<p>You are given a <strong>0-indexed</strong> <code>n x n</code> grid where <code>n</code> is odd, and <code>grid[r][c]</code> is <code>0</code>, <code>1</code>, or <code>2</code>.</p>

<p>We say that a cell belongs to the Letter <strong>Y</strong> if it belongs to one of the following:</p>

<ul>
	<li>The diagonal starting at the top-left cell and ending at the center cell of the grid.</li>
	<li>The diagonal starting at the top-right cell and ending at the center cell of the grid.</li>
	<li>The vertical line starting at the center cell and ending at the bottom border of the grid.</li>
</ul>

<p>The Letter <strong>Y</strong> is written on the grid if and only if:</p>

<ul>
	<li>All values at cells belonging to the Y are equal.</li>
	<li>All values at cells not belonging to the Y are equal.</li>
	<li>The values at cells belonging to the Y are different from the values at cells not belonging to the Y.</li>
</ul>

<p>Return <em>the <strong>minimum</strong> number of operations needed to write the letter Y on the grid given that in one operation you can change the value at any cell to</em> <code>0</code><em>,</em> <code>1</code><em>,</em> <em>or</em> <code>2</code><em>.</em></p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>
<img alt="" src="https://assets.leetcode.com/uploads/2024/01/22/y2.png" style="width: 461px; height: 121px;" />
<pre>
<strong>Input:</strong> grid = [[1,2,2],[1,1,0],[0,1,0]]
<strong>Output:</strong> 3
<strong>Explanation:</strong> We can write Y on the grid by applying the changes highlighted in blue in the image above. After the operations, all cells that belong to Y, denoted in bold, have the same value of 1 while those that do not belong to Y are equal to 0.
It can be shown that 3 is the minimum number of operations needed to write Y on the grid.
</pre>

<p><strong class="example">Example 2:</strong></p>
<img alt="" src="https://assets.leetcode.com/uploads/2024/01/22/y3.png" style="width: 701px; height: 201px;" />
<pre>
<strong>Input:</strong> grid = [[0,1,0,1,0],[2,1,0,1,2],[2,2,2,0,1],[2,2,2,2,2],[2,1,2,2,2]]
<strong>Output:</strong> 12
<strong>Explanation:</strong> We can write Y on the grid by applying the changes highlighted in blue in the image above. After the operations, all cells that belong to Y, denoted in bold, have the same value of 0 while those that do not belong to Y are equal to 2. 
It can be shown that 12 is the minimum number of operations needed to write Y on the grid.</pre>

<p>&nbsp;</p>
<p><strong>Constraints:</strong></p>

<ul>
	<li><code>3 &lt;= n &lt;= 49 </code></li>
	<li><code>n == grid.length == grid[i].length</code></li>
	<li><code>0 &lt;= grid[i][j] &lt;= 2</code></li>
	<li><code>n</code> is odd.</li>
</ul>

from collections import Counter
from typing import List

class Solution:
    def minimumOperationsToWriteY(self, matrix: List[List[int]]) -> int:
        def partOfY(r, c, n):
            # Check if the cell is part of the "Y" shape
            return (r <= n//2 and (r == c or r + c == n-1)) or (r >= n//2 and c == n//2)
                
        N = len(matrix)
        
        Ycells = []
        BGcells = []
        
        # Categorize cells into Y shape and background
        for r in range(N):
            for c in range(N):
                if partOfY(r, c, N):
                    Ycells.append(matrix[r][c])
                else:
                    BGcells.append(matrix[r][c])
        
        # Count occurrences of elements in Y cells and BG cells
        MY = Counter(Ycells) 
        MBG = Counter(BGcells)
        
        # Sort counts in descending order
        sortedY = MY.most_common()
        sortedBG = MBG.most_common()
        
        # Edge case handling: Check if there's a second dominant element in Y and BG
        dominant_Y = sortedY[0] if len(sortedY) > 0 else (None, 0)
        second_dominant_Y = sortedY[1] if len(sortedY) > 1 else (None, 0)
        
        dominant_BG = sortedBG[0] if len(sortedBG) > 0 else (None, 0)
        second_dominant_BG = sortedBG[1] if len(sortedBG) > 1 else (None, 0)
        
        # Case 1: If dominant elements in Y and BG are different
        if dominant_Y[0] != dominant_BG[0]:
            return len(Ycells) - dominant_Y[1] + len(BGcells) - dominant_BG[1]
        else:
            # Case 2: If dominant elements in Y and BG are the same, choose the best alternative
            option1 = len(Ycells) - dominant_Y[1] + len(BGcells) - second_dominant_BG[1]  # Change BG to 2nd dominant
            option2 = len(Ycells) - second_dominant_Y[1] + len(BGcells) - dominant_BG[1]  # Change Y to 2nd dominant
            return min(option1, option2)

16800