Given N points in 2 dimensional space. The task is to count the number of triplets pairs (A, B, C) such that point B is the midpoint of line segment formed by joining points A and C.
Examples:
Input: points = {{1, 1}, {2, 2}, {3, 3}}
Output: 1
The point (2, 2) is the midpoint of the line segment joining points (1, 1) and (3, 3).
Input: points = {{1, 1}, {1, 2}, {1, 5}}
Output: 0
Approach: Consider a pair of points A and C. The midpoint of the line segment joining these points will be ((A * X + C * X) / 2, (A * Y + C * Y) / 2)). If the point is present in the given list of points, we have found a triplet. To quickly check if a point is in our list of points we can use a set. Doing this for all pairs of points will give us the required count.
Below is the implementation of the above approach:
C++
// C++ implementation of the approach#include <bits/stdc++.h>using namespace std;// Function to return the count of possible tripletsint countTriplets(int n, vector<pair<int, int> > points){ set<pair<int, int> > pts; int ct = 0; // Insert all the points in a set for (int i = 0; i < n; i++) pts.insert(points[i]); for (int i = 0; i < n; i++) for (int j = i + 1; j < n; j++) { int x = points[i].first + points[j].first; int y = points[i].second + points[j].second; // If the mid point exists in the set if (x % 2 == 0 && y % 2 == 0) if (pts.find(make_pair(x / 2, y / 2)) != pts.end()) ct++; } // Return the count of valid triplets return ct;}// Driver codeint main(){ vector<pair<int, int> > points = { { 1, 1 }, { 2, 2 }, { 3, 3 } }; int n = points.size(); cout << countTriplets(n, points);} |
Java
// Java implementation of the approachimport java.util.*;class GFG{ static class pair{ int first,second; public pair(int first, int second) { this.first = first; this.second = second; } }// Function to return the count of possible tripletsstatic int countTriplets(int n, Vector<pair> points){ Set<pair> pts = new HashSet<pair>(); int ct = 0; // Insert all the points in a set for (int i = 0; i < n; i++) pts.add(points.get(i)); for (int i = 0; i < n; i++) for (int j = i + 1; j < n; j++) { int x = points.get(i).first + points.get(j).first; int y = points.get(i).second + points.get(j).second; // If the mid point exists in the set if (x % 2 == 0 && y % 2 == 0) if (!pts.contains(new pair(x / 2, y / 2))) ct++; } // Return the count of valid triplets return ct;}// Driver codepublic static void main(String args[]) { Vector<pair> points = new Vector<>(); points.add(new pair(1,1)); points.add(new pair(2,2)); points.add(new pair(3,3)); int n = points.size(); System.out.println(countTriplets(n, points));}}// This code is contributed by Princi Singh |
Python3
# Python3 implementation of the approach # Function to return the count # of possible triplets def countTriplets(n, points) : pts = [] ct = 0; # Insert all the points in a set for i in range(n) : pts.append(points[i]); for i in range(n) : for j in range(i + 1, n) : x = points[i][0] + points[j][0]; y = points[i][1] + points[j][1]; # If the mid point exists in the set if (x % 2 == 0 and y % 2 == 0) : if [x // 2, y // 2] in pts : ct += 1 # Return the count of valid triplets return ct # Driver code if __name__ == "__main__" : points = [[ 1, 1 ], [ 2, 2 ], [ 3, 3 ]] n = len(points) print(countTriplets(n, points))# This code is contributed by Ryuga |
C#
// C# implementation of the approachusing System; using System.Collections.Generic; class GFG{ public class pair{ public int first,second; public pair(int first, int second) { this.first = first; this.second = second; } }// Function to return the count of possible tripletsstatic int countTriplets(int n, List<pair> points){ HashSet<pair> pts = new HashSet<pair>(); int ct = 0; // Insert all the points in a set for (int i = 0; i < n; i++) pts.Add(points[i]); for (int i = 0; i < n; i++) for (int j = i + 1; j < n; j++) { int x = points[i].first + points[j].first; int y = points[i].second + points[j].second; // If the mid point exists in the set if (x % 2 == 0 && y % 2 == 0) if (!pts.Contains(new pair(x / 2, y / 2))) ct++; } // Return the count of valid triplets return ct;}// Driver codepublic static void Main(String []args) { List<pair> points = new List<pair>(); points.Add(new pair(1, 1)); points.Add(new pair(2, 2)); points.Add(new pair(3, 3)); int n = points.Count; Console.WriteLine(countTriplets(n, points));}}// This code is contributed by 29AjayKumar |
PHP
<?php// PHP implementation of the approach // Function to return the count // of possible triplets function countTriplets($n, $points){ $pts = array(); $ct = 0; // Insert all the points in a set for ($i = 0; $i < count($points); $i++) { for ($j = 0; $j < count($points[$i]); $j++) { $pts[] = $points[$i][$j]; } } for ($i = 0; $i < $n; $i++) for ($j = $i + 1; $j < $n; $j++) { $x = $points[$i][0] + $points[$j][0]; $y = $points[$i][1] + $points[$j][1]; // If the mid point exists in the set if ($x % 2 == 0 and $y % 2 == 0) if (in_array((int)($x / 2), $pts) and in_array((int)($y / 2), $pts)) $ct += 1; } // Return the count of valid triplets return $ct; }// Driver code $points = array(array( 1, 1 ), array( 2, 2 ), array( 3, 3 ));$n = count($points);print(countTriplets($n, $points));// This code is contributed by chandan_jnu?> |
Javascript
<script>// Javascript implementation of the approach// Function to return the count of possible tripletsfunction countTriplets(n, points){ var pts = new Set(); var ct = 0; // Insert all the points in a set for (var i = 0; i < n; i++) pts.add(points[i].toString()); for (var i = 0; i < n; i++) for (var j = i + 1; j < n; j++) { var x = points[i][0] + points[j][0]; var y = points[i][1] + points[j][1]; // If the mid point exists in the set if (x % 2 == 0 && y % 2 == 0) if (pts.has([(x / 2), (y / 2)].toString())) ct++; } // Return the count of valid triplets return ct;}// Driver codevar points = [ [ 1, 1 ], [ 2, 2 ], [ 3, 3 ] ];var n = points.length;document.write( countTriplets(n, points))// This code is contributed by famously.</script> |
Output:
1
Time Complexity: O(N2 logN), where N represents the size of the given vector.
Auxiliary Space: O(N), where N represents the size of the given vector.
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