<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>

class Solution:
    def minimumOperationsToWriteY(self, grid: List[List[int]]) -> int:

        
        count_map = {i: 0 for i in range(3)}
        count_map_y = {i: 0 for i in range(3)}

        n = len(grid)

        result = 0

        y_set = set()

        for i in range(n):
            for j in range(len(grid[0])):
                if (i == j and i <= n//2) or (j==n-i-1 and i<= n//2) or (j==n//2 and n//2<i<=n):
                    count_map_y[grid[i][j]] +=1
                    y_set.add((i,j))
                else:
                    count_map[grid[i][j]] +=1

        if len(count_map) == 1 and len(count_map_y) == 1 and count_map.keys() == count_map_y.keys():
            return sum(count_map_y.values())

        min_change = float('inf')
        total_elements = n * n

        count_map_sum = sum(count_map.values())
        count_map_y_sum = sum(count_map_y.values())

        for i in count_map_y:
            for j in count_map:
                if i != j:
                    min_change = min(min_change, count_map_y_sum - count_map_y[i] + count_map_sum - count_map[j])

        return min_change



                
        
16500