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Max Points on a Line – LeetCode Solution Java , Python 3, Python 2 , C , C++, Best and Optimal Solutions , All you need.
Given an array of points where points[i] = [xi, yi] represents a point on the X-Y plane, return the maximum number of points that lie on the same straight line.
Example 1:
Input: points = [[1,1],[2,2],[3,3]] Output: 3
Example 2:
Input: points = [[1,1],[3,2],[5,3],[4,1],[2,3],[1,4]] Output: 4
Constraints:
1 <= points.length <= 300points[i].length == 2-104 <= xi, yi <= 104- All the
pointsare unique.
C++ Max Points on a Line LeetCode Solution
class Solution {
public:
int maxPoints(vector<Point>& points) {
int n = points.size(), ans = 0;
for (int i = 0; i < n; i++) {
unordered_map<string, int> counter;
int dup = 1;
for (int j = i + 1; j < n; j++) {
if (points[j].x == points[i].x && points[j].y == points[i].y) {
dup++;
} else {
int dx = points[j].x - points[i].x, dy = points[j].y - points[i].y, g = gcd(dx, dy);
counter[to_string(dx / g) + '_' + to_string(dy / g)]++;
}
}
ans = max(ans, dup);
for (auto p : counter) {
ans = max(ans, p.second + dup);
}
}
return ans;
}
private:
int gcd(int a, int b) {
while (b) {
a = a % b;
swap(a, b);
}
return a;
}
};
Java Max Points on a Line LeetCode Solution
public class Solution{
public int maxPoints(Point[] points) {
if (points==null) return 0;
if (points.length<=2) return points.length;
Map<Integer,Map<Integer,Integer>> map = new HashMap<Integer,Map<Integer,Integer>>();
int result=0;
for (int i=0;i<points.length;i++){
map.clear();
int overlap=0,max=0;
for (int j=i+1;j<points.length;j++){
int x=points[j].x-points[i].x;
int y=points[j].y-points[i].y;
if (x==0&&y==0){
overlap++;
continue;
}
int gcd=generateGCD(x,y);
if (gcd!=0){
x/=gcd;
y/=gcd;
}
if (map.containsKey(x)){
if (map.get(x).containsKey(y)){
map.get(x).put(y, map.get(x).get(y)+1);
}else{
map.get(x).put(y, 1);
}
}else{
Map<Integer,Integer> m = new HashMap<Integer,Integer>();
m.put(y, 1);
map.put(x, m);
}
max=Math.max(max, map.get(x).get(y));
}
result=Math.max(result, max+overlap+1);
}
return result;
}
private int generateGCD(int a,int b){
if (b==0) return a;
else return generateGCD(b,a%b);
}
}
Python 3 Max Points on a Line LeetCode Solution
def maxPoints(self, points):
l = len(points)
m = 0
for i in range(l):
dic = {'i': 1}
same = 0
for j in range(i+1, l):
tx, ty = points[j].x, points[j].y
if tx == points[i].x and ty == points[i].y:
same += 1
continue
if points[i].x == tx: slope = 'i'
else:slope = (points[i].y-ty) * 1.0 /(points[i].x-tx)
if slope not in dic: dic[slope] = 1
dic[slope] += 1
m = max(m, max(dic.values()) + same)
return m
Array-1180
String-562
Hash Table-412
Dynamic Programming-390
Math-368
Sorting-264
Greedy-257
Depth-First Search-256
Database-215
Breadth-First Search-200
Tree-195
Binary Search-191
Matrix-176
Binary Tree-160
Two Pointers-151
Bit Manipulation-140
Stack-133
Heap (Priority Queue)-117
Design-116
Graph-108
Simulation-103
Prefix Sum-96
Backtracking-92
Counting-86
Sliding Window-73
Linked List-69
Union Find-66
Ordered Set-48
Monotonic Stack-47
Recursion-43
Trie-41
Binary Search Tree-40
Divide and Conquer-40
Enumeration-39
Bitmask-37
Queue-33
Memoization-32
Topological Sort-31
Geometry-30
Segment Tree-27
Game Theory-24
Hash Function-24
Binary Indexed Tree-21
Interactive-18
Data Stream-17
String Matching-17
Rolling Hash-17
Shortest Path-16
Number Theory-16
Combinatorics-15
Randomized-12
Monotonic Queue-9
Iterator-9
Merge Sort-9
Concurrency-9
Doubly-Linked List-8
Brainteaser-8
Probability and Statistics-7
Quickselect-7
Bucket Sort-6
Suffix Array-6
Minimum Spanning Tree-5
Counting Sort-5
Shell-4
Line Sweep-4
Reservoir Sampling-4
Eulerian Circuit-3
Radix Sort-3
Strongly Connected Componen-t2
Rejection Sampling-2
Biconnected Component-1
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