<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:
        prefixArr = [[0] * len(grid[0]) for _ in range(len(grid))]
        res = 0
        xFound = [False] * len(grid[0])
       #from 0,0 not valid till the first one appears thus starting a new row sets it to 0 until a 1 is detected
        mapping = {'X' : 1, 'Y' : -1, '.' : 0}
        for i in range(len(grid)):
            for j in range(len(grid[i])):
                if i == 0 and j == 0:
                    prefixArr[0][0] = mapping[grid[0][0]]
                elif i == 0:
                    prefixArr[0][j] += prefixArr[0][j - 1] + mapping[grid[0][j]]
                elif j == 0:
                    prefixArr[i][0] += prefixArr[i - 1][0] + mapping[grid[i][0]]
                else:
                    prefixArr[i][j] += prefixArr[i - 1][j] + prefixArr[i][j - 1] + mapping[grid[i][j]] - prefixArr[i - 1][j -1 ]
                if grid[i][j] != '.':
                    xFound[j] = True
                xFound[j] = xFound[j - 1] if not xFound[j] and j != 0  else xFound[j]
                if xFound[j] == True and prefixArr[i][j] == 0:
                    res += 1
                # print(i,j,res,xFound,prefixArr)
        return res

                     
                

        return 0
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