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Count Total Number of Colored Cells

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There exists an infinitely large two-dimensional grid of uncolored unit cells. You are given a positive integer n, indicating that you must do the following routine for n minutes:

At the first minute, color any arbitrary unit cell blue.
Every minute thereafter, color blue every uncolored cell that touches a blue cell.

Below is a pictorial representation of the state of the grid after minutes 1, 2, and 3.

Return the number of colored cells at the end of n minutes.

Example 1:

Input: n = 1
Output: 1
Explanation: After 1 minute, there is only 1 blue cell, so we return 1.

Example 2:

Input: n = 2
Output: 5
Explanation: After 2 minutes, there are 4 colored cells on the boundary and 1 in the center, so we return 5.

Constraints:

1 <= n <= 10⁵

Try coding it in LeetCode

Count Total Number of Colored Cells

Medium

Asked by:

17 views

There exists an infinitely large two-dimensional grid of uncolored unit cells. You are given a positive integer n, indicating that you must do the following routine for n minutes:

At the first minute, color any arbitrary unit cell blue.
Every minute thereafter, color blue every uncolored cell that touches a blue cell.

Below is a pictorial representation of the state of the grid after minutes 1, 2, and 3.

Return the number of colored cells at the end of n minutes.

Example 1:

Input: n = 1
Output: 1
Explanation: After 1 minute, there is only 1 blue cell, so we return 1.

Example 2:

Input: n = 2
Output: 5
Explanation: After 2 minutes, there are 4 colored cells on the boundary and 1 in the center, so we return 5.

Constraints:

1 <= n <= 10⁵

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 is the maximum possible value for `n`? I want to be mindful of potential integer overflow issues.
Can `n` be zero or negative?
By 'colored cells', do you mean all cells within the outermost square, or just the cells that are part of the concentric squares themselves (i.e., the cells that form the perimeter of each square)?
Is there a specific return type expected (e.g., `int`, `long`)?
Could you provide a small example (e.g., n=3) and the corresponding expected output to ensure I fully understand the problem?

Brute Force Solution

Approach

The brute force way to count colored cells involves simulating the coloring process step-by-step. We'll essentially 'draw' the grid for each size increment and count the colored cells directly. This method guarantees finding the answer by explicitly constructing the pattern.

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

Start with a completely empty grid.
For the first step (size 1), color the single center cell.
For the next step (size 2), color the appropriate cells around the existing colored cells to expand the pattern.
Count all the colored cells in the grid at this step.
Repeat the coloring and counting process for each subsequent step, always building upon the colored cells from the previous step.
Continue this process until you reach the desired size.
The final count of colored cells is your answer.

Code Implementation

def count_colored_cells_brute_force(size):
    grid_dimension = 2 * size - 1
    grid = [[0] * grid_dimension for _ in range(grid_dimension)]

    center_row = size - 1
    center_column = size - 1
    grid[center_row][center_column] = 1
    total_colored_cells = 1

    for current_size in range(2, size + 1):
        # Expand the colored cells outwards from the center

        for i in range(center_row - (current_size - 1), center_row + current_size):
            if 0 <= i < grid_dimension and grid[i][center_column - (current_size - 1)] == 0:
                grid[i][center_column - (current_size - 1)] = 1
                total_colored_cells += 1

        for i in range(center_row - (current_size - 1), center_row + current_size):
            if 0 <= i < grid_dimension and grid[i][center_column + (current_size - 1)] == 0:
                grid[i][center_column + (current_size - 1)] = 1
                total_colored_cells += 1

        for j in range(center_column - (current_size - 2), center_column + (current_size - 1)):
            if 0 <= j < grid_dimension and grid[center_row - (current_size - 1)][j] == 0:
                grid[center_row - (current_size - 1)][j] = 1
                total_colored_cells += 1

        for j in range(center_column - (current_size - 2), center_column + (current_size - 1)):
            if 0 <= j < grid_dimension and grid[center_row + (current_size - 1)][j] == 0:
                grid[center_row + (current_size - 1)][j] = 1
                total_colored_cells += 1

    return total_colored_cells

Big(O) Analysis

Time Complexity

O(n³) – The brute force approach involves simulating the coloring process on a grid. For each size 'n', we iterate 'n' times. In each iteration, we potentially need to examine and update cells within a square grid of size (2n-1) x (2n-1) representing the colored area. Examining each cell in this square grid takes O(n²) operations. Since we repeat this for 'n' steps, the total time complexity is O(n * n²) which simplifies to O(n³).

