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

3

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 
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)

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.

// C++ program to implement Klee's algorithm
#include<bits/stdc++.h>
using namespace std;

// Returns sum of lengths covered by union of given
// segments
int 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;

    // Trvaerse 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 code
int main()
{
    vector< pair <int,int> > segments;
    segments.push_back(make_pair(3, 15));
    segments.push_back(make_pair(2, 5));
    segments.push_back(make_pair(4, 8));
    segments.push_back(make_pair(9, 12));
    cout << "Length of Union of All segments = ";
    cout << segmentUnionLength(segments) << endl;
    return 0;
}

Output:

Length of Union of All segments = 9

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 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;

Time Complexity : O(n * log n)

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

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