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Strange Printer II

Hard

Asked by:

23 views

Topics:

ArraysRecursion

There is a strange printer with the following two special requirements:

On each turn, the printer will print a solid rectangular pattern of a single color on the grid. This will cover up the existing colors in the rectangle.
Once the printer has used a color for the above operation, the same color cannot be used again.

You are given a m x n matrix targetGrid, where targetGrid[row][col] is the color in the position (row, col) of the grid.

Return true if it is possible to print the matrix targetGrid, otherwise, return false.

Example 1:

Input: targetGrid = [[1,1,1,1],[1,2,2,1],[1,2,2,1],[1,1,1,1]]
Output: true

Example 2:

Input: targetGrid = [[1,1,1,1],[1,1,3,3],[1,1,3,4],[5,5,1,4]]
Output: true

Example 3:

Input: targetGrid = [[1,2,1],[2,1,2],[1,2,1]]
Output: false
Explanation: It is impossible to form targetGrid because it is not allowed to print the same color in different turns.

Constraints:

m == targetGrid.length
n == targetGrid[i].length
1 <= m, n <= 60
1 <= targetGrid[row][col] <= 60

Try coding it in LeetCode

Strange Printer II

Hard

Asked by:

23 views

Topics:

ArraysRecursion

There is a strange printer with the following two special requirements:

On each turn, the printer will print a solid rectangular pattern of a single color on the grid. This will cover up the existing colors in the rectangle.
Once the printer has used a color for the above operation, the same color cannot be used again.

You are given a m x n matrix targetGrid, where targetGrid[row][col] is the color in the position (row, col) of the grid.

Return true if it is possible to print the matrix targetGrid, otherwise, return false.

Example 1:

Input: targetGrid = [[1,1,1,1],[1,2,2,1],[1,2,2,1],[1,1,1,1]]
Output: true

Example 2:

Input: targetGrid = [[1,1,1,1],[1,1,3,3],[1,1,3,4],[5,5,1,4]]
Output: true

Example 3:

Input: targetGrid = [[1,2,1],[2,1,2],[1,2,1]]
Output: false
Explanation: It is impossible to form targetGrid because it is not allowed to print the same color in different turns.

Constraints:

m == targetGrid.length
n == targetGrid[i].length
1 <= m, n <= 60
1 <= targetGrid[row][col] <= 60

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 (rows and columns) of the `targetGrid` matrix, and what is the maximum possible value for any color in the grid?
Can the color values in the `targetGrid` be zero?
If it's impossible to print the `targetGrid` using any sequence of painting operations, what should the function return (e.g., `false`, throw an exception)?
Are the color values guaranteed to be within a specific range (e.g., 1 to 100)?
If there are multiple possible sequences of painting operations that can produce the target grid, is any valid sequence acceptable, or is there a specific criterion to select the best sequence (e.g., shortest sequence, lexicographically smallest)?

Brute Force Solution

Approach

The problem asks if we can print a rectangular grid of colors using a special printer that can only print rectangles of a single color at a time. The brute force approach involves trying to 'unpaint' each color rectangle one at a time, from all possible colors present in the grid.

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

First, list all the different colors that appear in the grid.
Then, for each color, we assume that the last time the printer ran, it painted all locations of that color as a single rectangle.
Check if this assumption is valid by virtually removing that color from the grid, seeing if other colors were painted over.
If removing the color creates inconsistencies (like breaks in other colors that should have been solid), then that color was NOT the last color printed.
If removing the color is valid, we temporarily remove it from the grid and continue this process with the remaining colors.
We continue removing colors one by one, if possible, until all colors are removed without inconsistencies. If we can do this, the answer is yes. If we find a color we can't remove, we try a different order by backing up and picking a different 'last' color.
If after trying all possible orders of unpainting colors we can't remove all the colors without inconsistencies, the answer is no.

