Gitt noen punkter på et plan som er forskjellige og ingen tre av dem ligger på samme linje. Vi må finne antall parallellogrammer med toppunktene som de gitte punktene. Eksempler:
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. Vi kan løse dette problemet ved å bruke en spesiell egenskap ved parallellogrammer at diagonaler i et parallellogram skjærer hverandre i midten. Så hvis vi får et slikt midtpunkt som er midtpunkt av mer enn ett linjestykke, kan vi konkludere med at et parallellogram eksisterer mer nøyaktig hvis et midtpunkt forekommer x ganger, kan diagonaler av mulige parallellogrammer velges ixC2måter, dvs. det vil være x*(x-1)/2 parallellogrammer som tilsvarer dette spesielle midtpunktet med en frekvens x. Så vi itererer over alle par av punkter og vi beregner deres midtpunkt og øker frekvensen til midtpunktet med 1. På slutten teller vi antall parallellogrammer i henhold til frekvensen til hvert distinkte midtpunkt som forklart ovenfor. Siden vi bare trenger frekvensen av midtpunktsdivisjon med 2 ignoreres mens du beregner midtpunktet for enkelhets skyld.
CPP// C++ program to get number of Parallelograms we // can make by given points of the plane #include
/*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 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)
Produksjon
2
Tidskompleksitet: På2logn) mens vi itererer gjennom to løkker opp til n og bruker også et kart som tar logn.
Hjelpeplass: På)
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