Problem Description

You have an infinitely large 2D grid of uncolored unit cells. Given a positive integer n, you need to simulate a coloring process for n minutes following these rules:

• Minute 1: Color any one arbitrary unit cell blue
• Every subsequent minute: Color all uncolored cells that are adjacent (share an edge) with any blue cell

The coloring spreads outward from the initial cell, forming a diamond-like pattern. Each minute, the blue region expands by one layer in all four directions (up, down, left, right).

For example:

• After minute 1: Only 1 cell is colored (the initial cell)
• After minute 2: 5 cells are colored (initial cell + 4 adjacent cells)
• After minute 3: 13 cells are colored (previous 5 + 8 new adjacent cells)

The task is to return the total number of colored cells after n minutes.

The pattern forms a diamond shape where:

• After n minutes, the grid has 2n - 1 columns
• The number of colored cells in each column follows the pattern: 1, 3, 5, ..., 2n-1, 2n-3, ..., 3, 1
• This creates two arithmetic progressions (ascending then descending) with the total sum equal to 2 * n * (n - 1) + 1

Intuition

When we color a cell and then expand outward each minute, the pattern forms a diamond shape. Let's trace through what happens:

Starting with one cell at minute 1, we can think of this as the center of a coordinate system at position (0, 0). After each minute, the coloring spreads exactly one unit in all four directions (up, down, left, right).

After n minutes, the colored region extends n-1 units from the center in each direction. This means:

• The topmost colored cell is at row n-1
• The bottommost colored cell is at row -(n-1)
• The leftmost colored cell is at column -(n-1)
• The rightmost colored cell is at column n-1

If we count the colored cells column by column, we notice a pattern. The middle column (column 0) has the most cells: 2n - 1 cells (from row -(n-1) to row n-1). As we move away from the center column in either direction, each column has 2 fewer cells than the previous one.

This gives us the sequence of cells per column: 1, 3, 5, ..., 2n-3, 2n-1, 2n-3, ..., 5, 3, 1

We can split this into two parts:

• Left side (including center): 1 + 3 + 5 + ... + (2n-1) which is the sum of the first n odd numbers = n²
• Right side: 1 + 3 + 5 + ... + (2n-3) which is the sum of the first n-1 odd numbers = (n-1)²

Total = n² + (n-1)² = n² + n² - 2n + 1 = 2n² - 2n + 1 = 2n(n-1) + 1

Therefore, the formula 2 * n * (n - 1) + 1 gives us the total number of colored cells after n minutes.

Learn more about Math patterns.

Solution Approach

The solution uses a direct mathematical formula based on the pattern we discovered. Since we know that after n minutes, the colored cells form a diamond shape with specific dimensions, we can calculate the total count directly without simulation.

The implementation is straightforward:

1. Pattern Recognition: After analyzing the diamond formation, we identified that the grid has 2n - 1 columns after n minutes.

2. Column-wise Counting: Each column contains an odd number of cells:

   • The center column has 2n - 1 cells
   • Moving outward, each column has 2 fewer cells
   • The sequence is: 1, 3, 5, ..., 2n-1, 2n-3, ..., 5, 3, 1

3. Mathematical Formula: Instead of iterating through each column and summing, we use the derived formula:

   return 2 * n * (n - 1) + 1

This formula comes from recognizing that:

• The left half (including center) forms an arithmetic progression: 1 + 3 + 5 + ... + (2n-1) = n²
• The right half forms another arithmetic progression: 1 + 3 + 5 + ... + (2n-3) = (n-1)²
• Total = n² + (n-1)² = 2n² - 2n + 1 = 2n(n-1) + 1

The beauty of this solution is its O(1) time complexity and O(1) space complexity, as we compute the result directly without any loops or additional data structures.

Example Walkthrough

Let's walk through the solution with n = 3 minutes to see how the diamond pattern forms and verify our formula.

