<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:
        newtemp = [[(1 if (grid[0][0] =='X') else 0, 1 if (grid[0][0] =='Y') else 0, grid[0][0] == 'X')]]
        count = 0
        for j in range(1,len(grid[0])):
            contain = grid[0][j] == 'X' or newtemp[0][j-1][2]
            x = newtemp[0][j-1][0] + (grid[0][j] == 'X')
            y = newtemp[0][j-1][1] + (grid[0][j] == 'Y')
            if x == y and contain:
                count +=1
            newtemp[0].append((x,y,contain))

        for i in range(1,len(grid)):
            contain = grid[i][0] == 'X' or newtemp[i-1][0][2]
            x = newtemp[i-1][0][0] + (grid[i][0] == 'X')
            y = newtemp[i-1][0][1] + (grid[i][0] == 'Y')
            if x == y and contain:
                count +=1
            newtemp.append([(x,y,contain)])

        print(newtemp)
        for i in range(1,len(grid)):
            for j in range(1,len(grid[0])):
                contain = (grid[i][j] == 'X') or newtemp[i][j-1][2] or newtemp[i-1][j][2]
                x = newtemp[i][j-1][0] + newtemp[i-1][j][0] - newtemp[i-1][j-1][0] + (grid[i][j] == 'X') 
                y = newtemp[i][j-1][1] + newtemp[i-1][j][1] - newtemp[i-1][j-1][1] + (grid[i][j] == 'Y')
                if x == y and contain:
                    count +=1
                newtemp[i].append((x,y,contain))
        
        return count
161259