# Check if four segments form a rectangle

• Difficulty Level : Medium
• Last Updated : 23 Jun, 2022

We are given four segments as a pair of coordinates of their end points. We need to tell whether those four line segments make a rectangle or not.
Examples:

```Input : segments[] =  [(4, 2), (7, 5),
(2, 4), (4, 2),
(2, 4), (5, 7),
(5, 7), (7, 5)]
Output : Yes
Given these segment make a rectangle of length 3X2.

Input : segment[] = [(7, 0), (10, 0),
(7, 0), (7, 3),
(7, 3), (10, 2),
(10, 2), (10, 0)]
Output : Not
These segments do not make a rectangle.

Above examples are shown in below diagram.```

This problem is mainly an extension of How to check if given four points form a square

We can solve this problem by using properties of a rectangle. First, we check total unique end points of segments, if count of these points is not equal to 4 then the line segment can’t make a rectangle. Then we check distances between all pair of points, there should be at most 3 different distances, one for diagonal and two for sides and at the end we will check the relation among these three distances, for line segments to make a rectangle these distance should satisfy Pythagorean relation because sides and diagonal of rectangle makes a right angle triangle. If they satisfy mentioned conditions then we will flag polygon made by line segment as rectangle otherwise not.

## CPP

 `// C++ program to check whether it is possible``// to make a rectangle from 4 segments``#include ``using` `namespace` `std;``#define N 4` `// structure to represent a segment``struct` `Segment``{``    ``int` `ax, ay;``    ``int` `bx, by;``};` `// Utility method to return square of distance``// between two points``int` `getDis(pair<``int``, ``int``> a, pair<``int``, ``int``> b)``{``    ``return` `(a.first - b.first)*(a.first - b.first) +``        ``(a.second - b.second)*(a.second - b.second);``}` `// method returns true if line Segments make``// a rectangle``bool` `isPossibleRectangle(Segment segments[])``{``    ``set< pair<``int``, ``int``> > st;` `    ``// putting all end points in a set to``    ``// count total unique points``    ``for` `(``int` `i = 0; i < N; i++)``    ``{``        ``st.insert(make_pair(segments[i].ax, segments[i].ay));``        ``st.insert(make_pair(segments[i].bx, segments[i].by));``    ``}` `    ``// If total unique points are not 4, then``    ``// they can't make a rectangle``    ``if` `(st.size() != 4)``        ``return` `false``;` `    ``// dist will store unique 'square of distances'``    ``set<``int``> dist;` `    ``// calculating distance between all pair of``    ``// end points of line segments``    ``for` `(``auto` `it1=st.begin(); it1!=st.end(); it1++)``        ``for` `(``auto` `it2=st.begin(); it2!=st.end(); it2++)``            ``if` `(*it1 != *it2)``                ``dist.insert(getDis(*it1, *it2));` `    ``// if total unique distance are more than 3,``    ``// then line segment can't make a rectangle``    ``if` `(dist.size() > 3)``        ``return` `false``;` `    ``// copying distance into array. Note that set maintains``    ``// sorted order.``    ``int` `distance;``    ``int` `i = 0;``    ``for` `(``auto` `it = dist.begin(); it != dist.end(); it++)``        ``distance[i++] = *it;` `    ``// If line seqments form a square``    ``if` `(dist.size() == 2)``    ``return` `(2*distance == distance);` `    ``// distance of sides should satisfy pythagorean``    ``// theorem``    ``return` `(distance + distance == distance);``}` `// Driver code to test above methods``int` `main()``{``    ``Segment segments[] =``    ``{``        ``{4, 2, 7, 5},``        ``{2, 4, 4, 2},``        ``{2, 4, 5, 7},``        ``{5, 7, 7, 5}``    ``};` `    ``(isPossibleRectangle(segments))?cout << ``"Yes\n"``:cout << ``"No\n"``;``}`

## Javascript

 `// JavaScript program to check whether it is possible``// to make a rectangle from 4 segments` `const N = 4;` `// Utility method to return square of distance``// between two points``function` `getDis(a, b)``{``    ``return` `(parseInt(a) - parseInt(b))*(parseInt(a) - parseInt(b)) + (parseInt(a) - parseInt(b))*(parseInt(a) - parseInt(b));``}` `// method returns true if line Segments make``// a rectangle``function` `isPossibleRectangle(segments)``{  ``    ``let st = ``new` `Set();` `    ``// putting all end points in a set to``    ``// count total unique points``    ``for` `(let i = 0; i < N; i++)``    ``{``        ``let tmp1 = [segments[i], segments[i]];``        ``let tmp2 = [segments[i], segments[i]];``        ``st.add(tmp1.join(``''``));``        ``st.add(tmp2.join(``''``));``    ``}` `    ``// If total unique points are not 4, then``    ``// they can't make a rectangle``    ``if` `(st.size != 4)``    ``{``        ``return` `false``;``    ``}``        ` `    ``// dist will store unique 'square of distances'``    ``let dist = ``new` `Set();` `    ``// calculating distance between all pair of``    ``// end points of line segments``    ``for``(let it1 of st)``    ``{``        ``for``(let it2 of st)``        ``{``            ``if``(it1 !== it2)``            ``{``                ``dist.add(getDis(it1.split('``'), it2.split('``')));``            ``}``        ``}``    ``}` `    ``// if total unique distance are more than 3,``    ``// then line segment can'``t make a rectangle``    ``if` `(dist.size > 3)``    ``{``        ``return` `false``;``    ``}``        ` `    ``// copying distance into array. Note that set maintains``    ``// sorted order.``    ``let distance = ``new` `Array();``    ``for` `(let x of dist)``    ``{``        ``distance.push(x);``    ``}``        ` `    ``// If line seqments form a square``    ``if` `(dist.size === 2)``    ``{``        ``return` `(2*distance == distance);``    ``}` `    ``// distance of sides should satisfy pythagorean``    ``// theorem``    ``return` `(distance + distance == distance);``}` `// Driver code to test above methods``{``    ``let segments = [``        ``[4, 2, 7, 5],``        ``[2, 4, 4, 2],``        ``[2, 4, 5, 7],``        ``[5, 7, 7, 5] ]` `    ``if``(isPossibleRectangle(segments)){``        ``console.log(``"Yes"``);``    ``}``    ``else``{``        ``console.log(``"No"``);``    ``}``}` `// The code is contributed by  Nidhi Goel`

Output:

`Yes`

Time Complexity: O(n2logn)
Auxiliary Space: O(n)

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