Space Complexity

O(N^2) – The brute force method constructs a grid to simulate the coloring process. In the worst-case scenario, for a size N, the grid will have dimensions proportional to N x N to accommodate the colored cells. Therefore, the algorithm uses auxiliary space to store this grid, which requires space proportional to N^2. This results in a space complexity of O(N^2).

Optimal Solution

Approach

The most efficient way to count colored cells is to recognize a pattern that emerges as the size increases. Instead of simulating the cell coloring process, we leverage this pattern to directly calculate the number of colored cells.

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

Observe that the number of colored cells follows a pattern related to squares.
Specifically, realize that for a given size, the total number of colored cells can be calculated using a simple formula involving squaring the size and then adjusting it.
The size of the board can be plugged directly into the derived formula to get the result.
Apply the formula to determine the total number of colored cells efficiently.

Code Implementation

def count_colored_cells(size):
    # For size 1, only 1 cell is colored.
    if size == 1:
        return 1

    # The total number of colored cells is size * size + (size - 1) * (size - 1)
    total_colored_cells = size * size + (size - 1) * (size - 1)

    return total_colored_cells

def count_total_number_colored_cells(size_of_board):
    # If the size is 0 or negative, there are no colored cells
    if size_of_board <= 0:
        return 0

    # Formula: n^2 + (n-1)^2 = 2n^2 - 2n + 1
    total_colored_cells = 2 * size_of_board * size_of_board - 2 * size_of_board + 1

    # Return the calculated number of colored cells.
    return total_colored_cells

Big(O) Analysis

Time Complexity

O(1) – The solution involves a direct calculation using a formula based on the input size n. There are no loops or recursive calls that depend on n. The number of operations remains constant regardless of the size of n, since it's a simple arithmetic calculation. Therefore, the time complexity is O(1).

Space Complexity

O(1) – The described solution focuses on applying a formula directly to the input size (N), without creating any auxiliary data structures or making recursive calls. No temporary lists, hash maps, or significant variables that scale with the input size are used. Therefore, the space complexity is constant, independent of the input size N.

Edge Cases

Case	How to Handle
n is zero	Return 0 as there are no squares and thus no colored cells.
n is one	Return 1, corresponding to the single colored cell of the innermost square.
Large value of n that could lead to integer overflow	Use a larger integer type (e.g., long) to avoid potential overflow during calculation.
Negative input	Throw an IllegalArgumentException or return an error value because the number of squares cannot be negative.
Maximum integer value for n	Check for possible overflow issues when computing `(2 * n * n) - (2 * n) + 1`, possibly requiring arbitrary-precision arithmetic.
n is close to the maximum integer value, where n^2 might overflow.	Use long long or a language that supports large integers.
n is a very large number, leading to memory issues	The formula-based solution has a constant memory footprint and handles arbitrarily large n (within the data type limits).
The calculated value might exceed the maximum value representable by the integer type.	Use a long integer to avoid integer overflow issues.

n is zero

How to Handle:

Return 0 as there are no squares and thus no colored cells.

n is one

How to Handle:

Return 1, corresponding to the single colored cell of the innermost square.

Large value of n that could lead to integer overflow

How to Handle:

Use a larger integer type (e.g., long) to avoid potential overflow during calculation.

Negative input

How to Handle:

Throw an IllegalArgumentException or return an error value because the number of squares cannot be negative.

Maximum integer value for n

How to Handle:

Check for possible overflow issues when computing `(2 * n * n) - (2 * n) + 1`, possibly requiring arbitrary-precision arithmetic.

n is close to the maximum integer value, where n^2 might overflow.

How to Handle:

Use long long or a language that supports large integers.

n is a very large number, leading to memory issues

How to Handle:

The formula-based solution has a constant memory footprint and handles arbitrarily large n (within the data type limits).

The calculated value might exceed the maximum value representable by the integer type.

How to Handle:

Use a long integer to avoid integer overflow issues.

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 is the maximum possible value for `n`? I want to be mindful of potential integer overflow issues.
Can `n` be zero or negative?
By 'colored cells', do you mean all cells within the outermost square, or just the cells that are part of the concentric squares themselves (i.e., the cells that form the perimeter of each square)?
Is there a specific return type expected (e.g., `int`, `long`)?
Could you provide a small example (e.g., n=3) and the corresponding expected output to ensure I fully understand the problem?

Brute Force Solution

Approach

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

Start with a completely empty grid.
For the first step (size 1), color the single center cell.
For the next step (size 2), color the appropriate cells around the existing colored cells to expand the pattern.
Count all the colored cells in the grid at this step.
Repeat the coloring and counting process for each subsequent step, always building upon the colored cells from the previous step.
Continue this process until you reach the desired size.
The final count of colored cells is your answer.

