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.
CPP
// 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 pointsint 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 methodsint 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;} |
Java
/*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. |
Python3
# python program to get number of Parallelograms we# can make by given points of the plane# Returns count of Parallelograms possible# from given pointsdef 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 methodsx = [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. |
C#
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
// JavaScript program to get number of Parallelograms we// can make by given points of the plane// Returns count of Parallelograms possible// from given pointsfunction 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 methodslet 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)
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