<p>You are given a <strong>0-indexed</strong> integer array <code>arr</code>, and an <code>m x n</code> integer <strong>matrix</strong> <code>mat</code>. <code>arr</code> and <code>mat</code> both contain <strong>all</strong> the integers in the range <code>[1, m * n]</code>.</p>

<p>Go through each index <code>i</code> in <code>arr</code> starting from index <code>0</code> and paint the cell in <code>mat</code> containing the integer <code>arr[i]</code>.</p>

<p>Return <em>the smallest index</em> <code>i</code> <em>at which either a row or a column will be completely painted in</em> <code>mat</code>.</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>
<img alt="" src="image explanation for example 1" /><img alt="image explanation for example 1" src="https://assets.leetcode.com/uploads/2023/01/18/grid1.jpg" style="width: 321px; height: 81px;" />
<pre>
<strong>Input:</strong> arr = [1,3,4,2], mat = [[1,4],[2,3]]
<strong>Output:</strong> 2
<strong>Explanation:</strong> The moves are shown in order, and both the first row and second column of the matrix become fully painted at arr[2].
</pre>

<p><strong class="example">Example 2:</strong></p>
<img alt="image explanation for example 2" src="https://assets.leetcode.com/uploads/2023/01/18/grid2.jpg" style="width: 601px; height: 121px;" />
<pre>
<strong>Input:</strong> arr = [2,8,7,4,1,3,5,6,9], mat = [[3,2,5],[1,4,6],[8,7,9]]
<strong>Output:</strong> 3
<strong>Explanation:</strong> The second column becomes fully painted at arr[3].
</pre>

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

<ul>
	<li><code>m == mat.length</code></li>
	<li><code>n = mat[i].length</code></li>
	<li><code>arr.length == m * n</code></li>
	<li><code>1 &lt;= m, n &lt;= 10<sup>5</sup></code></li>
	<li><code>1 &lt;= m * n &lt;= 10<sup>5</sup></code></li>
	<li><code>1 &lt;= arr[i], mat[r][c] &lt;= m * n</code></li>
	<li>All the integers of <code>arr</code> are <strong>unique</strong>.</li>
	<li>All the integers of <code>mat</code> are <strong>unique</strong>.</li>
</ul>

class Solution:
    def firstCompleteIndex(self, arr: List[int], mat: List[List[int]]) -> int:
        
        num_to_ind = {}
        for (i, num) in enumerate(arr):
            num_to_ind[num] = i
        
        final_ind = len(arr) - 1

        # Loop through each row and find max ind
        for r in range(len(mat)):
            max_row_ind = -1
            for c in range(len(mat[r])):
                num_ind = num_to_ind[mat[r][c]]
                max_row_ind = max(max_row_ind, num_ind)
            final_ind = min(max_row_ind, final_ind)

        # Loop through each column and find max ind
        for c in range(len(mat[0])):
            max_col_ind = -1
            for r in range(len(mat)):
                num_ind = num_to_ind[mat[r][c]]
                max_col_ind = max(max_col_ind, num_ind)
            final_ind = min(max_col_ind, final_ind)

        return final_ind

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