[Editorial] Max points on a line

content/5_Plat/Geo_Pri.problems.json

@@ -145,12 +145,25 @@
         "kind": "internal"
       }
     },
+    {
+      "uniqueId": "leetcode-149",
+      "name": "Max Points on a Line",
+      "url": "https://leetcode.com/problems/max-points-on-a-line/description/",
+      "source": "Leetcode",
+      "difficulty": "Easy",
+      "isStarred": false,
+      "tags": [],
+      "solutionMetadata": {
+        "kind": "internal"
+      }
+    },
     {
       "uniqueId": "ioi-02-frog",
       "name": "The Troublesome Frog",
       "url": "https://ioi.contest.codeforces.com/group/32KGsXgiKA/contest/103699/problem/A",
       "source": "IOI",
       "difficulty": "Normal",
+
       "isStarred": false,
       "tags": [],
       "solutionMetadata": {

solutions/platinum/leetcode-149.mdx

@@ -0,0 +1,56 @@
+---
+id: leetcode-149
+source: Leetcode
+title: Max Points on a Line
+author: Mihnea Brebenel
+---
+
+## Explanation
+
+Because $n \le 300$ we can naively go through all pairs of points and check how many other points are collinear with them.
+Three points $A$, $B$, and $C$ are collinear if and only if $AB$ has the same slope as $BC$:
+
+$$
+\frac{A_y-B_y}{A_x-B_x}=\frac{B_y-C_y}{B_x-C_x}
+$$
+
+In practice you'll find this formula in a product form to avoid dealing with floating-point values.
+
+## Implementation
+
+**Time Complexity:** $\mathcal{O}(n^3)$
+
+<LanguageSection>
+<CPPSection>
+
+```cpp
+class Solution {
+  public:
+	int maxPoints(vector<vector<int>> &points) {
+		if (points.size() <= 2) { return points.size(); }
+
+		int ans = 0;
+		for (int i = 0; i < points.size(); i++) {
+			for (int j = i + 1; j < points.size(); j++) {
+				int p = 2;  // the 2 points are collinear with themselves
+				for (int k = j + 1; k < points.size(); k++) {
+					int dx1 = points[i][0] - points[k][0],
+					    dx2 = points[j][0] - points[i][0];
+					int dy1 = points[i][1] - points[k][1],
+					    dy2 = points[j][1] - points[i][1];
+					// Check if dy1 / dx1 = dy2 / dx2
+					// Which is the same as: dy1 * dx2 = dy2 * dx1
+					if (dy1 * dx2 == dy2 * dx1) { p++; }
+				}
+
+				ans = max(ans, p);
+			}
+		}
+
+		return ans;
+	}
+};
+```
+
+</CPPSection>
+</LanguageSection>