<p>You are given a <strong>0-indexed</strong> 2D integer matrix <code><font face="monospace">grid</font></code> of size <code>n * n</code> with values in the range <code>[1, n<sup>2</sup>]</code>. Each integer appears <strong>exactly once</strong> except <code>a</code> which appears <strong>twice</strong> and <code>b</code> which is <strong>missing</strong>. The task is to find the repeating and missing numbers <code>a</code> and <code>b</code>.</p>

<p>Return <em>a <strong>0-indexed </strong>integer array </em><code>ans</code><em> of size </em><code>2</code><em> where </em><code>ans[0]</code><em> equals to </em><code>a</code><em> and </em><code>ans[1]</code><em> equals to </em><code>b</code><em>.</em></p>

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

<pre>
<strong>Input:</strong> grid = [[1,3],[2,2]]
<strong>Output:</strong> [2,4]
<strong>Explanation:</strong> Number 2 is repeated and number 4 is missing so the answer is [2,4].
</pre>

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

<pre>
<strong>Input:</strong> grid = [[9,1,7],[8,9,2],[3,4,6]]
<strong>Output:</strong> [9,5]
<strong>Explanation:</strong> Number 9 is repeated and number 5 is missing so the answer is [9,5].
</pre>

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

<ul>
	<li><code>2 &lt;= n == grid.length == grid[i].length &lt;= 50</code></li>
	<li><code>1 &lt;= grid[i][j] &lt;= n * n</code></li>
	<li>For all <code>x</code> that <code>1 &lt;= x &lt;= n * n</code> there is exactly one <code>x</code> that is not equal to any of the grid members.</li>
	<li>For all <code>x</code> that <code>1 &lt;= x &lt;= n * n</code> there is exactly one <code>x</code> that is equal to exactly two of the grid members.</li>
	<li>For all <code>x</code> that <code>1 &lt;= x &lt;= n * n</code> except two of them there is exatly one pair of <code>i, j</code> that <code>0 &lt;= i, j &lt;= n - 1</code> and <code>grid[i][j] == x</code>.</li>
</ul>

class Solution:
    def findMissingAndRepeatedValues(self, grid: List[List[int]]) -> List[int]:
        d = {}
        n = len(grid)**2
        s = int((2*1 + 1*(n-1))/2 * n)
        n = 0 
        for i in grid:
            for j in i:
                if j in d: 
                    n = j
                else:
                    d[j] = True
                    s -= j
        return [n, s]
         
        
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