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Sum of squares of distances between all pairs from given points

  • Last Updated : 24 Nov, 2021

Given an array arr[] consisting of coordinates of N points on an XY-plane, the task is to find the sum of squared distances between all pairs of points, i.e. (Xi – Xj)2 + (Yi – Yj)2 for every distinct pair (i, j).

Examples:

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Input: arr[][] = {{1, 1}, {-1, -1}, {1, -1}, {-1, 1}}
Output: 32
Explanation: 
Distance of 1st point (1, 1) from the 2nd, 3rd and 4th points are 8, 4 and 4 respectively.
Distance of 2nd point from the 3rd and 4th points are 4 and 4 respectively.
Distance of 3rd point from the 4th point is 8.
Therefore, the total distance = (8 + 4 + 4) + (4 + 4) + (8) = 32



Input: arr[][] = {{1, 1}, {1, 1}, {0, 0}}
Output: 4
Explanation:
Distance of 1st point from the 2nd and 3rd points are 0 and 2 respectively.
Distance of 2nd point from the 3rd point is 2.
Therefore, the total distance = (0 + 2) + (2) = 4

Naive Approach: The simplest approach to solve the problem is to generate all possible distinct pairs of the given array arr[][] and calculate the sum of squares of distances between all pairs of points (Xi, Yj) and (Xj, Yj), i.e. (Xi – Xj)2 + (Yi – Yj)2, for every distinct pair (i, j)

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

Efficient Approach: To optimize the above approach, the idea is to regroup the sum and split the sum of squares of distances into two sums. Follow the steps below to solve the problem:

  • Initialize variables, say xq, yq, xs, and ys.
  • Initialize a variable, say res, with zero, to store the resultant sum.
  • Traverse the given array and for each point {x, y}, perform the following steps:
    • Add the value of (i*x2 + i*y2) in the variable res, which corresponds to adding of squared distance.
    • Add the value (xq – 2 * xs * a) and (yq – 2 * ys * b) to res to nullify the effect of the 2 * X * Y in the expansion of (a – b)2.
    • Add the values a2 and b2 to variables xq and yq respectively.
    • Add the values a and b to variables xs and ys respectively.
    • Add the values xs and yq to variables a2 and b2 respectively.
  • After completing the above steps, print the value of res as the result.

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 the sum of squares
// of distance between all distinct pairs
void findSquareSum(
    int Coordinates[][2], int N)
{
    long long xq = 0, yq = 0;
    long long xs = 0, ys = 0;
 
    // Stores final answer
    long long res = 0;
 
    // Traverse the array
    for (int i = 0; i < N; i++) {
        int a, b;
 
        a = Coordinates[i][0];
        b = Coordinates[i][1];
 
        res += xq;
        res -= 2 * xs * a;
 
        // Adding the effect of this
        // point for all the previous
        // x - points
        res += i * (long long)(a * a);
 
        // Temporarily add the
        // square of x-coordinate
        xq += a * a;
        xs += a;
        res += yq;
        res -= 2 * ys * b;
        res += i * (long long)b * b;
 
        // Add the effect of this point
        // for all the previous y - points
        yq += b * b;
        ys += b;
    }
 
    // Print the desired answer
    cout << res;
}
 
// Driver Code
int main()
{
    int arr[][2] = { { 1, 1 },
                     { -1, -1 },
                     { 1, -1 },
                     { -1, 1 } };
    int N = sizeof(arr) / sizeof(arr[0]);
    findSquareSum(arr, N);
 
    return 0;
}

Java




// Java program for the above approach
import java.io.*;
import java.util.*;
class GFG
{
 
  // Function to find the sum of squares
  // of distance between all distinct pairs
  static void findSquareSum(
    int Coordinates[][], int N)
  {
    long xq = 0, yq = 0;
    long xs = 0, ys = 0;
 
    // Stores final answer
    long res = 0;
 
    // Traverse the array
    for (int i = 0; i < N; i++) {
      int a, b;
 
      a = Coordinates[i][0];
      b = Coordinates[i][1];
 
      res += xq;
      res -= 2 * xs * a;
 
      // Adding the effect of this
      // point for all the previous
      // x - points
      res += i * (long)(a * a);
 
      // Temporarily add the
      // square of x-coordinate
      xq += a * a;
      xs += a;
      res += yq;
      res -= 2 * ys * b;
      res += i * (long)b * b;
 
      // Add the effect of this point
      // for all the previous y - points
      yq += b * b;
      ys += b;
    }
 
    // Print the desired answer
    System.out.println(res);
  }
 
 
  // Driver Code
  public static void main(String[] args)
  {
    int arr[][] = { { 1, 1 },
                   { -1, -1 },
                   { 1, -1 },
                   { -1, 1 } };
    int N = arr.length;
    findSquareSum(arr, N);
  }
}
 
// This code is contributed by code_hunt.

