<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:
        
        n = len(grid)
        center_r = n // 2
        center_c = n // 2

        counts_y = {0:0, 1:0, 2:0}
        counts_ny = {0:0, 1:0, 2:0}

        total_y = 0
        total_ny = 0

        for r in range(n):
            for c in range(n):
                v = grid[r][c]
                is_y_cell = False
                if (r <= center_r) and (r == c):
                    is_y_cell = True
                elif (r <= center_r) and (r + c == n -1):
                    is_y_cell = True
                elif (r >= center_r) and (c == center_c):
                    is_y_cell = True

                if is_y_cell:
                    counts_y[v] += 1
                    total_y +=1
                else:
                    counts_ny[v] +=1
                    total_ny +=1

                    
                    
        min_changes = inf
        for a in [0,1,2]:
            for b in [0,1,2]:
                if a == b:
                    continue
                changes_y = total_y - counts_y[a]
                changes_ny = total_ny - counts_ny[b]
                total_changes = changes_y + changes_ny
                
                min_changes = min(min_changes,total_changes )
                # if (min_changes is None) or (total_changes < min_changes):
                #     min_changes = total_changes

        return min_changes



    '''
    
        0. 1. 1
        2. 2  0
        2  0. 0
    '''
   
    
            

        
     

            
   
        
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