Code Implementation

def is_printable(grid):
    rows = len(grid)
    cols = len(grid[0])
    colors = set()
    for row in range(rows):
        for col in range(cols):
            colors.add(grid[row][col])
    colors = list(colors)

    def check_remove_color(current_grid, color_to_remove):
        rows = len(current_grid)
        cols = len(current_grid[0])
        min_row = rows
        max_row = -1
        min_col = cols
        max_col = -1
        for row in range(rows):
            for col in range(cols):
                if current_grid[row][col] == color_to_remove:
                    min_row = min(min_row, row)
                    max_row = max(max_row, row)
                    min_col = min(min_col, col)
                    max_col = max(max_col, col)

        # Verify if color_to_remove can be removed
        for row in range(rows):
            for col in range(cols):
                if (min_row <= row <= max_row and min_col <= col <= max_col
                        and current_grid[row][col] != color_to_remove
                        and current_grid[row][col] != 0):
                    return False
        return True

    def solve(current_grid, remaining_colors):
        if not remaining_colors:
            return True

        for color_to_remove in remaining_colors:
            # Check if the current color can be removed
            if check_remove_color(current_grid, color_to_remove):

                # Creating a new grid to 'unpaint' the color
                new_grid = [row[:] for row in current_grid]
                rows = len(new_grid)
                cols = len(new_grid[0])
                for row in range(rows):
                    for col in range(cols):
                        if new_grid[row][col] == color_to_remove:
                            new_grid[row][col] = 0

                new_remaining_colors = remaining_colors[:]
                new_remaining_colors.remove(color_to_remove)

                # Recursively check
                if solve(new_grid, new_remaining_colors):
                    return True

        return False

    # Begin the solving process with the original grid
    return solve([row[:] for row in grid], colors)

Big(O) Analysis

Time Complexity

O(m * n * k) – Let m be the number of distinct colors, n be the number of rows in the grid, and k be the number of columns in the grid. The algorithm iterates through each of the m colors. For each color, it examines the entire n x k grid to simulate removing the color and checking for inconsistencies in other colors. Therefore, the time complexity for each color is O(n * k). Since we may need to repeat this for all colors in the worst case (backtracking), the overall time complexity becomes O(m * n * k).

Space Complexity

O(C) – The space complexity is primarily determined by the number of distinct colors, C, in the grid. We store a list of these colors, and during the 'unpainting' process, we modify the grid by virtually removing colors, potentially requiring extra space to keep track of the modified state, or to store rectangles coordinates for each color. In the worst-case, if all cells have different colors, C could be proportional to the number of cells in the grid, but the plain English explanation focuses on number of colors. Backtracking and trying different orders might add some overhead, but it is bounded by the total number of colors. Therefore, the auxiliary space complexity is O(C).

Optimal Solution

Approach

This puzzle asks if you can achieve a specific colorful image by printing rectangular regions on top of each other. The trick is to work backward by determining which colors must have been printed last and then removing those regions to see if the remaining image is valid.

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

First, identify the outermost layer of colors by finding which colors aren't covered by any other colors.
For each of these outermost colors, determine the smallest rectangle that fully contains all instances of that color in the image.
Check if the color is present only within its determined rectangle. If it extends outside the rectangle, then it could not have been the last printed color and the image is not valid. Return that it is not valid.
If the color meets all checks, conceptually remove the rectangle filled with this color from the canvas. You don't actually have to modify the image; just pretend it's gone.
Repeat the process on the modified (imagined) canvas. Find the new outermost colors and their bounding rectangles and validate the colors.
Keep going until the canvas is completely clear (all colors have been removed) or you determine a problem. If the canvas is cleared then the result is valid; otherwise invalid.