Python3




# Python3 program for the above approach
 
# Function to find the sum of squares
# of distance between all distinct pairs
def findSquareSum(Coordinates, N):
    xq , yq = 0, 0
    xs , ys = 0, 0
 
    # Stores final answer
    res = 0
 
    # Traverse the array
    for i in range(N):
 
        a = Coordinates[i][0]
        b = Coordinates[i][1]
 
        res += xq
        res -= 2 * xs * a
 
        # Adding the effect of this
        # point for all the previous
        # x - points
        res += i * (a * a)
 
        # Temporarily add the
        # square of x-coordinate
        xq += a * a
        xs += a
        res += yq
        res -= 2 * ys * b
        res += i * b * b
 
        # Add the effect of this point
        # for all the previous y - points
        yq += b * b
        ys += b
 
    # Print the desired answer
    print (res)
 
# Driver Code
if __name__ == '__main__':
    arr = [ [ 1, 1 ],
         [ -1, -1 ],
         [ 1, -1 ],
         [ -1, 1 ] ]
 
    N = len(arr)
 
    findSquareSum(arr, N)
 
# This code is contributed by mohit kumar 29.

C#




// C# program for the above approach
using System;
 
class GFG{
     
// Function to find the sum of squares
// of distance between all distinct pairs
static void findSquareSum(int[,] Coordinates, int N)
{
    long xq = 0, yq = 0;
    long xs = 0, ys = 0;
     
    // Stores final answer
    long res = 0;
     
    // Traverse the array
    for(int i = 0; i < N ; i++)
    {
        int a, b;
         
        a = Coordinates[i, 0];
        b = Coordinates[i, 1];
         
        res += xq;
        res -= 2 * xs * a;
         
        // Adding the effect of this
        // point for all the previous
         
        // x - points
        res += i * (long)(a * a);
         
        // Temporarily add the
        // square of x-coordinate
        xq += a * a;
        xs += a;
        res += yq;
        res -= 2 * ys * b;
        res += i * (long)b * b;
         
        // Add the effect of this point
        // for all the previous y - points
        yq += b * b;
        ys += b;
    }
     
    // Print the desired answer
    Console.Write(res);
}
 
// Driver code
static void Main()
{
    int[,] arr = { { 1, 1 },
                   { -1, -1 },
                   { 1, -1 },
                   { -1, 1 } };
    int N = arr.GetLength(0);
     
    findSquareSum(arr, N);
}
}
 
// This code is contributed by code_hunt

Javascript




<script>
 
// Javascript program for the above approach
 
  // Function to find the sum of squares
  // of distance between all distinct pairs
  function findSquareSum(
    Coordinates, N)
  {
    let xq = 0, yq = 0;
    let xs = 0, ys = 0;
 
    // Stores final answer
    let res = 0;
 
    // Traverse the array
    for (let i = 0; i < N; i++) {
      let a, b;
 
      a = Coordinates[i][0];
      b = Coordinates[i][1];
 
      res += xq;
      res -= 2 * xs * a;
 
      // Adding the effect of this
      // point for all the previous
      // x - points
      res += i * (a * a);
 
      // Temporarily add the
      // square of x-coordinate
      xq += a * a;
      xs += a;
      res += yq;
      res -= 2 * ys * b;
      res += i * b * b;
 
      // Add the effect of this point
      // for all the previous y - points
      yq += b * b;
      ys += b;
    }
 
    // Print the desired answer
    document.write(res);
  }
 
// Driver code
     
    let arr = [[ 1, 1 ],
               [ -1, -1 ],
               [ 1, -1 ],
               [ -1, 1 ]];
    let N = arr.length;
    findSquareSum(arr, N);
     
</script>
Output: 
32

 

Time Complexity: O(N)
Auxiliary Space: O(1)




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