<p>Given a 2D character matrix <code>grid</code>, where <code>grid[i][j]</code> is either <code>&#39;X&#39;</code>, <code>&#39;Y&#39;</code>, or <code>&#39;.&#39;</code>, return the number of <span data-keyword="submatrix">submatrices</span> that contain:</p>

<ul>
	<li><code>grid[0][0]</code></li>
	<li>an <strong>equal</strong> frequency of <code>&#39;X&#39;</code> and <code>&#39;Y&#39;</code>.</li>
	<li><strong>at least</strong> one <code>&#39;X&#39;</code>.</li>
</ul>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>

<div class="example-block">
<p><strong>Input:</strong> <span class="example-io">grid = [[&quot;X&quot;,&quot;Y&quot;,&quot;.&quot;],[&quot;Y&quot;,&quot;.&quot;,&quot;.&quot;]]</span></p>

<p><strong>Output:</strong> <span class="example-io">3</span></p>

<p><strong>Explanation:</strong></p>

<p><strong><img alt="" src="https://assets.leetcode.com/uploads/2024/06/07/examplems.png" style="padding: 10px; background: rgb(255, 255, 255); border-radius: 0.5rem; width: 175px; height: 350px;" /></strong></p>
</div>

<p><strong class="example">Example 2:</strong></p>

<div class="example-block">
<p><strong>Input:</strong> <span class="example-io">grid = [[&quot;X&quot;,&quot;X&quot;],[&quot;X&quot;,&quot;Y&quot;]]</span></p>

<p><strong>Output:</strong> <span class="example-io">0</span></p>

<p><strong>Explanation:</strong></p>

<p>No submatrix has an equal frequency of <code>&#39;X&#39;</code> and <code>&#39;Y&#39;</code>.</p>
</div>

<p><strong class="example">Example 3:</strong></p>

<div class="example-block">
<p><strong>Input:</strong> <span class="example-io">grid = [[&quot;.&quot;,&quot;.&quot;],[&quot;.&quot;,&quot;.&quot;]]</span></p>

<p><strong>Output:</strong> <span class="example-io">0</span></p>

<p><strong>Explanation:</strong></p>

<p>No submatrix has at least one <code>&#39;X&#39;</code>.</p>
</div>

<p>&nbsp;</p>
<p><strong>Constraints:</strong></p>

<ul>
	<li><code>1 &lt;= grid.length, grid[i].length &lt;= 1000</code></li>
	<li><code>grid[i][j]</code> is either <code>&#39;X&#39;</code>, <code>&#39;Y&#39;</code>, or <code>&#39;.&#39;</code>.</li>
</ul>

class Solution:
    def numberOfSubmatrices(self, grid: List[List[str]]) -> int:
        n = len(grid)
        m = len(grid[0])
        extra = [[0] * (m) for _ in range(n)]
        matrix = [[0] * (m + 1) for _ in range(n+1)]
        for i in range(1, n+1):
            for j in range(1, m+1):
                if grid[i-1][j-1] == "X":
                    matrix[i][j] = 1
                elif grid[i-1][j-1] == "Y": 
                    matrix[i][j] = -1

        for j in range(1, m+1): 
            for i in range(1, n+1):
                matrix[i][j] += matrix[i-1][j]
        for i in range(1, n+1): 
            for j in range(1, m+1):
                matrix[i][j] += matrix[i][j-1]
        #print(matrix)
        arr = []
        for i in range(1, n+1):
            for j in range(1, m+1):
                if matrix[i][j] == 0:
                    arr.append((i-1, j-1))
        #print(len(arr))

        for i in range(n):
            for j in range(m):
                if grid[i][j] == "X": extra[i][j] = 1
                else:
                    extra[i][j] = max((extra[i][j-1] if j != 0 else 0), (extra[i-1][j] if i != 0 else 0))
        
        ans = 0
        for i, j in arr:
            if extra[i][j] != 0:
                ans += 1
        return ans

                
        
        
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