Given some points on a plane, which are distinct and no three of them lie on the same line. We need to find number of Parallelograms with the vertices as the given points. Examples:
Input : points[] = {(0, 0), (0, 2), (2, 2), (4, 2), (1, 4), (3, 4)} Output : 2 Two Parallelograms are possible by choosing above given point as vertices, which are shown in below diagram.
We can solve this problem by using a special property of parallelograms that diagonals of a parallelogram intersect each other in the middle. So if we get such a middle point which is middle point of more than one line segment, then we can conclude that a parallelogram exists, more accurately if a middle point occurs x times, then diagonals of possible parallelograms can be chosen in xC2 ways, i.e. there will be x*(x-1)/2 parallelograms corresponding to this particular middle point with a frequency x. So we iterate over all pair of points and we calculate their middle point and increase frequency of middle point by 1. At the end, we count number of parallelograms according to the frequency of each distinct middle point as explained above. As we just need frequency of middle point, division by 2 is ignored while calculating middle point for simplicity.
// C++ program to get number of Parallelograms we // can make by given points of the plane #include <bits/stdc++.h> using namespace std;
// Returns count of Parallelograms possible // from given points int countOfParallelograms( int x[], int y[], int N)
{ // Map to store frequency of mid points
map<pair< int , int >, int > cnt;
for ( int i=0; i<N; i++)
{
for ( int j=i+1; j<N; j++)
{
// division by 2 is ignored, to get
// rid of doubles
int midX = x[i] + x[j];
int midY = y[i] + y[j];
// increase the frequency of mid point
cnt[make_pair(midX, midY)]++;
}
}
// Iterating through all mid points
int res = 0;
for ( auto it = cnt.begin(); it != cnt.end(); it++)
{
int freq = it->second;
// Increase the count of Parallelograms by
// applying function on frequency of mid point
res += freq*(freq - 1)/2;
}
return res;
} // Driver code to test above methods int main()
{ int x[] = {0, 0, 2, 4, 1, 3};
int y[] = {0, 2, 2, 2, 4, 4};
int N = sizeof (x) / sizeof ( int );
cout << countOfParallelograms(x, y, N) << endl;
return 0;
} |
/*package whatever //do not write package name here */ import java.io.*;
import java.util.*;
public class GFG {
// Returns count of Parallelograms possible
// from given points
public static int countOfParallelograms( int [] x, int [] y, int N)
{
// Map to store frequency of mid points
HashMap<String, Integer> cnt = new HashMap<>();
for ( int i= 0 ; i<N; i++)
{
for ( int j=i+ 1 ; j<N; j++)
{
// division by 2 is ignored, to get
// rid of doubles
int midX = x[i] + x[j];
int midY = y[i] + y[j];
// increase the frequency of mid point
String temp = String.join( " " , String.valueOf(midX), String.valueOf(midY));
if (cnt.containsKey(temp)){
cnt.put(temp, cnt.get(temp) + 1 );
}
else {
cnt.put(temp, 1 );
}
}
}
// Iterating through all mid points
int res = 0 ;
for (Map.Entry<String, Integer> it : cnt.entrySet()) {
int freq = it.getValue();
// Increase the count of Parallelograms by
// applying function on frequency of mid point
res = res + freq*(freq - 1 )/ 2 ;
}
return res;
}
public static void main(String[] args) {
int [] x = { 0 , 0 , 2 , 4 , 1 , 3 };
int [] y = { 0 , 2 , 2 , 2 , 4 , 4 };
int N = x.length;
System.out.println(countOfParallelograms(x, y, N));
}
} // The code is contributed by Nidhi goel. |
# python program to get number of Parallelograms we # can make by given points of the plane # Returns count of Parallelograms possible # from given points def countOfParallelograms(x, y, N):
# Map to store frequency of mid points
cnt = {}
for i in range (N):
for j in range (i + 1 , N):
# division by 2 is ignored, to get
# rid of doubles
midX = x[i] + x[j];
midY = y[i] + y[j];
# increase the frequency of mid point
if ((midX, midY) in cnt):
cnt[(midX, midY)] + = 1
else :
cnt[(midX, midY)] = 1
# Iterating through all mid points
res = 0
for key in cnt:
freq = cnt[key]
# Increase the count of Parallelograms by
# applying function on frequency of mid point
res + = freq * (freq - 1 ) / 2
return res
# Driver code to test above methods x = [ 0 , 0 , 2 , 4 , 1 , 3 ]
y = [ 0 , 2 , 2 , 2 , 4 , 4 ]
N = len (x);
print ( int (countOfParallelograms(x, y, N)))
# The code is contributed by Gautam goel. |
using System;
using System.Collections.Generic;
public class GFG
{ // Returns count of Parallelograms possible
// from given points
public static int CountOfParallelograms( int [] x, int [] y, int N)
{
// Map to store frequency of mid points
Dictionary< string , int > cnt = new Dictionary< string , int >();
for ( int i = 0; i < N; i++)
{
for ( int j = i + 1; j < N; j++)
{
// division by 2 is ignored, to get
// rid of doubles
int midX = x[i] + x[j];
int midY = y[i] + y[j];
// increase the frequency of mid point
string temp = string .Join( " " , midX.ToString(), midY.ToString());
if (cnt.ContainsKey(temp))
{
cnt[temp]++;
}
else
{
cnt.Add(temp, 1);
}
}
}
// Iterating through all mid points
int res = 0;
foreach (KeyValuePair< string , int > it in cnt)
{
int freq = it.Value;
// Increase the count of Parallelograms by
// applying function on frequency of mid point
res += freq * (freq - 1) / 2;
}
return res;
}
public static void Main( string [] args)
{
int [] x = { 0, 0, 2, 4, 1, 3 };
int [] y = { 0, 2, 2, 2, 4, 4 };
int N = x.Length;
Console.WriteLine(CountOfParallelograms(x, y, N));
}
} |
// JavaScript program to get number of Parallelograms we // can make by given points of the plane // Returns count of Parallelograms possible // from given points function countOfParallelograms(x, y, N)
{ // Map to store frequency of mid points
// map<pair<int, int>, int> cnt;
let cnt = new Map();
for (let i=0; i<N; i++)
{
for (let j=i+1; j<N; j++)
{
// division by 2 is ignored, to get
// rid of doubles
let midX = x[i] + x[j];
let midY = y[i] + y[j];
// increase the frequency of mid point
let make_pair = [midX, midY];
if (cnt.has(make_pair.join( '' ))){
cnt.set(make_pair.join( '' ), cnt.get(make_pair.join( '' )) + 1);
}
else {
cnt.set(make_pair.join( '' ), 1);
}
}
}
// Iterating through all mid points
let res = 0;
for (const [key, value] of cnt)
{
let freq = value;
// Increase the count of Parallelograms by
// applying function on frequency of mid point
res = res + Math.floor(freq*(freq - 1)/2);
}
return res;
} // Driver code to test above methods let x = [0, 0, 2, 4, 1, 3]; let y = [0, 2, 2, 2, 4, 4]; let N = x.length; console.log(countOfParallelograms(x, y, N)); // The code is contributed by Gautam goel (gautamgoel962) |
2
Time Complexity: O(n2logn), as we are iterating through two loops up to n and using a map as well which takes logn.
Auxiliary Space: O(n)