<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>

#include <vector>
#include <unordered_map>
#include <utility>

class Solution {
public:
    int largestOverlap(std::vector<std::vector<int>>& img1, std::vector<std::vector<int>>& img2) {
        int n = img1.size();
        std::vector<std::pair<int,int>> vp1, vp2;
        
        for(int i = 0; i < n; i++) {
            for(int j = 0; j < n; j++) {
                if(img1[i][j] == 1) {
                    vp1.push_back({i, j});
                }
                if(img2[i][j] == 1) {
                    vp2.push_back({i, j});
                }
            }
        }
        
        int ans = 0;
        std::unordered_map<std::pair<int, int>, int, pair_hash> mp;
        
        for(auto& it1 : vp1) {
            for(auto& it2 : vp2) {
                int a = it1.first - it2.first;
                int b = it1.second - it2.second;
                mp[{a, b}]++;
                ans = std::max(ans, mp[{a, b}]);

                // Early exit if perfect overlap is found
                if (ans == vp1.size()) return ans;
            }
        }
        
        return ans;
    }
    
private:
    struct pair_hash {
        std::size_t operator()(const std::pair<int, int>& p) const {
            return std::hash<int>{}(p.first) ^ std::hash<int>{}(p.second);
        }
    };
};

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