Max Points on a Line in C++

Suppose we have a 2D plane. We have to find the maximum number of points that reside on the same straight line. So if the points are like −

Then there are 4 points

To solve this, we will follow these steps −

• n := number of points, if n < 3, then return n

• ans := 2

• for i in range 1 to n – 1

  • count := 0

  • take two points from index i and i – 1, these are p1, p2

  • if p1 and p2 points are same, then

    • for j in range 0 to n – 1

      • if points[j].x = p1.x and points[j].y = p1.y, then increase count by 1

  • otherwise −

    • for j in range 0 to n – 1

      • p3 := point from index j

      • if p3.y – p2.y * p2.x – p1.x = p2.y – p1.y * p3.x – p2.x, then increase count by 1

  • ans := max of ans and count

• return ans

Example

Let us see the following implementation to get a better understanding −

 Live Demo

#include <bits/stdc++.h>
using namespace std;
typedef long long int lli;
class Solution {
   public:
   int maxPoints(vector<vector<int>>& points) {
      int n = points.size();
      if(n<3)return n;
      int ans = 2;
      for(int i = 1;i<n;i++){
         int count = 0;
         lli x1 = points[i-1][0];
         lli x2 = points[i][0];
         lli y1 = points[i-1][1];
         lli y2 = points[i][1];
         if(x1 == x2 && y1 == y2){
            for(int j =0;j<n;j++){
               if(points[j][0] ==x1 && points[j][1] == y1)count++;
            }
         } else {
            for(int j =0;j<n;j++){
               int x3 = points[j][0];
               int y3 = points[j][1];
               if((y3-y2)*(x2-x1) == (y2-y1)*(x3-x2))count++ ;
            }
         }
         ans = max(ans, count);
      }
      return ans;
   }
};
main(){
   Solution ob;
   vector<vector<int>> v = {{1,1},{3,2},{5,3},{4,1},{2,3},{1,4}};
   cout << (ob.maxPoints(v));
}

Input

[{1,1},{3,2},{5,3},{4,1},{2,3},{1,5}]

Output

4

Arnab Chakraborty

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Updated on: 26-May-2020

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