Minimum Sum of Euclidean Distances to all given Points
Last Updated :
13 May, 2024
Given a matrix mat[][] consisting of N pairs of the form {x, y} each denoting coordinates of N points, the task is to find the minimum sum of the Euclidean distances to all points.
Examples:
Input: mat[][] = { { 0, 1}, { 1, 0 }, { 1, 2 }, { 2, 1 }}
Output: 4
Explanation:
Average of the set of points, i.e. Centroid = ((0+1+1+2)/4, (1+0+2+1)/4) = (1, 1).
Euclidean distance of each point from the centroid are {1, 1, 1, 1}
Sum of all distances = 1 + 1 + 1 + 1 = 4
Input: mat[][] = { { 1, 1}, { 3, 3 }}
Output: 2.82843
Approach:
Since the task is to minimize the Euclidean Distance to all points, the idea is to calculate the Median of all the points. Geometric Median generalizes the concept of median to higher dimensions
Follow the steps below to solve the problem:
- Calculate the centroid of all the given coordinates, by getting the average of the points.
- Find the Euclidean distance of all points from the centroid.
- Calculate the sum of these distance and print as the answer.
Below is the implementation of above approach:
C++
// C++ Program to implement
// the above approach
#include <bits/stdc++.h>
using namespace std;
// Function to calculate Euclidean distance
double find(double x, double y,
vector<vector<int> >& p)
{
double mind = 0;
for (int i = 0; i < p.size(); i++) {
double a = p[i][0], b = p[i][1];
mind += sqrt((x - a) * (x - a)
+ (y - b) * (y - b));
}
return mind;
}
// Function to calculate the minimum sum
// of the euclidean distances to all points
double getMinDistSum(vector<vector<int> >& p)
{
// Calculate the centroid
double x = 0, y = 0;
for (int i = 0; i < p.size(); i++) {
x += p[i][0];
y += p[i][1];
}
x = x / p.size();
y = y / p.size();
// Calculate distance of all
// points
double mind = find(x, y, p);
return mind;
}
// Driver Code
int main()
{
// Initializing the points
vector<vector<int> > vec
= { { 0, 1 }, { 1, 0 }, { 1, 2 }, { 2, 1 } };
double d = getMinDistSum(vec);
cout << d << endl;
return 0;
}
// Java program to implement
// the above approach
class GFG{
// Function to calculate Euclidean distance
static double find(double x, double y,
int [][] p)
{
double mind = 0;
for(int i = 0; i < p.length; i++)
{
double a = p[i][0], b = p[i][1];
mind += Math.sqrt((x - a) * (x - a) +
(y - b) * (y - b));
}
return mind;
}
// Function to calculate the minimum sum
// of the euclidean distances to all points
static double getMinDistSum(int [][]p)
{
// Calculate the centroid
double x = 0, y = 0;
for(int i = 0; i < p.length; i++)
{
x += p[i][0];
y += p[i][1];
}
x = x / p.length;
y = y / p.length;
// Calculate distance of all
// points
double mind = find(x, y, p);
return mind;
}
// Driver Code
public static void main(String[] args)
{
// Initializing the points
int [][]vec = { { 0, 1 }, { 1, 0 },
{ 1, 2 }, { 2, 1 } };
double d = getMinDistSum(vec);
System.out.print(d + "\n");
}
}
// This code is contributed by Amit Katiyar
# Python3 program to implement
# the above approach
from math import sqrt
# Function to calculate Euclidean distance
def find(x, y, p):
mind = 0
for i in range(len(p)):
a = p[i][0]
b = p[i][1]
mind += sqrt((x - a) * (x - a) +
(y - b) * (y - b))
return mind
# Function to calculate the minimum sum
# of the euclidean distances to all points
def getMinDistSum(p):
# Calculate the centroid
x = 0
y = 0
for i in range(len(p)):
x += p[i][0]
y += p[i][1]
x = x // len(p)
y = y // len(p)
# Calculate distance of all
# points
mind = find(x, y, p)
return mind
# Driver Code
if __name__ == '__main__':
# Initializing the points
vec = [ [ 0, 1 ], [ 1, 0 ],
[ 1, 2 ], [ 2, 1 ] ]
d = getMinDistSum(vec)
print(int(d))
# This code is contributed by mohit kumar 29
// C# program to implement
// the above approach
using System;
class GFG{
// Function to calculate Euclidean distance
static double find(double x, double y,
int [,] p)
{
double mind = 0;
for(int i = 0; i < p.GetLength(0); i++)
{
double a = p[i,0], b = p[i,1];
mind += Math.Sqrt((x - a) * (x - a) +
(y - b) * (y - b));
}
return mind;
}
// Function to calculate the minimum sum
// of the euclidean distances to all points
static double getMinDistSum(int [,]p)
{
// Calculate the centroid
double x = 0, y = 0;
for(int i = 0; i < p.GetLength(0); i++)
{
x += p[i,0];
y += p[i,1];
}
x = x / p.Length;
y = y / p.Length;
// Calculate distance of all
// points
double mind = find(x, y, p);
return mind;
}
// Driver Code
public static void Main(String[] args)
{
// Initializing the points
int [,]vec = { { 0, 1 }, { 1, 0 },
{ 1, 2 }, { 2, 1 } };
int d = (int)getMinDistSum(vec);
Console.Write(d + "\n");
}
}
// This code is contributed by Rohit_ranjan
<script>
// JavaScript program for the above approach
// Function to calculate Euclidean distance
function find(x, y, p)
{
let mind = 0;
for(let i = 0; i < p.length; i++)
{
let a = p[i][0], b = p[i][1];
mind += Math.sqrt((x - a) * (x - a) +
(y - b) * (y - b));
}
return mind;
}
// Function to calculate the minimum sum
// of the euclidean distances to all points
function getMinDistSum(p)
{
// Calculate the centroid
let x = 0, y = 0;
for(let i = 0; i < p.length; i++)
{
x += p[i][0];
y += p[i][1];
}
x = x / p.length;
y = y / p.length;
// Calculate distance of all
// points
let mind = find(x, y, p);
return mind;
}
// Driver Code
// Initializing the points
let vec = [[ 0, 1 ], [ 1, 0 ],
[ 1, 2 ], [ 2, 1 ]];
let d = getMinDistSum(vec);
document.write(d);
</script>
Time Complexity: O(N)
Auxiliary Space: O(1)
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