<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:
        """
        BFS while changing the grid to two colrs
        1. We need to get freq of the Ygrid, Non Y grid 
        ygrid = {1:3, 2:1}
        rest = {0:3, 1:1, 2:1}
        
        ygrid:{2:3, 1:2, 0:2}
        rest:{2:15, 1:4, 0:3}
        """
        #get the y grid 
        
        rows, cols = len(grid), len(grid[0])
        y_grid = defaultdict(int)
        others = defaultdict(int)
        visited = set()
        # anti diagonal and the vertical
        r,c = 0,0
        while r < rows and c < cols:
            # if (r,c) not in visited:
            y_grid[grid[r][c]]+=1
            visited.add((r,c))
            if r >= rows//2:
                r+=1
            else:
                r+=1
                c+=1
                
        # diagonal and the vertical      
        r,c = 0,cols-1
        while r < rows and c < cols:
            if (r,c) == (rows//2, cols//2):
                break
            
            # if (r,c) not in visited:
            y_grid[grid[r][c]]+=1
            visited.add((r,c))
            r+=1
            c-=1
                
        #others
        for r in range(rows):
            for c in range(cols):
                if (r,c) not in visited:
                    visited.add((r,c))
                    others[grid[r][c]]+=1
        
        
        #double for loop to find minimum cost
        total = rows*cols
        res = float("inf")
        for a,b in itertools.permutations(list(range(3)), 2):
            # print(a,b)
            sum_a = sum(y_grid[x] if x!=a else 0 for x in range(3))
            sum_b = sum(others[x] if x!=b else 0 for x in range(3))
            total_changes = sum_a + sum_b
            res = min(total_changes, res)
        # print(others)
        # print(y_grid)
        return res
                            
                                          
                    
        
                    
        
                    
            
400