Code Implementation

def strange_printer_two(target_grid):
    number_of_rows = len(target_grid)
    number_of_columns = len(target_grid[0])

    color_bounds = {}
    for row_index in range(number_of_rows):
        for column_index in range(number_of_columns):
            color = target_grid[row_index][column_index]
            if color != 0:
                if color not in color_bounds:
                    color_bounds[color] = [row_index, column_index, row_index, column_index]
                else:
                    color_bounds[color][0] = min(color_bounds[color][0], row_index)
                    color_bounds[color][1] = min(color_bounds[color][1], column_index)
                    color_bounds[color][2] = max(color_bounds[color][2], row_index)
                    color_bounds[color][3] = max(color_bounds[color][3], column_index)

    while color_bounds:
        erasable_colors = []
        for color in list(color_bounds.keys()):
            is_erasable = True
            row_start, column_start, row_end, column_end = color_bounds[color]
            for row_index in range(row_start, row_end + 1):
                for column_index in range(column_start, column_end + 1):
                    current_color = target_grid[row_index][column_index]
                    if current_color != 0 and current_color != color:
                        if current_color in color_bounds:
                            is_erasable = False
                            break
                if not is_erasable:
                    break
            if is_erasable:
                erasable_colors.append(color)

        if not erasable_colors:
            return False

        # Simulate removing the colors
        for color in erasable_colors:
            row_start, column_start, row_end, column_end = color_bounds[color]
            for row_index in range(row_start, row_end + 1):
                for column_index in range(column_start, column_end + 1):
                    if target_grid[row_index][column_index] == color:
                        target_grid[row_index][column_index] = 0

            # Remove the color from bounds
            del color_bounds[color]

    # If all colors were removed, the image is valid
    return True

Big(O) Analysis

Time Complexity

O(m*n*k) – The algorithm iterates through each color (up to k, the number of unique colors) and for each color, it scans the entire image (m rows by n columns) to find its bounding rectangle and to verify if the color exists only within this rectangle. The bounding rectangle calculation and the validation check are both O(m*n) operations each potentially, and this occurs at most k times. Therefore, the overall time complexity is O(k * (m*n + m*n)) which simplifies to O(k * m * n). Since the maximum possible k is limited by m*n, the complexity could be considered O((m*n)*(m*n)), simplifying it to O(m²n²). However, assuming k is smaller than m*n it's best represented as O(kmn).

Space Complexity

O(C) – The algorithm uses auxiliary space primarily for storing the bounding rectangles for each color. In the worst case, where C is the number of distinct colors in the image, we store information for each color's rectangle (top, left, bottom, right coordinates). Additionally, there might be a set or list used to keep track of the outermost colors. Since the number of distinct colors 'C' is the dominant factor, the space complexity is O(C). The actual dimensions of the image do not significantly affect the auxiliary space used.

Edge Cases

Case	How to Handle
Null or empty input matrix	Return true immediately, as no printing is required, thus 'valid'.
Single color in the entire matrix	Printing that single color is always a valid solution, return true.
Matrix with only one row or one column	Analyze the row/column sequentially and ensure any color printed covers a contiguous segment.
Maximum matrix size (e.g., 60x60 as specified in problem constraints) with distinct colors	Ensure the algorithm's time complexity (likely involving cycle detection and topological sort) does not exceed the time limit for this large input.
Two colors overlap such that neither can be printed before the other (cyclic dependency)	The topological sort should detect cycles, indicating an invalid print sequence, and return false.
Integer overflow when calculating area or counts of color regions	Use appropriate data types (e.g., long) to avoid overflow when dealing with large matrix sizes.
Color values outside the specified range [1, 60]	Check for invalid color values at the start and return false, or treat out-of-range values as a distinct ignored color.
A color completely encloses another color	The dependency graph should correctly represent this, and the topological sort should handle printing the enclosing color first.

Null or empty input matrix

How to Handle:

Return true immediately, as no printing is required, thus 'valid'.

Single color in the entire matrix

How to Handle:

Printing that single color is always a valid solution, return true.

Matrix with only one row or one column

How to Handle:

Analyze the row/column sequentially and ensure any color printed covers a contiguous segment.

Maximum matrix size (e.g., 60x60 as specified in problem constraints) with distinct colors

How to Handle:

Ensure the algorithm's time complexity (likely involving cycle detection and topological sort) does not exceed the time limit for this large input.

Two colors overlap such that neither can be printed before the other (cyclic dependency)

How to Handle:

The topological sort should detect cycles, indicating an invalid print sequence, and return false.

Integer overflow when calculating area or counts of color regions

How to Handle:

Use appropriate data types (e.g., long) to avoid overflow when dealing with large matrix sizes.

