Count Total Number of Colored Cells

1. Handle the base case: if n is 1, return 1.
2. Initialize a long variable totalCells to 1 (for the cell colored at minute 1).
3. Loop with a variable i from 2 to n.
4. In each iteration, calculate the number of new cells added at minute i, which is 4 * (i - 1).
5. Add this number to totalCells.
6. After the loop completes, return the final totalCells.

Let's analyze the number of new cells added at each minute:

• Minute 1: 1 cell is colored. Total = 1.
• Minute 2: 4 new cells are added around the first one. Total = 1 + 4 = 5.
• Minute 3: The shape at minute 2 is a diamond. The new cells are added along its perimeter. The number of new cells is 8. Total = 5 + 8 = 13.
• Minute 4: 12 new cells are added. Total = 13 + 12 = 25.

The number of new cells added at minute i (for i > 1) is 4 * (i-1). This gives us a recurrence relation for the total number of cells C(n): C(n) = C(n-1) + 4 * (n-1) with the base case C(1) = 1.

We can implement this by starting with a total of 1 and iterating from i = 2 to n, adding 4 * (i-1) at each step. It is crucial to use a 64-bit integer type (like long in Java) for the total count to prevent overflow, as the result can exceed the capacity of a 32-bit integer for large n.

1class Solution {2    public long coloredCells(int n) {3        // Start with 1 cell at n=14        long totalCells = 1;5        6        // For each minute from 2 to n, add 4*(i-1) new cells7        for (int i = 2; i <= n; i++) {8            totalCells += 4L * (i - 1);9        }10        11        return totalCells;12    }13}

1. Define a Point class or use a similar structure to represent cell coordinates.
2. Initialize a Set<Point> called coloredCells and add the origin (0,0).
3. Loop for minute from 2 to n.
4. Create a temporary Set<Point> called newlyColored to store cells to be colored in the current minute.
5. For each point currently in coloredCells:
6. Check its four neighbors: `(x+1, y)`, `(x-1, y)`, `(x, y+1)`, and `(x, y-1)`.
7. For each `neighbor`, if it is not already in `coloredCells`, add it to the `newlyColored` set.
8. Add all points from newlyColored to the main coloredCells set.
9. After the loops complete, the answer is the final size of coloredCells.

We can use a Set of objects representing coordinates (e.g., a custom Point class with x and y attributes) to keep track of the colored cells. The Set data structure is crucial here to automatically handle duplicate cells and provide efficient lookups (average O(1) time).

The simulation starts with a single cell (0, 0) in the set at minute 1. Then, we loop from minute 2 to n. In each minute's iteration, we create a temporary set to hold the newly colored cells. We iterate through every cell that is already colored, find its four adjacent neighbors, and if a neighbor isn't in our main set of colored cells, we add it to the temporary set. After checking all existing colored cells, we merge the temporary set into the main set. This process is repeated n-1 times. The final answer is the total number of unique points in our main set.

1// This is a conceptual example that requires a Point class with proper equals() and hashCode() implementations.2// This code is too slow and will not pass the problem constraints.3import java.util.HashSet;4import java.util.Set;5import java.util.Objects;6 7class Point {8    int x, y;9    Point(int x, int y) { this.x = x; this.y = y; }10    @Override11    public boolean equals(Object o) {12        if (this == o) return true;13        if (o == null || getClass() != o.getClass()) return false;14        Point point = (Point) o;15        return x == point.x && y == point.y;16    }17    @Override18    public int hashCode() {19        return Objects.hash(x, y);20    }21}22 23public long coloredCells(int n) {24    if (n == 1) return 1;25    26    Set<Point> colored = new HashSet<>();27    colored.add(new Point(0, 0));28    29    int[] dx = {0, 0, 1, -1};30    int[] dy = {1, -1, 0, 0};31    32    // Loop from minute 2 to n33    for (int i = 2; i <= n; i++) {34        Set<Point> newlyColored = new HashSet<>();35        for (Point p : colored) {36            for (int j = 0; j < 4; j++) {37                Point neighbor = new Point(p.x + dx[j], p.y + dy[j]);38                if (!colored.contains(neighbor)) {39                    newlyColored.add(neighbor);40                }41            }42        }43        colored.addAll(newlyColored);44    }45    46    return colored.size();47}

1. Cast the input integer n to a 64-bit floating-point number (e.g., long in Java) to prevent potential overflow during intermediate calculations. Let's call it long_n.
2. Apply the derived mathematical formula: 1 + 2 * long_n * (long_n - 1).
3. Return the computed result.

The total number of colored cells, C(n), can be seen as an arithmetic series. The number of new cells added at each step i (for i>1) is 4*(i-1). So the total is: C(n) = 1 (at minute 1) + 4*1 (at minute 2) + 4*2 (at minute 3) + ... + 4*(n-1) (at minute n)

We can factor out the 4: C(n) = 1 + 4 * (1 + 2 + 3 + ... + (n-1))

The sum of the first k integers is given by the formula k*(k+1)/2. Here, k = n-1. Substituting this into our equation: C(n) = 1 + 4 * ((n-1) * ((n-1)+1) / 2) C(n) = 1 + 4 * ((n-1) * n / 2) C(n) = 1 + 2 * n * (n-1)

Expanding this gives an alternative form: C(n) = 1 + 2n^2 - 2n = 2n^2 - 2n + 1.

This formula allows us to calculate the result directly from n in a single step. To avoid integer overflow when calculating n*n (since n can be up to 10^5, n*n can be 10^10), we must cast n to a long before the multiplication.

1class Solution {2    public long coloredCells(int n) {3        // Cast n to long to avoid integer overflow during multiplication.4        long long_n = n;5        6        // Formula: 1 + 2 * n * (n - 1)7        // or 2*n*n - 2*n + 18        return 1L + 2L * long_n * (long_n - 1);9    }10}