Minute 1: Color one cell (let's call it the center at position (0,0))

    B

Total colored cells: 1

Minute 2: Color all uncolored cells adjacent to blue cells

    B
   BBB
    B

Total colored cells: 5

Minute 3: Color all uncolored cells adjacent to any blue cell

    B
   BBB
  BBBBB
   BBB
    B

Total colored cells: 13

Now let's verify this matches our formula for n = 3:

• Using the formula: 2 * n * (n - 1) + 1 = 2 * 3 * 2 + 1 = 12 + 1 = 13 ✓

Let's also count column by column to see the pattern:

• Column -2: 1 cell
• Column -1: 3 cells
• Column 0 (center): 5 cells
• Column 1: 3 cells
• Column 2: 1 cell

The sequence is 1, 3, 5, 3, 1 which follows our expected pattern of odd numbers rising to 2n-1 = 5 then falling back down.

Sum of cells = 1 + 3 + 5 + 3 + 1 = 13 ✓

Breaking it down into our two arithmetic progressions:

• Left side (including center): 1 + 3 + 5 = 9 = 3²
• Right side: 1 + 3 = 4 = 2²
• Total: 9 + 4 = 13 ✓

This confirms our mathematical approach works correctly!

Solution Implementation

Time and Space Complexity

The time complexity is O(1) because the function performs a fixed number of arithmetic operations (two multiplications, one subtraction, and one addition) regardless of the input size n. These operations are computed directly using a mathematical formula without any loops or recursive calls.

The space complexity is O(1) because the function only uses a constant amount of extra memory. It stores the input parameter n and returns a single integer value. No additional data structures, arrays, or variables that scale with the input size are created.

Learn more about how to find time and space complexity quickly.

Common Pitfalls

1. Misunderstanding the Growth Pattern

Many developers initially assume the pattern grows as a square or follows a simple geometric progression. The diamond pattern is actually more nuanced - it expands in all four directions equally, creating a rotated square (diamond) shape.

Wrong Approach:

def coloredCells(self, n: int) -> int:
    # Incorrect: Assuming square growth
    return (2 * n - 1) ** 2  # This would give n=3 -> 25, but correct is 13

Solution: Visualize the pattern for small values of n to understand the diamond formation. Draw it out on paper or trace through the first few minutes manually.

2. Off-by-One Errors in Formula Derivation

When deriving the formula, it's easy to miscalculate the number of rows/columns or the cells in each row.

Wrong Approach:

def coloredCells(self, n: int) -> int:
    # Incorrect: Wrong formula due to miscounting
    return 2 * n * n - 1  # Missing the proper adjustment

Solution: Verify your formula with known test cases:

• n=1 should give 1
• n=2 should give 5
• n=3 should give 13

3. Attempting Complex Simulation

Some developers try to simulate the actual spreading process using BFS or maintaining a grid, which is unnecessary and inefficient.

Wrong Approach:

def coloredCells(self, n: int) -> int:
    # Inefficient: Simulating the actual spreading
    colored = set()
    colored.add((0, 0))
  
    for minute in range(2, n + 1):
        new_colored = set()
        for x, y in colored:
            for dx, dy in [(0,1), (0,-1), (1,0), (-1,0)]:
                if (x+dx, y+dy) not in colored:
                    new_colored.add((x+dx, y+dy))
        colored.update(new_colored)
  
    return len(colored)

Solution: Recognize that this is a mathematical pattern problem, not a simulation problem. The direct formula approach is both simpler and more efficient.

4. Integer Overflow Concerns (in Other Languages)

While Python handles large integers automatically, in languages like C++ or Java, the multiplication 2 * n * (n - 1) could overflow for large n.

Solution for Other Languages:

// C++ example with proper type handling
long long coloredCells(int n) {
    return 2LL * n * (n - 1) + 1;  // Use long long to prevent overflow
}

5. Not Handling Edge Cases

Forgetting to test or handle the minimum case where n=1.

Solution: Always verify that your formula works for the base case. When n=1, only the initial cell is colored, so the result should be 1. The formula 2 * 1 * (1 - 1) + 1 = 1 correctly handles this.