<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:
        n = len(points)
        if n == 1:
            return 1
        elif n == 2:
            return 2
        ans = 2
        # Get the points with same x value, since they cannot get K
        x_freq = {}
        for p in points:
            (x, y) = p
            if x not in x_freq:
                x_freq[x] = 1
            else:
                x_freq[x] += 1
                ans = max(ans, x_freq[x])
        k_freq = {}
        for i in range(n):
            x1, y1 = points[i]
            for j in range(i+1, n):
                x2, y2 = points[j]
                if x2 != x1:
                    k = (y2 - y1) / (x2 - x1)
                    b = y2 - k * x2
                    id = f"{k}#{b}"
                    if id not in k_freq:
                        k_freq[id] = set([x1, x2])
                    else:
                        k_freq[id].add(x1)
                        k_freq[id].add(x2)
                        ans = max(ans, len(k_freq[id]))
        return ans
        
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