<p>You are given two images, <code>img1</code> and <code>img2</code>, represented as binary, square matrices of size <code>n x n</code>. A binary matrix has only <code>0</code>s and <code>1</code>s as values.</p>

<p>We <strong>translate</strong> one image however we choose by sliding all the <code>1</code> bits left, right, up, and/or down any number of units. We then place it on top of the other image. We can then calculate the <strong>overlap</strong> by counting the number of positions that have a <code>1</code> in <strong>both</strong> images.</p>

<p>Note also that a translation does <strong>not</strong> include any kind of rotation. Any <code>1</code> bits that are translated outside of the matrix borders are erased.</p>

<p>Return <em>the largest possible overlap</em>.</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>
<img alt="" src="https://assets.leetcode.com/uploads/2020/09/09/overlap1.jpg" style="width: 450px; height: 231px;" />
<pre>
<strong>Input:</strong> img1 = [[1,1,0],[0,1,0],[0,1,0]], img2 = [[0,0,0],[0,1,1],[0,0,1]]
<strong>Output:</strong> 3
<strong>Explanation:</strong> We translate img1 to right by 1 unit and down by 1 unit.
<img alt="" src="https://assets.leetcode.com/uploads/2020/09/09/overlap_step1.jpg" style="width: 450px; height: 105px;" />
The number of positions that have a 1 in both images is 3 (shown in red).
<img alt="" src="https://assets.leetcode.com/uploads/2020/09/09/overlap_step2.jpg" style="width: 450px; height: 231px;" />
</pre>

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

<pre>
<strong>Input:</strong> img1 = [[1]], img2 = [[1]]
<strong>Output:</strong> 1
</pre>

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

<pre>
<strong>Input:</strong> img1 = [[0]], img2 = [[0]]
<strong>Output:</strong> 0
</pre>

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

<ul>
	<li><code>n == img1.length == img1[i].length</code></li>
	<li><code>n == img2.length == img2[i].length</code></li>
	<li><code>1 &lt;= n &lt;= 30</code></li>
	<li><code>img1[i][j]</code> is either <code>0</code> or <code>1</code>.</li>
	<li><code>img2[i][j]</code> is either <code>0</code> or <code>1</code>.</li>
</ul>

class Solution:
    def largestOverlap(self, img1: List[List[int]], img2: List[List[int]]) -> int:
        N = len(img1)
        
        def get_overlap(offx, offy):
            res = 0
            six1, siy1, six2, siy2 = 0, 0, 0, 0

            if offx < 0:
                six1 = abs(offx)
            elif offx > 0:
                six2 = offx

            if offy < 0:
                siy1 = abs(offy)
            elif offy > 0:
                siy2 = offy
                
            for dx in range(N):
                ix1 = six1 + dx 
                ix2 = six2 + dx

                if ix1 == N or ix2 == N:
                    break

                for dy in range(N):
                    iy1 = siy1 + dy
                    iy2 = siy2 + dy

                    if iy1 == N or iy2 == N:
                        break

                    res += int(img1[ix1][iy1] and img2[ix2][iy2])

            return res

        res = 0
        for offx in range(-N + 1, N):
            for offy in range(-N + 1, N):
                res = max(res, get_overlap(offx, offy))
        
        return res
16200