Color values outside the specified range [1, 60]

How to Handle:

Check for invalid color values at the start and return false, or treat out-of-range values as a distinct ignored color.

A color completely encloses another color

How to Handle:

The dependency graph should correctly represent this, and the topological sort should handle printing the enclosing color first.

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 (rows and columns) of the `targetGrid` matrix, and what is the maximum possible value for any color in the grid?
Can the color values in the `targetGrid` be zero?
If it's impossible to print the `targetGrid` using any sequence of painting operations, what should the function return (e.g., `false`, throw an exception)?
Are the color values guaranteed to be within a specific range (e.g., 1 to 100)?
If there are multiple possible sequences of painting operations that can produce the target grid, is any valid sequence acceptable, or is there a specific criterion to select the best sequence (e.g., shortest sequence, lexicographically smallest)?

Brute Force Solution

Approach

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

First, list all the different colors that appear in the grid.
Then, for each color, we assume that the last time the printer ran, it painted all locations of that color as a single rectangle.
Check if this assumption is valid by virtually removing that color from the grid, seeing if other colors were painted over.
If removing the color creates inconsistencies (like breaks in other colors that should have been solid), then that color was NOT the last color printed.
If removing the color is valid, we temporarily remove it from the grid and continue this process with the remaining colors.
We continue removing colors one by one, if possible, until all colors are removed without inconsistencies. If we can do this, the answer is yes. If we find a color we can't remove, we try a different order by backing up and picking a different 'last' color.
If after trying all possible orders of unpainting colors we can't remove all the colors without inconsistencies, the answer is no.

Code Implementation

def is_printable(grid):
    rows = len(grid)
    cols = len(grid[0])
    colors = set()
    for row in range(rows):
        for col in range(cols):
            colors.add(grid[row][col])
    colors = list(colors)

    def check_remove_color(current_grid, color_to_remove):
        rows = len(current_grid)
        cols = len(current_grid[0])
        min_row = rows
        max_row = -1
        min_col = cols
        max_col = -1
        for row in range(rows):
            for col in range(cols):
                if current_grid[row][col] == color_to_remove:
                    min_row = min(min_row, row)
                    max_row = max(max_row, row)
                    min_col = min(min_col, col)
                    max_col = max(max_col, col)

        # Verify if color_to_remove can be removed
        for row in range(rows):
            for col in range(cols):
                if (min_row <= row <= max_row and min_col <= col <= max_col
                        and current_grid[row][col] != color_to_remove
                        and current_grid[row][col] != 0):
                    return False
        return True

    def solve(current_grid, remaining_colors):
        if not remaining_colors:
            return True

        for color_to_remove in remaining_colors:
            # Check if the current color can be removed
            if check_remove_color(current_grid, color_to_remove):

                # Creating a new grid to 'unpaint' the color
                new_grid = [row[:] for row in current_grid]
                rows = len(new_grid)
                cols = len(new_grid[0])
                for row in range(rows):
                    for col in range(cols):
                        if new_grid[row][col] == color_to_remove:
                            new_grid[row][col] = 0

                new_remaining_colors = remaining_colors[:]
                new_remaining_colors.remove(color_to_remove)

                # Recursively check
                if solve(new_grid, new_remaining_colors):
                    return True

        return False

    # Begin the solving process with the original grid
    return solve([row[:] for row in grid], colors)

Big(O) Analysis

Time Complexity

Space Complexity

Optimal Solution

Approach

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

First, identify the outermost layer of colors by finding which colors aren't covered by any other colors.
For each of these outermost colors, determine the smallest rectangle that fully contains all instances of that color in the image.
Check if the color is present only within its determined rectangle. If it extends outside the rectangle, then it could not have been the last printed color and the image is not valid. Return that it is not valid.
If the color meets all checks, conceptually remove the rectangle filled with this color from the canvas. You don't actually have to modify the image; just pretend it's gone.
Repeat the process on the modified (imagined) canvas. Find the new outermost colors and their bounding rectangles and validate the colors.
Keep going until the canvas is completely clear (all colors have been removed) or you determine a problem. If the canvas is cleared then the result is valid; otherwise invalid.

