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Given n line segments, find if any two segments intersect
  • Difficulty Level : Hard
  • Last Updated : 10 Apr, 2021

We have discussed the problem to detect if two given line segments intersect or not. In this post, we extend the problem. Here we are given n line segments and we need to find out if any two line segments intersect or not. 

Naive Algorithm A naive solution to solve this problem is to check every pair of lines and check if the pair intersects or not. We can check two line segments in O(1) time. Therefore, this approach takes O(n2). 

Sweep Line Algorithm: We can solve this problem in O(nLogn) time using Sweep Line Algorithm. The algorithm first sorts the end points along the x axis from left to right, then it passes a vertical line through all points from left to right and checks for intersections. Following are detailed steps. 

1) Let there be n given lines. There must be 2n end points to represent the n lines. Sort all points according to x coordinates. While sorting maintain a flag to indicate whether this point is left point of its line or right point. 

2) Start from the leftmost point. Do following for every point 
…..a) If the current point is a left point of its line segment, check for intersection of its line segment with the segments just above and below it. And add its line to active line segments (line segments for which left end point is seen, but right end point is not seen yet).  Note that we consider only those neighbors which are still active. 
….b) If the current point is a right point, remove its line segment from active list and check whether its two active neighbors (points just above and below) intersect with each other. 

The step 2 is like passing a vertical line from all points starting from the leftmost point to the rightmost point. That is why this algorithm is called Sweep Line Algorithm. The Sweep Line technique is useful in many other geometric algorithms like calculating the 2D Voronoi diagram 

What data structures should be used for efficient implementation? 
In step 2, we need to store all active line segments. We need to do following operations efficiently: 
a) Insert a new line segment 
b) Delete a line segment 
c) Find predecessor and successor according to y coordinate values 
The obvious choice for above operations is Self-Balancing Binary Search Tree like AVL Tree, Red Black Tree. With a Self-Balancing BST, we can do all of the above operations in O(Logn) time. 
Also, in step 1, instead of sorting, we can use min heap data structure. Building a min heap takes O(n) time and every extract min operation takes O(Logn) time (See this). 

The following pseudocode doesn’t use heap. It simply sort the array. 

1. Sort Points[] from left to right (according to x coordinate)

2. Create an empty Self-Balancing BST T. It will contain 
  all active line Segments ordered by y coordinate.

// Process all 2n points 
3. for i = 0 to 2n-1

    // If this point is left end of its line  
    if (Points[i].isLeft) 
       T.insert(Points[i].line())  // Insert into the tree

       // Check if this points intersects with its predecessor and successor
       if ( doIntersect(Points[i].line(), T.pred(Points[i].line()) )
         return true
       if ( doIntersect(Points[i].line(), T.succ(Points[i].line()) )
         return true

    else  // If it's a right end of its line
       // Check if its predecessor and successor intersect with each other
       if ( doIntersect(T.pred(Points[i].line(), T.succ(Points[i].line()))
         return true
       T.delete(Points[i].line())  // Delete from tree

4. return False

Let us consider the following example taken from here.  There are 5 line segments 1, 2, 3, 4 and 5.  The dotted green lines show sweep lines. 


Following are steps followed by the algorithm.  All points from left to right are processed one by one. We maintain a self-balancing binary search tree. 

Left end point of line segment 1 is processed: 1 is inserted into the Tree. The tree contains 1.   No intersection. 

Left end point of line segment  2 is processed:  Intersection of 1 and 2 is checked. 2 is inserted into the Tree. No intersection. The tree contains 1, 2

Left end point of line segment 3 is processed:  Intersection of  3 with 1 is checked.  No intersection. 3 is inserted into the Tree. The tree contains  2, 1, 3

Right end point of line segment 1 is processed: 1 is deleted from the Tree.  Intersection of 2 and 3 is checked.  Intersection of 2 and 3 is reported.  The tree contains  2, 3. Note that the above pseudocode returns at this point. We can continue from here to report all intersection points. 

Left end point of line segment  4 is processed:   Intersections of  line 4 with lines 2 and 3 are checked.  No intersection. 4 is inserted into the Tree. The tree contains  2, 4, 3

Left end point of line segment 5 is processed:   Intersection of  5 with 3 is checked.  No intersection. 5 is inserted into the Tree. The tree contains   2, 4, 3, 5

Right end point of line segment 5 is processed: 5 is deleted from the Tree.   The tree contains  2, 4, 3

Right end point of line segment 4 is processed: 4 is deleted from the Tree.   The tree contains  2, 4, 3.   Intersection of 2 with 3 is checked.  Intersection of 2 with 3 is reported.  The tree contains  2, 3. Note that the intersection of 2 and 3 is reported again. We can add some logic to check for duplicates. 

Right end point of line segment  2 and 3 are processed:   Both are deleted from tree and tree becomes empty. 



