# Maximum distinct lines passing through a single point

Given lines represented by two points and . The task is to find maximum number of lines which can pass through a single point, without superimposing (or covering) any other line. We can move any line but not rotate it.
Examples:

Input : Line 1 : x1 = 1, y1 = 1, x2 = 2, y2 = 2
Line 2 : x2 = 2, y1 = 2, x2 = 4, y2 = 10
Output : 2
There are two lines. These two lines are not
parallel, so both of them will pass through
a single point.

Input : Line 1 : x1 = 1, y1 = 5, x2 = 1, y2 = 10
Line 2 : x2 = 5, y1 = 1, x2 = 10, y2 = 1
Output : 2

• Represent lines as pair where line can be given as , called line slope form. We can now see that we can change the c for any line, but cannot modify m.
• Lines having same value of m parallel, given that (c1 ? c2). Also no two parallel lines can pass through same point without superimposing to each other.
• So, our problem reduces to finding different values of slopes from given set of lines.

We can calculate slope of a line as , add them to a set and count the number of distinct values of slope in set. But we have to handle vertical lines separately.
So, if then, slope = INT_MAX
Otherwise, slope = .
Below is the implementation of the approach.

## C++

 // C++ program to find maximum number of lines // which can pass through a single point #include  using namespace std;   // function to find maximum lines which passes // through a single point int maxLines(int n, int x1[], int y1[],                  int x2[], int y2[]) {     unordered_set<double> s;       double slope;     for (int i = 0; i < n; ++i) {         if (x1[i] == x2[i])             slope = INT_MAX;         else             slope = (y2[i] - y1[i]) * 1.0                      / (x2[i] - x1[i]) * 1.0;           s.insert(slope);     }       return s.size(); }   // Driver program int main() {     int n = 2, x1[] = { 1, 2 }, y1[] = { 1, 2 },             x2[] = { 2, 4 }, y2[] = { 2, 10 };     cout << maxLines(n, x1, y1, x2, y2);     return 0; } // This code is written by // Sanjit_Prasad

## Java

 // Java program to find maximum number of lines // which can pass through a single point   import java.util.*; import java.lang.*; import java.io.*;   class GFG{   // function to find maximum lines which passes // through a single point static int maxLines(int n, int x1[], int y1[],                      int x2[], int y2[]) {     Set s=new HashSet();       double slope;     for (int i = 0; i < n; ++i) {         if (x1[i] == x2[i])             slope = Integer.MAX_VALUE;         else             slope = (y2[i] - y1[i]) * 1.0                     / (x2[i] - x1[i]) * 1.0;           s.add(slope);     }       return s.size(); }   // Driver program public static void main(String args[]) {     int n = 2, x1[] = { 1, 2 }, y1[] = { 1, 2 },             x2[] = { 2, 4 }, y2[] = { 2, 10 };     System.out.print(maxLines(n, x1, y1, x2, y2)); } } // This code is written by // Subhadeep

## Python3

 # Python3 program to find maximum number  # of lines which can pass through a  # single point import sys # function to find maximum lines  # which passes through a single point def maxLines(n, x1, y1, x2, y2):       s = [];       slope=sys.maxsize;     for i in range(n):         if (x1[i] == x2[i]):             slope = sys.maxsize;         else:             slope = (y2[i] - y1[i]) * 1.0 /(x2[i] - x1[i]) * 1.0;           s.append(slope);       return len(s);   # Driver Code n = 2; x1 = [ 1, 2 ]; y1 = [1, 2]; x2 = [2, 4]; y2 = [2, 10]; print(maxLines(n, x1, y1, x2, y2));   # This code is contributed by mits

## C#

 // C# program to find maximum number of lines // which can pass through a single point using System; using System.Collections.Generic;   class GFG {   // function to find maximum lines which passes // through a single point static int maxLines(int n, int []x1, int []y1,                      int []x2, int []y2) {     HashSet s = new HashSet();       double slope;     for (int i = 0; i < n; ++i)      {         if (x1[i] == x2[i])             slope = int.MaxValue;         else             slope = (y2[i] - y1[i]) * 1.0                     / (x2[i] - x1[i]) * 1.0;           s.Add(slope);     }       return s.Count; }   // Driver code public static void Main() {     int n = 2;     int []x1 = { 1, 2 }; int []y1 = { 1, 2 };     int []x2 = { 2, 4 }; int []y2 = { 2, 10 };     Console.Write(maxLines(n, x1, y1, x2, y2)); } }   /* This code contributed by PrinciRaj1992 */

## PHP

 

## Javascript

 

Output:

2

Time Complexity: Space Complexity: O(N) since using auxiliary space for set

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