Klee’s Algorithm (Length Of Union Of Segments of a line)

16 Jun, 2022
Given starting and ending positions of segments on a line, the task is to take the union of all given segments and find length covered by these segments.Examples:  
Input : segments[] = {{2, 5}, {4, 8}, {9, 12}}
Output : 9 
Explanation:
segment 1 = {2, 5}
segment 2 = {4, 8}
segment 3 = {9, 12}
If we take the union of all the line segments,
we cover distances [2, 8] and [9, 12]. Sum of 
these two distances is 9 (6 + 3)
Approach:
The algorithm was proposed by Klee in 1977. The time complexity of the algorithm is O (N log N). It has been proven that this algorithm is the fastest (asymptotically) and this problem can not be solved with a better complexity. 
Description : 1) Put all the coordinates of all the segments in an auxiliary array points[]. 2) Sort it on the value of the coordinates. 3) An additional condition for sorting – if there are equal coordinates, insert the one which is the left coordinate of any segment instead of a right one. 4) Now go through the entire array, with the counter “count” of overlapping segments. 5) If the count is greater than zero, then the result is added to the difference between the points[i] – points[i-1]. 6) If the current element belongs to the left end, we increase “count”, otherwise reduce it.Illustration: 
Lets take the example :
segment 1 : (2,5)
segment 2 : (4,8)
segment 3 : (9,12)

Counter = result = 0;
n = number of segments = 3;

for i=0,  points[0] = {2, false}
          points[1] = {5, true}
for i=1,  points[2] = {4, false}
          points[3] = {8, true}
for i=2,  points[4] = {9, false}
          points[5] = {12, true}

Therefore :
points = {2, 5, 4, 8, 9, 12}
         {f, t, f, t, f, t}

after applying sorting :
points = {2, 4, 5, 8, 9, 12}
         {f, f, t, t, f, t}

Now,
for i=0, result = 0;
         Counter = 1;

for i=1, result = 2;
         Counter = 2;

for i=2, result = 3;
         Counter = 1;

for i=3, result = 6;
         Counter = 0;

for i=4, result = 6;
         Counter = 1;

for i=5, result = 9;
         Counter = 0;

