# Finding the number of triangles amongst horizontal and vertical line segments

Prerequisites: BIT

Given ‘n’ line segments, each of them is either horizontal or vertical, find the maximum number of triangles(including triangles with zero area) that can be formed by joining the intersection points of the line segments.

No two horizontal line segments overlap, nor do two vertical line segments. A line is represented using two points(four integers, first two being the x and y coordinates, respectively for the first point and the other two being the x and y coordinates for the second point)

Examples:

```       |
---|-------|--
|       |    -----
|  --|--|-     |
|       |      |

For the above line segments, there are four points of intersection between
vertical and horizontal lines, every three out of which form a triangle,
so there can be 4C3 triangles.
```

## Recommended: Please try your approach on {IDE} first, before moving on to the solution.

The idea is based on Sweep Line Algorithm.

Building a solution in steps:

1. Store both points of all line segments with the corresponding event(described below) in a vector and sort all the points, in non-decreasing order of their x coordinates.
2. Let’s now imagine a vertical line that we sweep across all these points and describe 3 events, based on which point we currently are:
• in – leftmost point of a horizontal line segment
• out – rightmost point of a horizontal line segment
• a vertical line
3. We call the region “active” or the horizontal lines “active” that have had the first event but not second. We will have a BIT(Binary indexed tree) to store the ‘y’ coordinates of all active lines.
4. Once a line becomes inactive, we remove its ‘y’ from the BIT.
5. When an event of third type occurs, that is, when we are at a vertical line, we query the tree in range of its ‘y’ coordinates and add the result to the number of intersection points so far.
6. Finally, we will have the number of points of intersections, say m, then the number of triangles (including zero area) will be mC3.

Note: We need to carefully sort the points, look at the cmp() function in the implementation for clarification.

