Givet nogle punkter på et plan, som er forskellige, og ikke tre af dem ligger på samme linje. Vi skal finde antallet af Parallelogrammer med toppunkterne som de givne punkter. 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 problem ved at bruge en speciel egenskab ved parallelogrammer, at diagonaler af et parallelogram skærer hinanden i midten. Så hvis vi får et sådant midtpunkt, som er midtpunktet af mere end et linjestykke, kan vi konkludere, at et parallelogram eksisterer mere præcist, hvis et midterpunkt forekommer x gange, så kan diagonaler af mulige parallelogrammer vælges ixC2måder, dvs. der vil være x*(x-1)/2 parallelogrammer svarende til dette særlige midtpunkt med en frekvens x. Så vi itererer over alle par af punkter, og vi beregner deres midtpunkt og øger frekvensen af midtpunkt med 1. Til sidst tæller vi antallet af parallelogrammer i henhold til frekvensen af hvert enkelt midtpunkt som forklaret ovenfor. Da vi bare har brug for, ignoreres frekvensen af midtpunkts division med 2, mens midtpunktet beregnes for nemheds 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)
Produktion
2
Tidskompleksitet: På2logn), da vi itererer gennem to sløjfer op til n og bruger også et kort, som tager logn.
Hjælpeplads: På)
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