Final answer = 9;
C++
Java
Python3
Javascript
// C++ program to implement Klee's algorithm#include<bits/stdc++.h>using namespace std; // Returns sum of lengths covered by union of given// segmentsint segmentUnionLength(const vector<                          pair <int,int> > &seg){    int n = seg.size();     // Create a vector to store starting and ending    // points    vector <pair <int, bool> > points(n * 2);    for (int i = 0; i < n; i++)    {        points[i*2]     = make_pair(seg[i].first, false);        points[i*2 + 1] = make_pair(seg[i].second, true);    }     // Sorting all points by point value    sort(points.begin(), points.end());     int result = 0; // Initialize result     // To keep track of counts of    // current open segments    // (Starting point is processed,    // but ending point    // is not)    int Counter = 0;     // Traverse through all points    for (unsigned i=0; i<n*2; i++)    {        // If there are open points, then we add the        // difference between previous and current point.        // This is interesting as we don't check whether        // current point is opening or closing,        if (Counter)            result += (points[i].first -                        points[i-1].first);         // If this is an ending point, reduce, count of        // open points.        (points[i].second)? Counter-- : Counter++;    }    return result;} // Driver program for the above codeint main(){    vector< pair <int,int> > segments;    segments.push_back(make_pair(2, 5));    segments.push_back(make_pair(4, 8));    segments.push_back(make_pair(9, 12));    cout << segmentUnionLength(segments) << endl;    return 0;}
// Java program to implement Klee's algorithmimport java.io.*;import java.util.*; class GFG {   // to use create a pair of segments  static class SegmentPair  {    int x,y;    SegmentPair(int xx, int yy){      this.x = xx;      this.y = yy;    }  }   //to create a pair of points  static class PointPair{    int x;    boolean isEnding;    PointPair(int xx, boolean end){      this.x = xx;      this.isEnding = end;    }  }   // creates the comparator for comparing objects of PointPair class  static class Comp implements Comparator<PointPair>  {         // override the compare() method    public int compare(PointPair p1, PointPair p2)    {      if (p1.x < p2.x) {        return -1;      }      else {        if(p1.x == p2.x){          return 0;        }else{          return 1;        }      }    }  }   public static int segmentUnionLength(List<SegmentPair> segments){    int n = segments.size();     // Create a list to store    // starting and ending points    List<PointPair> points = new ArrayList<>();    for(int i = 0; i < n; i++){      points.add(new PointPair(segments.get(i).x,false));      points.add(new PointPair(segments.get(i).y,true));    }         // Sorting all points by point value    Collections.sort(points, new Comp());     int result = 0; // Initialize result     // To keep track of counts of    // current open segments    // (Starting point is processed,    // but ending point    // is not)    int Counter = 0;     // Traverse through all points    for(int i = 0; i < 2 * n; i++)    {             // If there are open points, then we add the      // difference between previous and current point.      // This is interesting as we don't check whether      // current point is opening or closing,      if (Counter != 0)      {        result += (points.get(i).x - points.get(i-1).x);      }       // If this is an ending point, reduce, count of      // open points.      if(points.get(i).isEnding)      {        Counter--;      }      else      {        Counter++;      }    }    return result;  }   // Driver Code  public static void main (String[] args) {    List<SegmentPair> segments = new ArrayList<>();    segments.add(new SegmentPair(2,5));    segments.add(new SegmentPair(4,8));    segments.add(new SegmentPair(9,12));    System.out.println(segmentUnionLength(segments));  }} // This code is contributed by shruti456rawal
# Python program for the above approach def segmentUnionLength(segments):       # Size of given segments list    n = len(segments)         # Initialize empty points container    points = [None] * (n * 2)         # Create a vector to store starting    # and ending points    for i in range(n):        points[i * 2] = (segments[i][0], False)        points[i * 2 + 1] = (segments[i][1], True)             # Sorting all points by point value    points = sorted(points, key=lambda x: x[0])         # Initialize result as 0    result = 0         # To keep track of counts of current open segments    # (Starting point is processed, but ending point    # is not)    Counter = 0         # Traverse through all points    for i in range(0, n * 2):               # If there are open points, then we add the        # difference between previous and current point.        if (i > 0) & (points[i][0] > points[i - 1][0]) &  (Counter > 0):            result += (points[i][0] - points[i - 1][0])                     # If this is an ending point, reduce, count of        # open points.        if points[i][1]:            Counter -= 1        else:            Counter += 1    return result  # Driver codeif __name__ == '__main__':    segments = [(2, 5), (4, 8), (9, 12)]    print(segmentUnionLength(segments))
// JavaScript program to implement Klee's algorithm // Returns sum of lengths covered by union of given// segmentsfunction segmentUnionLength(seg){    let n = seg.length;     // Create a vector to store starting and ending    // points    let points = new Array(2*n);    for (let i = 0; i < n; i++)    {        points[i*2] = [seg[i][0], false];        points[i*2 + 1] = [seg[i][1], true];    }     // Sorting all points by point value    points.sort(function(a, b){        return a[0] - b[0];    });         let result = 0; // Initialize result     // To keep track of counts of    // current open segments    // (Starting point is processed,    // but ending point    // is not)    let Counter = 0;     // Traverse through all points    for (let i=0; i<n*2; i++)    {        // If there are open points, then we add the        // difference between previous and current point.        // This is interesting as we don't check whether        // current point is opening or closing,        if (Counter)            result += (points[i][0] - points[i-1][0]);         // If this is an ending point, reduce, count of        // open points.        if(points[i][1]){            Counter = Counter - 1;        }        else{            Counter = Counter + 1;        }    }    return result;} let segments = new Array();segments.push([2, 5]);segments.push([4, 8]);segments.push([9, 12]);console.log(segmentUnionLength(segments)); // The code is contributed by Gautam goel (gautamgoel962)
9
Time Complexity : O(n * log n)Auxiliary Space: O(n)
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