<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 {
public:
    int largestOverlap(vector<vector<int>>& img1, vector<vector<int>>& img2) {
        
    vector<pair<int, int>> aPoints, bPoints;
        int n = img1.size();
        
        // Collect coordinates of all the 1s in both matrices
        for (int i = 0; i < n; ++i) {
            for (int j = 0; j < n; ++j) {
                if (img1[i][j] == 1) aPoints.emplace_back(i, j);
                if (img2[i][j] == 1) bPoints.emplace_back(i, j);
            }
        }
        
        unordered_map<pair<int, int>, int, hash_pair> count;
        int maxOverlap = 0;
        
        // Compare each pair of points from A and B
        for (auto& [ax, ay] : aPoints) {
            for (auto& [bx, by] : bPoints) {
                pair<int, int> delta = {ax - bx, ay - by};
                ++count[delta];
                maxOverlap = max(maxOverlap, count[delta]);
            }
        }
        
        return maxOverlap;
    }

private:
    // Custom hash function for pair<int, int>
    struct hash_pair { 
        template <class T1, class T2> 
        size_t operator()(const pair<T1, T2>& p) const { 
            auto hash1 = hash<T1>{}(p.first); 
            auto hash2 = hash<T2>{}(p.second); 
            return hash1 ^ hash2; 
        } 
    };
};
16600