<p>Given an array of <code>points</code> where <code>points[i] = [x<sub>i</sub>, y<sub>i</sub>]</code> represents a point on the <strong>X-Y</strong> plane, return <em>the maximum number of points that lie on the same straight line</em>.</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>
<img alt="" src="https://assets.leetcode.com/uploads/2021/02/25/plane1.jpg" style="width: 300px; height: 294px;" />
<pre>
<strong>Input:</strong> points = [[1,1],[2,2],[3,3]]
<strong>Output:</strong> 3
</pre>

<p><strong class="example">Example 2:</strong></p>
<img alt="" src="https://assets.leetcode.com/uploads/2021/02/25/plane2.jpg" style="width: 300px; height: 294px;" />
<pre>
<strong>Input:</strong> points = [[1,1],[3,2],[5,3],[4,1],[2,3],[1,4]]
<strong>Output:</strong> 4
</pre>

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

<ul>
	<li><code>1 &lt;= points.length &lt;= 300</code></li>
	<li><code>points[i].length == 2</code></li>
	<li><code>-10<sup>4</sup> &lt;= x<sub>i</sub>, y<sub>i</sub> &lt;= 10<sup>4</sup></code></li>
	<li>All the <code>points</code> are <strong>unique</strong>.</li>
</ul>

struct pair_hash {
    template <class T1, class T2>
    std::size_t operator () (const std::pair<T1,T2> &p) const {
        auto h1 = std::hash<T1>{}(p.first);
        auto h2 = std::hash<T2>{}(p.second);

        // Mainly for demonstration purposes, i.e. works but is overly simple
        // In the real world, use sth. like boost.hash_combine
        return h1 ^ h2;  
    }
};

class Solution {
public:
    int maxPoints(vector<vector<int>>& points)
    {
        int ans = 1;
        unordered_map<pair<double, double>, unordered_set<pair<int, int>, pair_hash>, pair_hash> lines;
        for (int i = 0; i < points.size(); i++)
        {
            for (int j = i + 1; j < points.size(); j++)
            {
                auto line = calcLine(points[i], points[j]);
                lines[line].insert({ points[i][0], points[i][1] });
                lines[line].insert({ points[j][0], points[j][1] });
                ans = max(ans, (int)lines[line].size());
            }
        }
        return ans;
    }

    pair<double, double> calcLine(vector<int>& A, vector<int>& B)
    {
        if (A[0] == B[0])
            return {DBL_MAX, A[0]};
        double m = (double)(A[1] - B[1]) / (A[0] - B[0]);
        return { m, A[1] - m * A[0] };
    }
};
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