1148MediumMath

Cells Inked After n Minutes

Tracked in this browser only

Trains the technique from

LeetCode 2579Count Total Number of Colored Cells

This is an original problem, written from a brief that listed the technique, the difficulty, the topics, the function shape and the input bounds — none of that problem's wording, examples, hints or editorials. The link is there so you can map your practice onto the standard set.

Same function shape, different story and different numbers.

An endless sheet of square cells starts blank. In the first minute a single cell is inked. In every minute after that, every blank cell sharing a side with an inked cell becomes inked as well.

Return how many cells are inked at the end of minute n.

Examples

Example 1

Input: n = 3
Output: 13

One cell after the first minute, its four side neighbours after the second for five in all, and the eight cells ringing those after the third for thirteen.

Example 2

Input: n = 4
Output: 25

A fourth minute adds the ring of twelve cells one step further out, bringing the total to twenty-five.

Example 3

Input: n = 7
Output: 85

After seven minutes the diamond reaches six steps from the middle in every direction.

Constraints

1 <= n <= 10^5

The signature

The editor is preloaded with this. It matches the parent problem's shape, so a solution that works here transfers to a judge unchanged.

Python

def colored_cells(n: int) -> int:

Java

public long coloredCells(int n)

The intended approach

Time O(1).Space O(1).

The inked shape is always a diamond centred on the first cell, and each minute adds the ring of cells exactly one step further out. After the first minute the shape reaches nowhere, and each later minute pushes the reach out by one, so at the end of minute `i` the shape reaches `i - 1` steps from the middle. The ring at reach `d` holds `4d` cells, because the diamond's four sides each carry `d` of them. So the total is one for the middle plus four times the reaches from one up to `n - 1`, which is one plus four times `(n - 1)n / 2`. That tidies to `2n^2 - 2n + 1`, so the whole thing is arithmetic with no simulation at all. The count grows with the square of the minutes, so it runs past a thirty-two-bit integer well inside the allowed size.

Pythonentry point: colored_cells

def colored_cells(n):
    return 2 * n * n - 2 * n + 1

The obvious version

Spread the ink out one minute at a time. Hold the cells inked in the previous minute, look at their four side neighbours, and ink every one not already inked, keeping that new set as the next minute's edge. Count the inked cells at the finish. Costs the square of the minutes in cells.

This is not filler. It was written independently of the fast solution and the two were run against thousands of random inputs to confirm they agree — which is how the answer above is known to be right. Read it when the intended approach feels like a rabbit pulled out of a hat.

Python

def bf_colored_cells(n):
    inked = set()
    inked.add((0, 0))
    edge = [(0, 0)]
    for _ in range(n - 1):
        fresh = []
        for y, x in edge:
            for dy, dx in ((1, 0), (-1, 0), (0, 1), (0, -1)):
                spot = (y + dy, x + dx)
                if spot not in inked:
                    inked.add(spot)
                    fresh.append(spot)
        edge = fresh
    return len(inked)

How this problem was checked

Reference solution executed against 15 test cases — 15 passed
3 worked examples confirmed by running the code
0 eight-word sequences shared with the parent problem's text
Agrees with an independent brute-force implementation on random inputs
Test set proven to catch deliberately broken solutions
Reviewed by a separate quality pass
Function arity matches the parent problem's

Create a free account to unlock the solution

See the step-by-step animation, the intuition, and clean code in every language — free, no credit card.

Create a free account Already have an account? Sign in