Given an array arr[] consisting of N integers, denoting N points lying on the X-axis, the task is to find the point which has the least sum of distances from all other points.
Example:
Input: arr[] = {4, 1, 5, 10, 2}
Output: (4, 0)
Explanation:
Distance of 4 from rest of the elements = |4 – 1| + |4 – 5| + |4 – 10| + |4 – 2| = 12
Distance of 1 from rest of the elements = |1 – 4| + |1 – 5| + |1 – 10| + |1 – 2| = 17
Distance of 5 from rest of the elements = |5 – 1| + |5 – 4| + |5 – 2| + |5 – 10| = 13
Distance of 10 from rest of the elements = |10 – 1| + |10 – 2| + |10 – 5| + |10 – 4| = 28
Distance of 2 from rest of the elements = |2 – 1| + |2 – 4| + |2 – 5| + |2 – 10| = 14
Input: arr[] = {3, 5, 7, 10}
Output: (5, 0)
Naive Approach:
The task is to iterate over the array, and for each array element, calculate the sum of its absolute difference with all other array elements. Finally, print the array element with the maximum sum of differences.
Below are the steps to implement the above approach:
- Initialize the minimum sum of distances with infinity.
- Initialize the index of the element with minimum sum of distances as -1.
- Iterate over each point in the array.
- For the current point, calculate the sum of distances from all other points in the array.
- If the sum of distances for the current point is less than the minimum sum of distances, update the minimum sum and the index of the point with the minimum sum.
- Return the index of the point with minimum sum of distances.
- Print the point with the minimum sum of distances as (point, 0).
Below is the code to implement the above approach:
// C++ implementation of the given approach #include <bits/stdc++.h> using namespace std;
// Function to find the point with the least sum of // distances from all other points int leastSumOfDistances( int arr[], int n) {
// Initialize the minimum sum of distances with infinity
int minSum = INT_MAX;
// Initialize the index of the element with minimum sum
// of distances as -1
int idx = -1;
// Iterate over each point
for ( int i = 0; i < n; i++) {
// Initialize the sum of distances for the current
// point to zero
int sum = 0;
// Calculate the sum of distances of the current
// point from all other points
for ( int j = 0; j < n; j++) {
sum += abs (arr[j] - arr[i]);
}
// If the sum of distances for the current point is
// less than the minimum sum of distances, update
// the minimum sum and the index of the point with
// minimum sum
if (sum < minSum) {
minSum = sum;
idx = i;
}
}
// Return the index of the point with minimum sum of
// distances
return idx;
} // Driver code int main()
{ int arr[] = { 4, 1, 5, 10, 2 };
int n = sizeof (arr) / sizeof ( int );
// Function call to find the point with the least sum of
// distances from all other points
int idx = leastSumOfDistances(arr, n);
// Printing the result
cout << "("
<< arr[idx] << ", " << 0 << ")" ;
return 0;
} |
// Java implementation of the given approach import java.util.*;
public class GFG {
// Function to find the point with the least sum of
// distances from all other points
public static int leastSumOfDistances( int [] arr, int n)
{
// Initialize the minimum sum of distances with
// infinity
int minSum = Integer.MAX_VALUE;
// Initialize the index of the element with minimum
// sum of distances as -1
int idx = - 1 ;
// Iterate over each point
for ( int i = 0 ; i < n; i++) {
// Initialize the sum of distances for the
// current point to zero
int sum = 0 ;
// Calculate the sum of distances of the current
// point from all other points
for ( int j = 0 ; j < n; j++) {
sum += Math.abs(arr[j] - arr[i]);
}
// If the sum of distances for the current point
// is less than the minimum sum of distances,
// update the minimum sum and the index of the
// point with minimum sum
if (sum < minSum) {
minSum = sum;
idx = i;
}
}
// Return the index of the point with minimum sum of
// distances
return idx;
}
// Driver code
public static void main(String[] args)
{
int [] arr = { 4 , 1 , 5 , 10 , 2 };
int n = arr.length;
// Function call to find the point with the least
// sum of distances from all other points
int idx = leastSumOfDistances(arr, n);
// Printing the result
System.out.print( "(" + arr[idx] + ", " + 0 + ")" );
}
} |
# Python3 implementation of the given approach # Function to find the point with the least sum of # distances from all other points def leastSumOfDistances(arr, n):
# Initialize the minimum sum of distances with infinity
minSum = float ( 'inf' )
# Initialize the index of the element with minimum sum
# of distances as -1
idx = - 1
# Iterate over each point
for i in range (n):
# Initialize the sum of distances for the current
# point to zero
sum = 0
# Calculate the sum of distances of the current
# point from all other points
for j in range (n):
sum + = abs (arr[j] - arr[i])
# If the sum of distances for the current point is
# less than the minimum sum of distances, update
# the minimum sum and the index of the point with
# minimum sum
if sum < minSum:
minSum = sum
idx = i
# Return the index of the point with minimum sum of
# distances
return idx
# Driver code arr = [ 4 , 1 , 5 , 10 , 2 ]
n = len (arr)
# Function call to find the point with the least sum of # distances from all other points idx = leastSumOfDistances(arr, n)
# Printing the result print ( "({}, {})" . format (arr[idx], 0 ))
|
// C# implementation of the given approach using System;
class Program
{ // Function to find the point with the least sum of distances from all other points
