<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 defaultdict
from typing import List

class Solution:
    def minimumOperationsToWriteY(self, grid: List[List[int]]) -> int:
        frequency = defaultdict(int)  # Frequency of numbers in the "Y"
        n = len(grid) 
        center = n // 2  # Middle point of the grid
        visited = set()
        y_total, non_y_total = 0, 0
        
        # Top half: diagonal arms of the "Y"
        for r in range(center):
            # Main diagonal (left arm)
            frequency[grid[r][r]] += 1
            visited.add((r, r))
            y_total += 1
            
            # Secondary diagonal (right arm)
            frequency[grid[r][n - r - 1]] += 1
            visited.add((r, n - r - 1))
            y_total += 1
        
        # Center cell (the middle point of the Y)
        frequency[grid[center][center]] += 1
        visited.add((center, center))
        y_total += 1
        
        # Bottom half: vertical stem of the "Y"
        for row in range(center + 1, n):
            frequency[grid[row][center]] += 1
            visited.add((row, center))
            y_total += 1

        # Collect the frequencies of elements not part of the "Y"
        frequency_else = defaultdict(int)
        for row in range(n):
            for col in range(n):
                if (row, col) not in visited:
                    frequency_else[grid[row][col]] += 1
                    non_y_total += 1
        
        # Now, we need to calculate the minimum number of operations to make all cells in the "Y" have the same type
        ans = float("inf")
        for y_type in [0, 1, 2]:
            for non_y_type in [0, 1, 2]:
                if y_type != non_y_type:
            # Operations to make all Y cells of type `y_type` plus non-Y cells to another type
                    operations = (y_total - frequency[y_type]) + (non_y_total - frequency_else[non_y_type])
                    ans = min(ans, operations)
        
        return ans

16700