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Find intersection point of lines inside a section

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  • Last Updated : 22 Jun, 2022
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Given N lines in two-dimensional space in y = mx + b form and a vertical section. We need to find out whether there is an intersection point inside the given section or not. 
Examples: 
 

In below diagram four lines are there,
L1 :    y = x + 2
L2 :    y = -x + 7
L3 :    y = -3
L4 :    y = 2x – 7
and vertical section is given from x = 2 to x = 4

intersection point of lines inside a section

We can see that in above diagram, the 
intersection point of line L1 and L2 
lies between the section.

We can solve this problem using sorting. First, we will calculate intersection point of each line with both the boundaries of vertical section and store that as a pair. We just need to store y-coordinates of intersections as a pair because x-coordinates are equal to boundary itself. Now we will sort these pairs on the basis of their intersection with left boundary. After that, we will loop over these pairs one by one and if for any two consecutive pairs, the second value of the current pair is less than that of the second value of the previous pair then there must be an intersection in the given vertical section. 
The possible orientation of two consecutive pairs can be seen in above diagram for L1 and L2. We can see that when the second value is less, intersection lies in vertical section. 
Total time complexity of solution will be O(n logn) 
 

CPP




// C++ program to check an intersection point
// inside a given vertical section
#include <bits/stdc++.h>
using namespace std;
 
// structure to represent a line
struct line {
    int m, b;
    line()  { }
    line(int m, int b) : m(m), b(b)  { }
};
 
// Utility method to get Y-coordinate
// corresponding to x in line l
int getYFromLine(line l, int x)
{
    return (l.m * x + l.b);
}
 
// method returns true if two line cross
// each other between xL and xR range
bool isIntersectionPointInsideSection(line lines[],
                            int xL, int xR, int N)
{
    pair<int, int> yBoundary[N];
 
    // first calculating y-values and putting
    // in boundary pair
    for (int i = 0; i < N; i++)
        yBoundary[i] = make_pair(getYFromLine(lines[i], xL),
                               getYFromLine(lines[i], xR));
     
 
    // sorting the pair on the basis of first
    // boundary intersection
    sort(yBoundary, yBoundary + N);
 
    // looping over sorted pairs for comparison
    for (int i = 1; i < N; i++) {
 
        // if current pair's second value is smaller
        // than previous pair's then return true
        if (yBoundary[i].second < yBoundary[i - 1].second)
            return true;       
    }
 
    return false;
}
 
// Driver code to test above methods
int main()
{
    int N = 4;
    int m[] = { 1, -1, 0, 2 };
    int b[] = { 2, 7, -3, -7 };
 
    // copy values in line struct
    line lines[N];
    for (int i = 0; i < N; i++) {
        lines[i] = line(m[i], b[i]);
    }
   
    int xL = 2;
    int xR = 4;
 
    if (isIntersectionPointInsideSection(lines, xL, xR, N)) {
        cout << "Intersection point lies between "
             << xL << " and " << xR << endl;
    } else {
        cout << "No Intersection point lies between "
             << xL << " and " << xR << endl;
    }
}

Java




//Java program to check an intersection point
// inside a given vertical section
 
import java.io.*;
import java.util.*;
 
class GFG {
     
    // a structure to represent a line
    public static class line
    {
        int m,b;
        line(int s, int d){
            this.m = s;
            this.b = d;
        }
    }
     
    public static class Pair{
        int x,y;
        Pair(int xx, int yy){
            this.x = xx;
            this.y = yy;
        }
    }
     
    // Utility method to get Y-coordinate
    // corresponding to x in line l
    public static int getYFromLine(line l, int x)
    {
        return ((l.m * x) + l.b);
    }
      
    // method returns true if two line cross
    // each other between xL and xR range
    public static boolean isIntersectionPointInsideSection(line lines[], int xL, int xR, int N)
    {
        Pair[] yBoundary = new Pair[N];
      
        // first calculating y-values and putting
        // in boundary pair
        for (int i = 0; i < N; i++){
            yBoundary[i] = new Pair(getYFromLine(lines[i], xL),getYFromLine(lines[i], xR));
        }
      
        // sorting the pair on the basis of first
        // boundary intersection
        Arrays.sort(yBoundary,new Comparator<Pair>(){
            public int compare(Pair p1, Pair p2){
                return p1.x>p2.x?1:-1;
            }
        });
      
        // looping over sorted pairs for comparison
        for (int i = 1; i < N; i++) {
      
            // if current pair's y value is smaller
            // than previous pair's then return true
            if (yBoundary[i].y < yBoundary[i - 1].y){
                return true;
            }
                       
        }
      
        return false;
    }
     
    // Driver program to test above functions
    public static void main (String[] args) {
        int N = 4;
        int m[] = { 1, -1, 0, 2 };
        int b[] = { 2, 7, -3, -7 };
      
        // copy values in line struct
        line lines[] = new line[N];
        for (int i = 0; i < N; i++) {
            lines[i] = new line(m[i], b[i]);
        }
        
        int xL = 2;
        int xR = 4;
      
        if (isIntersectionPointInsideSection(lines, xL, xR, N)) {
            System.out.println("Intersection point lies between "+xL+ " and "+ xR);
        } else {
            System.out.println("No Intersection point lies between "+xL+ " and "+ xR);
        }
    }
}
//This code is contributed by shruti456rawal

Javascript




// JavaScript program to check an intersection point
// inside a given vertical section
 
// structure to represent a line
class line {
    constructor() {}
    constructor(m, b){
        this.m = m;
        this.b = b;
    }
};
 
// Utility method to get Y-coordinate
// corresponding to x in line l
function getYFromLine(l, x){
    return (l.m * x + l.b);
}
 
// method returns true if two line cross
// each other between xL and xR range
function isIntersectionPointInsideSection(lines, xL, xR, N){
    let yBoundary = new Array(N);
 
    // first calculating y-values and putting
    // in boundary pair
    for (let i = 0; i < N; i++){
        yBoundary[i] = [getYFromLine(lines[i], xL), getYFromLine(lines[i], xR)];
    }
 
    // sorting the pair on the basis of first
    // boundary intersection
    yBoundary.sort();
     
    // looping over sorted pairs for comparison
    for (let i = 1; i < N; i++) {      
        // if current pair's second value is smaller
        // than previous pair's then return true
        if (yBoundary[i][1] < yBoundary[i - 1][1]){
             return true;
        }
    }
 
    return false;
}
 
// Driver code to test above methods
{
    let N = 4;
    let m = [ 1, -1, 0, 2 ];
    let b = [2, 7, -3, -7 ];
 
    // copy values in line struct
    let lines = new Array();
    for (let i = 0; i < N; i++) {
        lines.push(new line(m[i], b[i]));
    }
    let xL = 2;
    let xR = 4;
 
    if (isIntersectionPointInsideSection(lines, xL, xR, N)) {
        console.log("Intersection point lies between ", xL, " and ", xR);
    } else {
        console.log("No Intersection point lies between ", xL, " and ", xR);
    }
}
 
// The code is contributed by Gautam goel (gautamgoel962)

Output

Intersection point lies between 2 and 4

Time Complexity : O(n*log(n))

Space Complexity : O(1)

This article is contributed by Utkarsh Trivedi. If you like GeeksforGeeks and would like to contribute, you can also write an article using write.geeksforgeeks.org or mail your article to review-team@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.
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