update LC-149.mdx

content/5_Plat/Geo_Pri.problems.json

@@ -75,7 +75,7 @@
       "uniqueId": "leetcode-1610",
       "name": "Maximum Number of Visible Points",
       "url": "https://leetcode.com/problems/maximum-number-of-visible-points/description/",
-      "source": "LeetCode",
+      "source": "LC",
       "difficulty": "Normal",
       "isStarred": false,
       "tags": ["Geometry", "Radians", "Sliding Windows"],
@@ -149,7 +149,7 @@
       "uniqueId": "leetcode-149",
       "name": "Max Points on a Line",
       "url": "https://leetcode.com/problems/max-points-on-a-line/description/",
-      "source": "Leetcode",
+      "source": "LC",
       "difficulty": "Easy",
       "isStarred": false,
       "tags": [],
@@ -188,7 +188,7 @@
       "uniqueId": "leetcode-1453",
       "name": "Maximum Number of Darts Inside of a Circular Dartboard",
       "url": "https://leetcode.com/problems/maximum-number-of-darts-inside-of-a-circular-dartboard/description/",
-      "source": "Leetcode",
+      "source": "LC",
       "difficulty": "Hard",
       "isStarred": false,
       "tags": [],

solutions/platinum/leetcode-1453.mdx

@@ -1,6 +1,6 @@
 ---
 id: leetcode-1453
-source: Leetcode
+source: LC
 title: Maximum Number of Darts Inside of a Circular Dartboard
 author: Mihnea Brebenel
 ---

solutions/platinum/leetcode-149.mdx

@@ -1,8 +1,9 @@
 ---
 id: leetcode-149
-source: Leetcode
+source: LC
 title: Max Points on a Line
 author: Mihnea Brebenel
+contributors: Rameez Parwez
 ---
 
 ## Explanation
@@ -84,4 +85,104 @@ class Solution {
 ```
 
 </JavaSection>
+<PySection>
+
+```py
+class Solution:
+	def maxPoints(self, points: List[List[int]]) -> int:
+		n = len(points)
+		if n <= 2:
+			return n
+
+		ans = 0
+		for i in range(n):
+			for j in range(i + 1, n):
+				p = 2  # the 2 points are collinear with themselves
+				for k in range(j + 1, n):
+					dx1 = points[i][0] - points[k][0]
+					dx2 = points[j][0] - points[i][0]
+					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 += 1
+
+				ans = max(ans, p)
+
+		return ans
+```
+
+</PySection>
+</LanguageSection>
+
+## Efficient Solution
+
+We can take a point $A$ and compute the slopes of all lines connecting $A$ to the other points. Two different points lie on the same line if they share the same slope. In this way, we can determine the maximum number of points lying on the same straight line wrt $A$.
+
+We perform this for all other points.
+
+## Implementation
+
+**Time Complexity:** $\mathcal{O}(n^2)$
+
+<LanguageSection>
+<CPPSection>
+
+```cpp
+class Solution {
+  public:
+	int maxPoints(vector<vector<int>> &points) {
+		int n = (int)points.size();
+		int res = 0;
+		for (int i = 0; i < n; i++) {
+			unordered_map<float, int> umap;
+			for (int j = i + 1; j < n; j++) {
+				float slope = 0.0;
+				if (points[i][0] - points[j][0] == 0) {  // avoid division by 0
+					slope = INT_MAX;
+				} else {
+					slope = float(points[i][1] - points[j][1]) /
+					        float(points[i][0] - points[j][0]);
+				}
+
+				umap[slope]++;
+			}
+			for (auto &[_, slope] : umap) { res = max(res, slope); }
+		}
+		return res + 1;
+	}
+};
+```
+
+</CPPSection>
+<PySection>
+
+```py
+from collections import defaultdict
+
+
+class Solution:
+	def maxPoints(self, points: List[List[int]]) -> int:
+		n = len(points)
+		res = 0
+		for i in range(n):
+			slope_count = defaultdict(int)
+			for j in range(i + 1, n):
+				if points[i][0] == points[j][0]:  # avoid division by zero
+					slope = float("inf")
+				else:
+					slope = (points[i][1] - points[j][1]) / (
+						points[i][0] - points[j][0]
+					)
+
+				slope_count[slope] += 1
+
+			res = max(res, max(slope_count.values(), default=0))
+
+		return res + 1
+```
+
+</PySection>
 </LanguageSection>