<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:
        @cache
        def count_y_ops(cell):
            counts = count_y()
            return count_ops(cell, counts)

        @cache
        def count_outside_ops(cell):
            counts = count_outside_y()
            return count_ops(cell, counts)
        
        def count_ops(cell, counts):
            ops_count = 0
            for k in counts:
                if k == cell:
                    continue
                ops_count += counts[k]

            return ops_count

        @cache
        def count_y():
            mid = len(grid) // 2
            counts = Counter()
            # count diagonal
            for i in range(mid):
                counts[grid[i][i]] += 1
                counts[grid[i][len(grid) - 1 - i]] += 1

            # count vertical
            for i in range(mid, len(grid)):
                counts[grid[i][mid]] += 1
            
            return counts
        
        @cache
        def count_outside_y():
            counts = Counter()
            for i in range(len(grid)):
                for j in range(len(grid)):
                    counts[grid[i][j]] += 1
            
            y_counts = count_y()
            for cell in counts:
                counts[cell] -= y_counts[cell]
            
            return counts
        
        ans = float('inf')
        
        y_counts = count_y()
        outside_counts = count_outside_y()

        for cell_y in range(3):
            for cell_outside in range(3):
                if cell_y == cell_outside:
                    continue
                
                candidate = count_y_ops(cell_y) + count_outside_ops(cell_outside)
                ans = min(ans, candidate)
        
        return ans

388