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Count triplet pairs (A, B, C) of points in 2-D space that satisfy the given condition

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  • Last Updated : 24 May, 2022
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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:
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:
 

 

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 triplets
int 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 code
int 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 approach
import 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 triplets
static 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 code
public 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 approach
using 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 triplets
static 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 code
public 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 triplets
function 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 code
var 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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