Code Implementation

def count_colored_cells_brute_force(size):
    grid_dimension = 2 * size - 1
    grid = [[0] * grid_dimension for _ in range(grid_dimension)]

    center_row = size - 1
    center_column = size - 1
    grid[center_row][center_column] = 1
    total_colored_cells = 1

    for current_size in range(2, size + 1):
        # Expand the colored cells outwards from the center

        for i in range(center_row - (current_size - 1), center_row + current_size):
            if 0 <= i < grid_dimension and grid[i][center_column - (current_size - 1)] == 0:
                grid[i][center_column - (current_size - 1)] = 1
                total_colored_cells += 1

        for i in range(center_row - (current_size - 1), center_row + current_size):
            if 0 <= i < grid_dimension and grid[i][center_column + (current_size - 1)] == 0:
                grid[i][center_column + (current_size - 1)] = 1
                total_colored_cells += 1

        for j in range(center_column - (current_size - 2), center_column + (current_size - 1)):
            if 0 <= j < grid_dimension and grid[center_row - (current_size - 1)][j] == 0:
                grid[center_row - (current_size - 1)][j] = 1
                total_colored_cells += 1

        for j in range(center_column - (current_size - 2), center_column + (current_size - 1)):
            if 0 <= j < grid_dimension and grid[center_row + (current_size - 1)][j] == 0:
                grid[center_row + (current_size - 1)][j] = 1
                total_colored_cells += 1

    return total_colored_cells

Big(O) Analysis

Time Complexity

Space Complexity

Optimal Solution

Approach

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

Observe that the number of colored cells follows a pattern related to squares.
Specifically, realize that for a given size, the total number of colored cells can be calculated using a simple formula involving squaring the size and then adjusting it.
The size of the board can be plugged directly into the derived formula to get the result.
Apply the formula to determine the total number of colored cells efficiently.

Code Implementation

def count_colored_cells(size):
    # For size 1, only 1 cell is colored.
    if size == 1:
        return 1

    # The total number of colored cells is size * size + (size - 1) * (size - 1)
    total_colored_cells = size * size + (size - 1) * (size - 1)

    return total_colored_cells

def count_total_number_colored_cells(size_of_board):
    # If the size is 0 or negative, there are no colored cells
    if size_of_board <= 0:
        return 0

    # Formula: n^2 + (n-1)^2 = 2n^2 - 2n + 1
    total_colored_cells = 2 * size_of_board * size_of_board - 2 * size_of_board + 1

    # Return the calculated number of colored cells.
    return total_colored_cells

Big(O) Analysis

Time Complexity

Space Complexity

Edge Cases

Case	How to Handle
n is zero	Return 0 as there are no squares and thus no colored cells.
n is one	Return 1, corresponding to the single colored cell of the innermost square.
Large value of n that could lead to integer overflow	Use a larger integer type (e.g., long) to avoid potential overflow during calculation.
Negative input	Throw an IllegalArgumentException or return an error value because the number of squares cannot be negative.
Maximum integer value for n	Check for possible overflow issues when computing `(2 * n * n) - (2 * n) + 1`, possibly requiring arbitrary-precision arithmetic.
n is close to the maximum integer value, where n^2 might overflow.	Use long long or a language that supports large integers.
n is a very large number, leading to memory issues	The formula-based solution has a constant memory footprint and handles arbitrarily large n (within the data type limits).
The calculated value might exceed the maximum value representable by the integer type.	Use a long integer to avoid integer overflow issues.

n is zero

How to Handle:

Return 0 as there are no squares and thus no colored cells.

n is one

How to Handle:

Return 1, corresponding to the single colored cell of the innermost square.

Large value of n that could lead to integer overflow

How to Handle:

Use a larger integer type (e.g., long) to avoid potential overflow during calculation.

Negative input

How to Handle:

Throw an IllegalArgumentException or return an error value because the number of squares cannot be negative.

Maximum integer value for n

How to Handle:

Check for possible overflow issues when computing `(2 * n * n) - (2 * n) + 1`, possibly requiring arbitrary-precision arithmetic.

n is close to the maximum integer value, where n^2 might overflow.

How to Handle:

Use long long or a language that supports large integers.

n is a very large number, leading to memory issues

How to Handle:

The formula-based solution has a constant memory footprint and handles arbitrarily large n (within the data type limits).

The calculated value might exceed the maximum value representable by the integer type.

How to Handle:

Use a long integer to avoid integer overflow issues.