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++
// 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 pointsint 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 codeint 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
// Java implementation of the given approachimport 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
# Python3 implementation of the given approach# Function to find the point with the least sum of# distances from all other pointsdef 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 codearr = [4, 1, 5, 10, 2]n = len(arr)# Function call to find the point with the least sum of# distances from all other pointsidx = leastSumOfDistances(arr, n)# Printing the resultprint("({}, {})".format(arr[idx], 0)) |
C#
// C# implementation of the given approachusing 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)"); }} |
Output
(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++
// C++ Program to implement// the above approach#include <bits/stdc++.h>using namespace std;// Function to find median of the arrayint 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 Codeint main(){ int A[] = { 4, 1, 5, 10, 2 }; int N = sizeof(A) / sizeof(A[0]); cout << "(" << findLeastDist(A, N) << ", " << 0 << ")"; return 0;} |
Java
// Java program to implement// the above approachimport java.util.*;class GFG{// Function to find median of the arraystatic 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 Codepublic 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
# Python3 program to implement# the above approach# Function to find median of the arraydef 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 CodeA = [4, 1, 5, 10, 2];N = len(A);print("(" , findLeastDist(A, N), ", " , 0 , ")");# This code is contributed by PrinciRaj1992 |
C#
// C# program to implement// the above approachusing System;class GFG{// Function to find median of the arraystatic 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 Codepublic 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 |
Javascript
<script>// Javascript Program to implement// the above approach// Function to find median of the arrayfunction 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 Codevar A = [ 4, 1, 5, 10, 2 ];var N = A.length;document.write( "(" + findLeastDist(A, N) + ", " + 0 + ")");</script> |
Output
(4, 0)
Time Complexity: O(Nlog(N))
Auxiliary Space: O(1)
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