<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>

class Solution:
    def maxPoints(self, points: List[List[int]]) -> int:
        if len(points) == 1:
            return 1
        
        x = {}
        for i in range(len(points)):
            for j in range(i+1, len(points)):
                k, b = self.line(points[i], points[j])
                if k not in x:
                    x[k] = {}
                
                if b not in x[k]:
                    x[k][b] = [i, j]
                else:
                    temp = x[k][b]
                    if i not in temp:
                        temp.append(i)
                    if j not in temp:
                        temp.append(j) 
                    x[k][b] = temp
        
        print(x)
        maxl = 1
        for k in x:
            for b in x[k]:
                maxl = max(maxl, len(x[k][b]))
        return maxl
        
    def line(self, a, b):
        if a[1] == b[1]:
            return 'NA', a[1]
        if a[0] == b[0]:
            return a[0], 'NA'
        k = (b[1] - a[1])/(b[0] - a[0])
        return k, a[1] - k * a[0]
        
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