Code Implementation

def strange_printer_two(target_grid):
    number_of_rows = len(target_grid)
    number_of_columns = len(target_grid[0])

    color_bounds = {}
    for row_index in range(number_of_rows):
        for column_index in range(number_of_columns):
            color = target_grid[row_index][column_index]
            if color != 0:
                if color not in color_bounds:
                    color_bounds[color] = [row_index, column_index, row_index, column_index]
                else:
                    color_bounds[color][0] = min(color_bounds[color][0], row_index)
                    color_bounds[color][1] = min(color_bounds[color][1], column_index)
                    color_bounds[color][2] = max(color_bounds[color][2], row_index)
                    color_bounds[color][3] = max(color_bounds[color][3], column_index)

    while color_bounds:
        erasable_colors = []
        for color in list(color_bounds.keys()):
            is_erasable = True
            row_start, column_start, row_end, column_end = color_bounds[color]
            for row_index in range(row_start, row_end + 1):
                for column_index in range(column_start, column_end + 1):
                    current_color = target_grid[row_index][column_index]
                    if current_color != 0 and current_color != color:
                        if current_color in color_bounds:
                            is_erasable = False
                            break
                if not is_erasable:
                    break
            if is_erasable:
                erasable_colors.append(color)

        if not erasable_colors:
            return False

        # Simulate removing the colors
        for color in erasable_colors:
            row_start, column_start, row_end, column_end = color_bounds[color]
            for row_index in range(row_start, row_end + 1):
                for column_index in range(column_start, column_end + 1):
                    if target_grid[row_index][column_index] == color:
                        target_grid[row_index][column_index] = 0

            # Remove the color from bounds
            del color_bounds[color]

    # If all colors were removed, the image is valid
    return True

Big(O) Analysis

Time Complexity

Space Complexity

Edge Cases

Case	How to Handle
Null or empty input matrix	Return true immediately, as no printing is required, thus 'valid'.
Single color in the entire matrix	Printing that single color is always a valid solution, return true.
Matrix with only one row or one column	Analyze the row/column sequentially and ensure any color printed covers a contiguous segment.
Maximum matrix size (e.g., 60x60 as specified in problem constraints) with distinct colors	Ensure the algorithm's time complexity (likely involving cycle detection and topological sort) does not exceed the time limit for this large input.
Two colors overlap such that neither can be printed before the other (cyclic dependency)	The topological sort should detect cycles, indicating an invalid print sequence, and return false.
Integer overflow when calculating area or counts of color regions	Use appropriate data types (e.g., long) to avoid overflow when dealing with large matrix sizes.
Color values outside the specified range [1, 60]	Check for invalid color values at the start and return false, or treat out-of-range values as a distinct ignored color.
A color completely encloses another color	The dependency graph should correctly represent this, and the topological sort should handle printing the enclosing color first.

Null or empty input matrix

How to Handle:

Return true immediately, as no printing is required, thus 'valid'.

Single color in the entire matrix

How to Handle:

Printing that single color is always a valid solution, return true.

Matrix with only one row or one column

How to Handle:

Analyze the row/column sequentially and ensure any color printed covers a contiguous segment.

Maximum matrix size (e.g., 60x60 as specified in problem constraints) with distinct colors

How to Handle:

Ensure the algorithm's time complexity (likely involving cycle detection and topological sort) does not exceed the time limit for this large input.

Two colors overlap such that neither can be printed before the other (cyclic dependency)

How to Handle:

The topological sort should detect cycles, indicating an invalid print sequence, and return false.

Integer overflow when calculating area or counts of color regions

How to Handle:

Use appropriate data types (e.g., long) to avoid overflow when dealing with large matrix sizes.

Color values outside the specified range [1, 60]

How to Handle:

Check for invalid color values at the start and return false, or treat out-of-range values as a distinct ignored color.

A color completely encloses another color

How to Handle:

The dependency graph should correctly represent this, and the topological sort should handle printing the enclosing color first.