<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
class Solution:
    # Time: O()
    # Space: O(), where n = len(grid)
    def minimumOperationsToWriteY(self, grid: List[List[int]]) -> int:
        
        n = len(grid)
        n_half = n//2

        def is_Y(i, j):
            if i >= n_half and j == n_half:
                return True
            if i == j and i <= n_half and j <= n_half:
                return True
            if i == n-j-1 and i <= n_half and j >= n_half:
                return True
            return False
        
        count_in, count_out = defaultdict(int), defaultdict(int)
        n_in, n_out = 0, 0
        for i in range(n):
            for j in range(n):
                if is_Y(i, j):
                    count_in[grid[i][j]] += 1
                    n_in += 1
                    # print(f'{i}{j}')
                else:
                    count_out[grid[i][j]] += 1
                    n_out += 1
        sorted_in = count_in # sorted([[num, f] for num, f in count_in.items()], key = lambda x: -x[1])
        sorted_out = count_out # sorted([[num, f] for num, f in count_out.items()], key = lambda x: -x[1])
        # print(sorted_in, sorted_out)

        output = n**2
        for i in range(0, 3):
            res = n_out - sorted_out[i]
            for j in range(0, 3):
                if i != j:
                    output = min(output, res + n_in - sorted_in[j])
        return output
            
        

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