<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>

#define pb push_back
#define ll long long
#include <bits/stdc++.h>
using namespace std;

class Solution {
public:

    pair<ll, ll> f(int a, int b) {
        int gcd = __gcd(a, b);  // This is Euclid's algorithm
        return {a / gcd, b / gcd};  // Return the reduced form of the fraction
    }

    int maxPoints(vector<vector<int>>& points) {
        int pl = points.size();
        if(pl == 1) return 1;
        int ans = 0;
        vector<vector<pair<ll,ll>>> s(pl, vector<pair<ll,ll>>(pl, {LONG_MAX-1, LONG_MAX-1}));

        for(int i=0;i<pl-1;++i) {
            for(int j=i+1;j<pl;++j) {
                if(points[j][0] - points[i][0] == 0)
                    s[i][j] = s[j][i] = {LONG_MAX,LONG_MAX};
                else    
                    s[i][j] = s[j][i] = f(points[j][1] - points[i][1], points[j][0] - points[i][0]);
            }
        }

        for(int i=0;i<pl;++i) {
            map<pair<ll,ll>,int> counter;
            int maxO = 0;
            for(int j=0;j<pl;++j) {
                if(i != j && s[i][j].first != LONG_MAX - 1) {
                    counter[s[i][j]]++; 
                    if(counter[s[i][j]]+1 > maxO) {
                        maxO = counter[s[i][j]]+1;
                    }
                }
            }
            ans = max(ans, maxO);
        }

        return ans;
    }
};
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