2579. Count Total Number of Colored Cells - Detailed Explanation

Problem Statement

Imagine you start with one colored cell. Then, in every subsequent step (or “layer”), you add cells around the previously colored pattern. The way the pattern grows is that the first cell forms the center (layer 1), and for each new layer from 2 to n, you add cells in such a way that the number of new cells added is proportional to the layer. Through analysis, you can derive that the total number of colored cells after n layers is given by the formula:

This formula is equivalent to:

Example Inputs & Outputs:

1. Example 1:

   • Input: n = 1
   • Output: 1
   • Explanation:
     The pattern starts with a single cell. No additional layers are added when n = 1.

2. Example 2:

   • Input: n = 2
   • Output: 5
   • Explanation:
     • Layer 1: 1 cell
     • Layer 2: Adds 4 new cells (one on each side of the center)
     • Total = 1 + 4 = 5

3. Example 3:

   • Input: n = 3
   • Output: 13
   • Explanation:
     • Layer 1: 1 cell
     • Layer 2: Adds 4 cells → Total becomes 5
     • Layer 3: Adds 8 cells (each new layer adds 4 more cells than the previous layer) → Total becomes 5 + 8 = 13

Constraints:

• n is a positive integer (for example, (1 \leq n \leq 10^5)).
• The solution should work efficiently even for large values of n.

Hints Before the Full Solution

1. Hint 1:
   Think about the pattern layer by layer. The first layer is just one cell, and every new layer wraps around the previous pattern. Notice that the number of cells added at each new layer forms an arithmetic progression.

2. Hint 2:
   Realize that the number of cells added in layer ( i ) (for ( i \geq 2 )) is given by:

   4 \times (i - 1)

   Then, sum these values along with the initial 1 cell. This observation leads you to a summation that can be simplified into a closed-form formula.

Approaches to Solve the Problem

1. Brute Force / Simulation Approach

Idea:

• Simulate each layer by “drawing” it or counting the cells added step-by-step.
• Start with 1 cell, then for each subsequent layer, add ( 4 \times (i - 1) ) cells.

Step-by-Step:

• Layer 1: 1 cell.
• Layer 2: Add ( 4 \times (2-1) = 4 ) cells.
• Layer 3: Add ( 4 \times (3-1) = 8 ) cells.
• … and so on until layer n.

Pseudo-code:

def coloredCells(n):
    total = 1  # layer 1
    for i in range(2, n + 1):
        total += 4 * (i - 1)
    return total

Drawback:

• Although simple and intuitive, this simulation runs in O(n) time. For very large n, this is less efficient than necessary.

2. Optimal Mathematical Derivation

Observation:

• The total number of colored cells is the sum of cells from each layer:

  \text{Total} = 1 + \sum_{i=2}^{n} 4 \times (i - 1)

• Recognize that the sum:

  \sum_{i=2}^{n} (i - 1)

  is equivalent to the sum of the first ((n-1)) natural numbers:

  \sum_{j=1}^{n-1} j = \frac{(n-1)n}{2}

• Therefore, substitute back into the total formula:

  \text{Total} = 1 + 4 \times \frac{(n-1)n}{2}

  which simplifies to:

  \text{Total} = 1 + 2n(n-1) = 2n^2 - 2n + 1

• Note that this is equivalent to:

  \text{Total Colored Cells} = n^2 + (n-1)^2

Time Complexity:

• O(1) – The answer is computed directly using the formula.

Code Implementation

Python Code