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

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;

9

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