static int LeastSumOfDistances( int [] arr, int n)
{
// Initialize the minimum sum of distances with infinity
int minSum = int .MaxValue;
// Initialize the index of the element with minimum sum of distances as -1
int idx = -1;
// Iterate over each point
for ( int i = 0; i < n; i++)
{
// Initialize the sum of distances for the current point to zero
int sum = 0;
// Calculate the sum of distances of the current point from all other points
for ( int j = 0; j < n; j++)
{
sum += Math.Abs(arr[j] - arr[i]);
}
// If the sum of distances for the current point is less than the minimum sum of distances,
// update the minimum sum and the index of the point with minimum sum
if (sum < minSum)
{
minSum = sum;
idx = i;
}
}
// Return the index of the point with minimum sum of distances
return idx;
}
static void Main()
{
int [] arr = { 4, 1, 5, 10, 2 };
int n = arr.Length;
// Function call to find the point with the least sum of distances from all other points
int idx = LeastSumOfDistances(arr, n);
// Printing the result
Console.WriteLine($ "({arr[idx]}, 0)" );
}
} |
// Function to find the point with the least sum of distances from all other points function leastSumOfDistances(arr) {
// Initialize the minimum sum of distances with infinity
let minSum = Infinity;
// Initialize the index of the element with minimum sum of distances as -1
let idx = -1;
// Iterate over each point
for (let i = 0; i < arr.length; i++) {
// Initialize the sum of distances for the current point to zero
let sum = 0;
// Calculate the sum of distances of the current point from all other points
for (let j = 0; j < arr.length; j++) {
sum += Math.abs(arr[j] - arr[i]);
}
// If the sum of distances for the current point is less than the minimum sum of distances, update the minimum sum and the index of the point with minimum sum
if (sum < minSum) {
minSum = sum;
idx = i;
}
}
// Return the index of the point with the minimum sum of distances
return idx;
} // Driver code const arr = [4, 1, 5, 10, 2]; // Function call to find the point with the least sum of distances from all other points const idx = leastSumOfDistances(arr); // Printing the result console.log(`(${arr[idx]}, 0)`); |
(4, 0)
Time Complexity: O(N2)
Auxiliary Space: O(1)
Efficient Approach: To optimize the above approach, the idea is to find the median of the array. The median of the array will have the least possible total distance from other elements in the array. For an array with an even number of elements, there are two possible medians and both will have the same total distance, return the one with the lower index since it is closer to origin.
Follow the below steps to solve the problem:
- Sort the given array.
- If N is odd, return the (N + 1 / 2)th element.
- Otherwise, return the (N / 2)th element.
Below is the implementation of the above approach:
// C++ Program to implement // the above approach #include <bits/stdc++.h> using namespace std;
// Function to find median of the array int findLeastDist( int A[], int N)
{ // Sort the given array
sort(A, A + N);
// If number of elements are even
if (N % 2 == 0) {
// Return the first median
return A[(N - 1) / 2];
}
// Otherwise
else {
return A[N / 2];
}
} // Driver Code int main()
{ int A[] = { 4, 1, 5, 10, 2 };
int N = sizeof (A) / sizeof (A[0]);
cout << "(" << findLeastDist(A, N)
<< ", " << 0 << ")" ;
return 0;
} |
// Java program to implement // the above approach import java.util.*;
class GFG{
// Function to find median of the array static int findLeastDist( int A[], int N)
{ // Sort the given array
Arrays.sort(A);
// If number of elements are even
if (N % 2 == 0 )
{
// Return the first median
return A[(N - 1 ) / 2 ];
}
// Otherwise
else
{
return A[N / 2 ];
}
} // Driver Code public static void main(String[] args)
{ int A[] = { 4 , 1 , 5 , 10 , 2 };
int N = A.length;
System.out.print( "(" + findLeastDist(A, N) +
", " + 0 + ")" );
} } // This code is contributed by PrinciRaj1992 |
# Python3 program to implement # the above approach # Function to find median of the array def findLeastDist(A, N):
# Sort the given array
A.sort();
# If number of elements are even
if (N % 2 = = 0 ):
# Return the first median
return A[(N - 1 ) / / 2 ];
# Otherwise
else :
return A[N / / 2 ];
# Driver Code A = [ 4 , 1 , 5 , 10 , 2 ];
N = len (A);
print ( "(" , findLeastDist(A, N),
", " , 0 , ")" );
# This code is contributed by PrinciRaj1992 |
// C# program to implement // the above approach using System;
class GFG{
// Function to find median of the array static int findLeastDist( int []A, int N)
{ // Sort the given array
Array.Sort(A);
// If number of elements are even
if (N % 2 == 0)
{
// Return the first median
return A[(N - 1) / 2];
}
// Otherwise
else
{
return A[N / 2];
}
} // Driver Code public static void Main( string [] args)
{ int []A = { 4, 1, 5, 10, 2 };
int N = A.Length;
Console.Write( "(" + findLeastDist(A, N) +
", " + 0 + ")" );
} } // This code is contributed by rutvik_56 |
<script> // Javascript Program to implement // the above approach // Function to find median of the array function findLeastDist(A, N)
{ // Sort the given array
A.sort((a,b) => a-b);
console.log(A);
// If number of elements are even
if ((N % 2) == 0) {
// Return the first median
return A[parseInt((N - 1) / 2)];
}
// Otherwise
else {
return A[parseInt(N / 2)];
}
} // Driver Code var A = [ 4, 1, 5, 10, 2 ];
var N = A.length;
document.write( "(" + findLeastDist(A, N)
+ ", " + 0 + ")" );
</script> |
(4, 0)
Time Complexity: O(Nlog(N))
Auxiliary Space: O(1)