for column_index in range(column_start, column_end + 1):
                if grid[row_index][column_index] != ' ':
                    return False
        return True

    def place_value(row_start, column_start, row_end, column_end, value):
        for row_index in range(row_start, row_end + 1):
            for column_index in range(column_start, column_end + 1):
                grid[row_index][column_index] = value

    while values_to_place:
        best_row_start = -1
        best_column_start = -1
        best_row_end = -1
        best_column_end = -1
        best_value = None
        max_size = 0

        for value in list(values_to_place.keys()):
            for row_start in range(grid_rows):
                for column_start in range(grid_columns):

                    # Attempt placing horizontally
                    for column_end in range(column_start, grid_columns):
                        current_size = column_end - column_start + 1
                        if current_size > max_size and can_place(row_start, column_start, row_start, column_end, value):
                            max_size = current_size
                            best_row_start = row_start
                            best_column_start = column_start
                            best_row_end = row_start
                            best_column_end = column_end
                            best_value = value

                    # Attempt placing vertically
                    for row_end in range(row_start, grid_rows):
                        current_size = row_end - row_start + 1
                        if current_size > max_size and can_place(row_start, column_start, row_end, column_start, value):
                            max_size = current_size
                            best_row_start = row_start
                            best_column_start = column_start
                            best_row_end = row_end
                            best_column_end = column_start
                            best_value = value

        if best_value is None:
            break

        # Place the value that allows filling the largest block
        place_value(best_row_start, best_column_start, best_row_end, best_column_end, best_value)

        values_to_place[best_value] -= (abs(best_row_end - best_row_start) + abs(best_column_end - best_column_start) + 1)
        if values_to_place[best_value] == 0:
            del values_to_place[best_value]

    result = []
    for row in grid:
        result.append("".join(row))

    return result

Big(O) Analysis

Time Complexity

O(n^2) – The algorithm iterates through the grid, prioritizing filling from the edges inward. In the worst case, each cell in the n x n grid might be visited and potentially updated. Although the filling strategy attempts to optimize the placement, in scenarios with scattered placements or specific value distributions, the process might require revisiting and evaluating multiple potential placements for a significant portion of the grid cells. Therefore, the time complexity is dominated by the potential need to examine each of the n^2 cells, resulting in O(n^2) time complexity.

Space Complexity

O(1) – The provided plain English explanation outlines a strategy that primarily focuses on in-place modifications of the input grid. It doesn't explicitly mention creating any auxiliary data structures like lists, stacks, or queues for storing intermediate results or tracking visited locations. The steps involve iteratively filling the grid based on local constraints, suggesting a constant amount of extra memory is used for variables to track positions and values being placed, irrespective of the grid's size (N). Therefore, the auxiliary space complexity is O(1).

Edge Cases

Case	How to Handle
Null or invalid input grid dimensions (rows or cols <= 0)	Return an empty grid or an error code to indicate invalid input.
Grid dimensions are very large (e.g., exceeding memory limits)	Ensure the algorithm uses space efficiently, potentially using an iterative approach and considering data type sizes.
The special rule for filling the grid is impossible to satisfy.	Return an empty grid or an error indicator to signal no solution exists.
The filling rule involves arithmetic operations that may result in integer overflow.	Use appropriate data types (e.g., long) or modulo operations to prevent overflow.
Multiple valid grid fillings are possible.	The solution should consistently return one valid solution or have a documented strategy for choosing among multiple possible solutions.
The fill rule involves specific numbers (e.g., prime numbers, zero, negative numbers).	Handle the impact of these numbers on the rule without causing exceptions or errors.
The provided fill rule is computationally expensive, causing the algorithm to run slowly with larger grids.	Optimize the algorithm's time complexity by using efficient data structures or algorithmic strategies.
The filling requires some sort of random number generator, and the initial seed value is important.	Initialize the random number generator with a predictable seed value for consistent and testable behavior.

Null or invalid input grid dimensions (rows or cols <= 0)

How to Handle:

Return an empty grid or an error code to indicate invalid input.

Grid dimensions are very large (e.g., exceeding memory limits)

How to Handle:

Ensure the algorithm uses space efficiently, potentially using an iterative approach and considering data type sizes.

The special rule for filling the grid is impossible to satisfy.

How to Handle:

Return an empty grid or an error indicator to signal no solution exists.

The filling rule involves arithmetic operations that may result in integer overflow.

How to Handle:

Use appropriate data types (e.g., long) or modulo operations to prevent overflow.

Multiple valid grid fillings are possible.

How to Handle:

The solution should consistently return one valid solution or have a documented strategy for choosing among multiple possible solutions.

The fill rule involves specific numbers (e.g., prime numbers, zero, negative numbers).

How to Handle:

Handle the impact of these numbers on the rule without causing exceptions or errors.

The provided fill rule is computationally expensive, causing the algorithm to run slowly with larger grids.

How to Handle:

Optimize the algorithm's time complexity by using efficient data structures or algorithmic strategies.

The filling requires some sort of random number generator, and the initial seed value is important.

How to Handle:

Initialize the random number generator with a predictable seed value for consistent and testable behavior.