// Implementation of Sweep Line Algorithm
#include <bits/stdc++.h>
using namespace std;
// A point in 2D plane
struct Point
    int x, y;
// A line segment with left as Point
// with smaller x value and right with
// larger x value.
struct Segment
    Point left, right;
// An event for sweep line algorithm
// An event has a point, the position
// of point (whether left or right) and
// index of point in the original input
// array of segments.
struct Event {
    int x, y;
    bool isLeft;
    int index;
    Event(int x, int y, bool l, int i) : x(x), y(y), isLeft(l), index(i) {}
    // This is for maintaining the order in set.
    bool operator<(const Event& e) const {
            return y < e.y;
// Given three colinear points p, q, r, the function checks if
// point q lies on line segment 'pr'
bool onSegment(Point p, Point q, Point r)
    if (q.x <= max(p.x, r.x) && q.x >= min(p.x, r.x) &&
        q.y <= max(p.y, r.y) && q.y >= min(p.y, r.y))
       return true;
    return false;
// To find orientation of ordered triplet (p, q, r).
// The function returns following values
// 0 --> p, q and r are colinear
// 1 --> Clockwise
// 2 --> Counterclockwise
int orientation(Point p, Point q, Point r)
    // for details of below formula.
    int val = (q.y - p.y) * (r.x - q.x) -
              (q.x - p.x) * (r.y - q.y);
    if (val == 0) return 0;  // colinear
    return (val > 0)? 1: 2; // clock or counterclock wise
// The main function that returns true if line segment 'p1q1'
// and 'p2q2' intersect.
bool doIntersect(Segment s1, Segment s2)
    Point p1 = s1.left, q1 = s1.right, p2 = s2.left, q2 = s2.right; 
    // Find the four orientations needed for general and
    // special cases
    int o1 = orientation(p1, q1, p2);
    int o2 = orientation(p1, q1, q2);
    int o3 = orientation(p2, q2, p1);
    int o4 = orientation(p2, q2, q1);
    // General case
    if (o1 != o2 && o3 != o4)
        return true;
    // Special Cases
    // p1, q1 and p2 are colinear and p2 lies on segment p1q1
    if (o1 == 0 && onSegment(p1, p2, q1)) return true;
    // p1, q1 and q2 are colinear and q2 lies on segment p1q1
    if (o2 == 0 && onSegment(p1, q2, q1)) return true;
    // p2, q2 and p1 are colinear and p1 lies on segment p2q2
    if (o3 == 0 && onSegment(p2, p1, q2)) return true;
     // p2, q2 and q1 are colinear and q1 lies on segment p2q2
    if (o4 == 0 && onSegment(p2, q1, q2)) return true;
    return false; // Doesn't fall in any of the above cases
// Find predecessor of iterator in s.
auto pred(set<Event> &s, set<Event>::iterator it) {
    return it == s.begin() ? s.end() : --it;
// Find successor of iterator in s.
auto succ(set<Event> &s, set<Event>::iterator it) {
    return ++it;
// Returns true if any two lines intersect. 
bool isIntersect(Segment arr[], int n)
    // Pushing all points to a vector of events
    vector<Event> e;
    for (int i = 0; i < n; ++i) {
        e.push_back(Event(arr[i].left.x, arr[i].left.y, true, i));
        e.push_back(Event(arr[i].right.x, arr[i].right.y, false, i));
    // Sorting all events according to x coordinate.
    sort(e.begin(), e.end(), [](Event &e1, Event &e2) {return e1.x < e2.x;});
    // For storing active segments.
    set<Event> s;
    // Traversing through sorted points
    for (int i=0; i<2*n; i++)
        Event curr = e[i];
        int index = curr.index;
        // If current point is left of its segment
        if (curr.isLeft)
            // Get above and below points
            auto next = s.lower_bound(curr);
            auto prev = pred(s, next);
            // Check if current point intersects with 
            // any of its adjacent 
            if (next != s.end() && doIntersect(arr[next->index], arr[index]))
               return true;
            if (prev != s.end() && doIntersect(arr[prev->index], arr[index]))
               return true;
            // Insert current point (or event)   
        // If current point is right of its segment
            // Find the iterator
            auto it = s.find(curr);
            // Find above and below points
            auto next = succ(s, it);
            auto prev = pred(s, it);
            // If above and below point intersect
            if (next != s.end() && prev != s.end())
               if (doIntersect(arr[prev->index], arr[next->index]))
                  return true;
            // Return current point
    return false;
// Driver code
int main() {
    Segment arr[] = { {{0, 0}, {0, 4}}, {{1, 0}, {5, 0}}};
    int n = sizeof(arr)/sizeof(arr[0]);
    cout << isIntersect(arr, n);
    return 0;


Time Complexity: The first step is sorting which takes O(nLogn) time. The second step process 2n points and for processing every point, it takes O(Logn) time. Therefore, overall time complexity is O(nLogn) 


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