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Count pairs of points having distance between them equal to integral values in a K-dimensional space
  • Difficulty Level : Medium
  • Last Updated : 05 Feb, 2021

Given an array points[] representing N points in a K-dimensional space, the task is to find the count of pairs of points in the space such that the distance between the points of each pair is an integer value.

Examples:

Input: points[] = { {1, 2}, {5, 5}, {2, 8} }, K = 2
Output: 1
Explanation: 
Distance between points of the pair(points[0], points[1]) = 5 
Distance between points of the pair(points[1], points[2]) = sqrt(58) 
Distance between points of the pair(points[0], points[2]) = 3 * sqrt(5) 
Therefore, the required output is 1.

Input: points[] = { {-3, 7, 8, 2}, {-12, 1, 10, 2}, {-2, 8, 9, 3} }, K = 4
Output: 2.

 

Approach: The idea is to generate all possible pairs of the given array and find the distance between the points of each pair and check if it is an integer value or not. If found to be true, then increment the count. Finally, print the total count obtained. Follow the steps below to solve the problem:



  • Distance between the points of the pair({ a1, a2, …, aK }, { b1, b2, …, bK }) can be calculated using the below formula: 
     

Distance(a, b) = sqrt(((a1 – b1)2 + (a2 – b2)2 + …. + (aK – bK)2 ))
 

  • Traverse the array, and generate all possible pairs of the given array.
  • For each pair of points, check if the distance between the points of the pair is an integer or not. If found to be true, then increment the count.
  • Finally, print the count obtained.

Below is the implementation of the above approach:

C++




// C++ program for the above approach
#include <bits/stdc++.h>
using namespace std;
 
// Function to  find pairs whose distance between
// the points of is an integer value.
void cntPairs(vector<vector<int> > points, int n, int K)
{
 
    // Stores count of pairs whose distance
    // between points is an integer
    int ans = 0;
 
    // Traverse the array, points[]
    for (int i = 0; i < n; i++) {
 
        for (int j = i + 1; j < n; j++) {
 
            // Stores distance between
            // points(i, j)
            int dist = 0;
 
            // Traverse all the points of
            // current pair
            for (int k = 0; k < K; k++) {
 
                // Update temp
                int temp = (points[i][k]
                            - points[j][k]);
 
                // Update dist
                dist += temp * temp;
            }
 
            // If dist is a perfect square
            if (sqrt(dist) * sqrt(dist) == dist) {
 
                // Update ans
                ans += 1;
            }
        }
    }
 
    cout << ans << endl;
}
 
// Driver Code
int main()
{
 
    // Given value of K
    int K = 2;
 
    // Given points
    vector<vector<int> > points
        = { { 1, 2 }, { 5, 5 }, { -2, 8 } };
 
    // Given value of N
    int n = points.size();
 
    // Function Call
    cntPairs(points, n, K);
 
    return 0;
}

Java




// Java program for the above approach
class GFG
{
 
  // Function to  find pairs whose distance between
  // the points of is an integer value.
  static void cntPairs(int [][]points, int n, int K)
  {
 
    // Stores count of pairs whose distance
    // between points is an integer
    int ans = 0;
 
    // Traverse the array, points[]
    for (int i = 0; i < n; i++)
    {
      for (int j = i + 1; j < n; j++)
      {
 
        // Stores distance between
        // points(i, j)
        int dist = 0;
 
        // Traverse all the points of
        // current pair
        for (int k = 0; k < K; k++)
        {
 
          // Update temp
          int temp = (points[i][k]
                      - points[j][k]);
 
          // Update dist
          dist += temp * temp;
        }
 
        // If dist is a perfect square
        if (Math.sqrt(dist) * Math.sqrt(dist) == dist)
        {
 
          // Update ans
          ans += 1;
        }
      }
    }
    System.out.print(ans +"\n");
  }
 
  // Driver Code
  public static void main(String[] args)
  {
 
    // Given value of K
    int K = 2;
 
    // Given points
    int [][]points
      = { { 1, 2 }, { 5, 5 }, { -2, 8 } };
 
    // Given value of N
    int n = points.length;
 
    // Function Call
    cntPairs(points, n, K);
  }
}
 
// This code is contributed by shikhasingrajput

Python3




# Python program for the above approach
 
# Function to  find pairs whose distance between
# the points of is an integer value.
def cntPairs(points, n, K):
   
    # Stores count of pairs whose distance
    # between points is an integer
    ans = 0
 
    # Traverse the array, points[]
    for i in range(0, n):
 
        for j in range(i + 1, n):
 
            # Stores distance between
            # points(i, j)
            dist = 0
 
            # Traverse all the points of
            # current pair
            for k in range(K):
 
                # Update temp
                temp = (points[i][k] - points[j][k])
 
                # Update dist
                dist += temp * temp
 
            # If dist is a perfect square
            if (((dist)**(1/2)) * ((dist)**(1/2)) == dist):
 
                # Update ans
                ans += 1
    print(ans)
 
# Driver Code
# Given value of K
K = 2
 
# Given points
points = [ [ 1, 2 ], [ 5, 5 ], [ -2, 8 ]]
 
# Given value of N
n = len(points)
 
# Function Call
cntPairs(points, n, K)
 
# This code is contributed by rohitsingh07052.

C#




// C# program for the above approach
using System;
class GFG
{
 
  // Function to  find pairs whose distance between
  // the points of is an integer value.
  static void cntPairs(int[, ] points, int n, int K)
  {
 
    // Stores count of pairs whose distance
    // between points is an integer
    int ans = 0;
 
    // Traverse the array, points[]
    for (int i = 0; i < n; i++) {
 
      for (int j = i + 1; j < n; j++) {
 
        // Stores distance between
        // points(i, j)
        int dist = 0;
 
        // Traverse all the points of
        // current pair
        for (int k = 0; k < K; k++) {
 
          // Update temp
          int temp
            = (points[i, k] - points[j, k]);
 
          // Update dist
          dist += temp * temp;
        }
 
        // If dist is a perfect square
        if (Math.Sqrt(dist) * Math.Sqrt(dist)
            == dist) {
 
          // Update ans
          ans += 1;
        }
      }
    }
 
    Console.WriteLine(ans);
  }
 
  // Driver Code
  public static void Main()
  {
 
    // Given value of K
    int K = 2;
 
    // Given points
    int[, ] points = { { 1, 2 }, { 5, 5 }, { -2, 8 } };
 
    // Given value of N
    int n = points.GetLength(0);
 
    // Function Call
    cntPairs(points, n, K);
  }
}
 
// This code is contributed by chitranayal.
Output: 
1

 

Time Complexity: O(N2 * K)
Auxiliary Space: O(1)

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