 `// A C++ implementation of the above idea ` `#include ` `#define maxy 1000005 ` `#define maxn 10005 ` ` `  `using` `namespace` `std; ` ` `  `// structure to store point ` `struct` `point ` `{ ` `    ``int` `x, y; ` `    ``point(``int` `a, ``int` `b) ` `    ``{ ` `        ``x = a, y = b; ` `    ``} ` `}; ` ` `  `// Note: Global arrays are initially zero ` `// array to store BIT and vector to store ` `// the points and their corresponding event number, ` `// in the second field of the pair ` `int` `bit[maxy]; ` `vector > events; ` ` `  `// compare function to sort in order of non-decreasing ` `// x coordinate and if x coordinates are same then ` `// order on the basis of events on the points ` `bool` `cmp(pair &a, pair &b) ` `{ ` `    ``if` `( a.first.x != b.first.x ) ` `        ``return` `a.first.x < b.first.x; ` ` `  `    ``//if the x coordinates are same ` `    ``else` `    ``{ ` `        ``// both points are of the same vertical line ` `        ``if` `(a.second == 3 && b.second == 3) ` `        ``{ ` `            ``return` `true``; ` `        ``} ` ` `  `        ``// if an 'in' event occurs before 'vertical' ` `        ``// line event for the same x coordinate ` `        ``else` `if` `(a.second == 1 && b.second == 3) ` `        ``{ ` `            ``return` `true``; ` `        ``} ` ` `  `        ``// if a 'vertical' line comes before an 'in' ` `        ``// event for the same x coordinate, swap them ` `        ``else` `if` `(a.second == 3 && b.second == 1) ` `        ``{ ` `            ``return` `false``; ` `        ``} ` ` `  `        ``// if an 'out' event occurs before a 'vertical' ` `        ``// line event for the same x coordinate, swap. ` `        ``else` `if` `(a.second == 2 && b.second == 3) ` `        ``{ ` `            ``return` `false``; ` `        ``} ` ` `  `        ``//in all other situations ` `        ``return` `true``; ` `    ``} ` `} ` ` `  `// update(y, 1) inserts a horizontal line at y coordinate ` `// in an active region, while update(y, -1) removes it ` `void` `update(``int` `idx, ``int` `val) ` `{ ` `    ``while` `(idx < maxn) ` `    ``{ ` `        ``bit[idx] += val; ` `        ``idx += idx & (-idx); ` `    ``} ` `} ` ` `  `// returns the number of lines in active region whose y ` `// coordinate is between 1 and idx ` `int` `query(``int` `idx) ` `{ ` `    ``int` `res = 0; ` `    ``while` `(idx > 0) ` `    ``{ ` `        ``res += bit[idx]; ` `        ``idx -= idx & (-idx); ` `    ``} ` `    ``return` `res; ` `} ` ` `  `// inserts a line segment ` `void` `insertLine(point a, point b) ` `{ ` `    ``// if it is a horizontal line ` `    ``if` `(a.y == b.y) ` `    ``{ ` `        ``int` `beg = min(a.x, b.x); ` `        ``int` `end = max(a.x, b.x); ` ` `  `        ``// the second field in the pair is the event number ` `        ``events.push_back(make_pair(point(beg, a.y), 1)); ` `        ``events.push_back(make_pair(point(end, a.y), 2)); ` `    ``} ` ` `  `    ``//if it is a vertical line ` `    ``else` `    ``{ ` `        ``int` `up = max(b.y, a.y); ` `        ``int` `low = min(b.y, a.y); ` ` `  `        ``//the second field of the pair is the event number ` `        ``events.push_back(make_pair(point(a.x, up), 3)); ` `        ``events.push_back(make_pair(point(a.x, low), 3)); ` `    ``} ` `} ` ` `  `// returns the number of intersection points between all ` `// the lines, vertical and horizontal, to be run after the ` `// points have been sorted using the cmp() function ` `int` `findIntersectionPoints() ` `{ ` `    ``int` `intersection_pts = 0; ` `    ``for` `(``int` `i = 0 ; i < events.size() ; i++) ` `    ``{ ` `        ``//if the current point is on an 'in' event ` `        ``if` `(events[i].second == 1) ` `        ``{ ` `            ``//insert the 'y' coordinate in the active region ` `            ``update(events[i].first.y, 1); ` `        ``} ` ` `  `        ``// if current point is on an 'out' event ` `        ``else` `if` `(events[i].second == 2) ` `        ``{ ` `            ``// remove the 'y' coordinate from the active region ` `            ``update(events[i].first.y, -1); ` `        ``} ` ` `  `        ``// if the current point is on a 'vertical' line ` `        ``else` `        ``{ ` `            ``// find the range to be queried ` `            ``int` `low = events[i++].first.y; ` `            ``int` `up = events[i].first.y; ` `            ``intersection_pts += query(up) - query(low); ` `        ``} ` `    ``} ` `    ``return` `intersection_pts; ` `} ` ` `  `// returns (intersection_pts)C3 ` `int` `findNumberOfTriangles() ` `{ ` `    ``int` `pts = findIntersectionPoints(); ` `    ``if` `( pts >= 3 ) ` `        ``return` `( pts * (pts - 1) * (pts - 2) ) / 6; ` `    ``else` `        ``return` `0; ` `} ` ` `  ` `  `// driver code ` `int` `main() ` `{ ` `    ``insertLine(point(2, 1), point(2, 9)); ` `    ``insertLine(point(1, 7), point(6, 7)); ` `    ``insertLine(point(5, 2), point(5, 8)); ` `    ``insertLine(point(3, 4), point(6, 4)); ` `    ``insertLine(point(4, 3), point(4, 5)); ` `    ``insertLine(point(7, 6), point(9, 6)); ` `    ``insertLine(point(8, 2), point(8, 5)); ` ` `  `    ``// sort the points based on x coordinate ` `    ``// and event they are on ` `    ``sort(events.begin(), events.end(), cmp); ` ` `  `    ``cout << ``"Number of triangles are: "` `<< ` `         ``findNumberOfTriangles() << ````" "````; ` ` `  `    ``return` `0; ` `} `

Output:

```Number of triangles are: 4
```
```Time Complexity: O( n * log(n) + n * log(maximum_y) )
```

Please write comments if you find anything incorrect, or you want to share more information about the topic discussed above.

This article is attributed to GeeksforGeeks.org

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