Count of Triplets

Given N points in a plane in the form of 2D array such that every row consist of two integer L and R where L belongs to x-ordinate and R belongs to y-ordinate. The task is to count the triplets of points(say a, b & c) such that distance between a & b is equals to the distance between a & c.
Note: The order of triplets matters.

Examples:

Input: arr[] = { { 0, 0 }, { 1, 0 }, { 2, 0 } }
Output: 2
Explanation:
The possible triplets are: {{1, 0}, {0, 0}, {2, 0}} and {{1, 0}, {2, 0}, {0, 0}}

Input: arr[] = { {1, 0}, {1, -1}, {2, 3}, {4, 3}, {4, 4} }
Output: 0
Explanation:
There is no such triplets exists.

Approach:



  1. For each point calculate it’s distance to every other points.
  2. Store intermidiate distances(say d) for point to other points in a Map.
  3. If Map has already same distance then count of triplets is twice the value stored for d in Map.
  4. Update the count of current distance in the Map.

Below is the implementation of the above:

C++

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// C++ program for the above appproach
#include <bits/stdc++.h>
using namespace std;
  
// Function to count the triplets
int countTriplets(vector<vector<int> >& p)
{
  
    // Intialise count
    int count = 0;
  
    // Traverse the arr[]
    for (int i = 0; i < p.size(); i++) {
  
        // Map to store the distance between
        // every pairs p[i] and p[j]
        unordered_map<int, int> d;
  
        for (int j = 0; j < p.size(); j++) {
  
            // Find the distance
            int dist = pow(p[j][1] - p[i][1], 2)
                       + pow(p[j][0] - p[i][0], 2);
  
            // If count of distance is greater
            // than 0, then find the count
            if (d[dist] > 0) {
                count += 2 * d[dist];
            }
  
            // Update the current count of the
            // distance
            d[dist]++;
        }
    }
  
    // Return the count of triplets
    return count;
}
  
// Driver Code
int main()
{
  
    // Set of points in plane
    vector<vector<int> > arr = { { 0, 0 },
                                 { 1, 0 },
                                 { 2, 0 } };
  
    // Function call
    cout << countTriplets(arr);
    return 0;
}

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Java

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// Java program for the above appproach
import java.util.*;
class GFG{
  
    // Function to count the triplets
    static int countTriplets(int p[][])
    {
      
        // Intialise count
        int count = 0;
      
        // Traverse the arr[]
        for (int i = 0; i < p.length; i++) {
      
            // Map to store the distance between
            // every pairs p[i] and p[j]
            HashMap<Integer, Integer> d = new HashMap<Integer,Integer>();
      
            for (int j = 0; j < p.length; j++) {
      
                // Find the distance
                int dist = (int)(Math.pow(p[j][1] - p[i][1], 2)+ Math.pow(p[j][0] - p[i][0], 2));
      
                // If count of distance is greater
                // than 0, then find the count
                  
                if (d.containsKey(dist) && d.get(dist) > 0) {
                    count += 2 * d.get(dist);
                }
      
                // Update the current count of the
                // distance
                if (d.containsKey(dist)){
                    d.put(dist,d.get(dist)+1);
                }
                else
                    d.put(dist,1);
            }
        }
      
        // Return the count of triplets
        return count;
    }
      
    // Driver Code
    public static void main(String args[])
    {
      
        // Set of points in plane
        int arr[][] = { { 0, 0 },
                                    { 1, 0 },
                                    { 2, 0 } };
      
        // Function call
        System.out.println(countTriplets(arr));
          
    }
}
  
// This code is contributed by AbhiThakur

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Python3

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# Python3 program for the above appproach
  
# Function to count the triplets
def countTriplets(p) :
  
    # Intialise count
    count = 0;
  
    # Traverse the arr[]
    for i in range(len(p)) :
  
        # Map to store the distance between
        # every pairs p[i] and p[j]
        d = {};
  
        for j in range(len(p)) :
  
              
            # Find the distance
            dist = pow(p[j][1] - p[i][1], 2) + \
                    pow(p[j][0] - p[i][0], 2);
  
            if dist not in d :
                d[dist] = 0;
                  
            # If count of distance is greater
            # than 0, then find the count
            if (d[dist] > 0) :
                count += 2 * d[dist];
  
            # Update the current count of the
            # distance
            d[dist] += 1;
      
    # Return the count of triplets
    return count;
  
# Driver Code
if __name__ == "__main__" :
  
    # Set of points in plane
    arr = [ [ 0, 0 ],
            [ 1, 0 ],
            [ 2, 0 ] ];
  
    # Function call
    print(countTriplets(arr));
      
# This code is contributed by Yash_R

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Output:

2

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