# Check if four segments form a rectangle

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 <bits/stdc++.h> ` `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[3]; ` ` ` `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[0] == distance[1]); ` ` ` `// distance of sides should satisfy pythagorean ` ` ` `// theorem ` ` ` `return` `(distance[0] + distance[1] == distance[2]); ` `} ` `// 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[0]) - parseInt(b[0]))*(parseInt(a[0]) - parseInt(b[0])) + (parseInt(a[1]) - parseInt(b[1]))*(parseInt(a[1]) - parseInt(b[1])); ` `} ` `// 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][0], segments[i][1]];` ` ` `let tmp2 = [segments[i][2], segments[i][3]];` ` ` `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[0] == distance[1]); ` ` ` `}` ` ` `// distance of sides should satisfy pythagorean ` ` ` `// theorem ` ` ` `return` `(distance[0] + distance[1] == distance[2]); ` `} ` `// 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(n^{2}logn) **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.

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