Sort an Array of Points by their distance from a reference Point

Given an array arr[] containing N points and a reference point P, the task is to sort these points according to their distance from the given point P.

Examples:

Input: arr[] = {{5, 0}, {4, 0}, {3, 0}, {2, 0}, {1, 0}}, P = (0, 0) 
Output: (1, 0) (2, 0) (3, 0) (4, 0) (5, 0) 
Explanation: 
Distance between (0, 0) and (1, 0) = 1 
Distance between (0, 0) and (2, 0) = 2 
Distance between (0, 0) and (3, 0) = 3 
Distance between (0, 0) and (4, 0) = 4 
Distance between (0, 0) and (5, 0) = 5 

Hence, the sorted array of points will be: {(1, 0) (2, 0) (3, 0) (4, 0) (5, 0)}
Input: arr[] = {{5, 0}, {0, 4}, {0, 3}, {2, 0}, {1, 0}}, P = (0, 0) 
Output: (1, 0) (2, 0) (0, 3) (0, 4) (5, 0) 
Explanation: 
Distance between (0, 0) and (1, 0) = 1 
Distance between (0, 0) and (2, 0) = 2 
Distance between (0, 0) and (0, 3) = 3 
Distance between (0, 0) and (0, 4) = 4 
Distance between (0, 0) and (5, 0) = 5 
Hence, the sorted array of points will be: {(1, 0) (2, 0) (0, 3) (0, 4) (5, 0)}

Approach: The idea is to store each element at its distance from the given point P in a pair and then sort all the elements of the vector according to the distance stored.

• For each of the given points: 
  
  • Find the distance of the point from the reference point P formula below:

Distance = \sqrt{(p2-x1)^{2} + (p2-y1)^{2}}

• Append the distance in an array
• Sort the array of distance and print the points based on the sorted distance.
  • Time Complexity: As in the above approach, there is sorting of an array of length N, which takes O(N*logN) time in the worst case. Hence, the Time Complexity will be O(N*log N).
  • Auxiliary Space Complexity: As in the above approach, there is extra space used to store the distance and the points as pairs. Hence, the auxiliary space complexity will be O(N).

// C++ program

#include <bits/stdc++.h>
using namespace std;

bool compare(pair<int, pair<int, int> > a,
             pair<int, pair<int, int> > b)
{
    if (a.first == b.first) {
        return 0;
    }
    else {
        return (a.first < b.first) ? -1 : 1;
    }
}

// Function to sort the array of
// points by its distance from P
void sortArr(vector<vector<int> > arr, int n, vector<int> p)
{
    // Vector to store the distance
    // with respective elements

    vector<pair<int, pair<int, int> > > vp;
    // Storing the distance with its
    // distance in the vector array
    for (int i = 0; i < n; i++) {

        int dist = pow((p[0] - arr[i][0]), 2)
                   + pow((p[1] - arr[i][1]), 2);
        vp.push_back(make_pair(
            dist, make_pair(arr[i][0], arr[i][1])));
    }

    // Sorting the array with
    // respect to its distance
    sort(vp.begin(), vp.end(), compare);

    // Output
    for (int i = 0; i < n; i++) {
        cout << "(" << vp[i].second.first << ", "
             << vp[i].second.second << ") " << endl;
    }
}

int main()
{
    vector<vector<int> > arr
        = { { 5, 5 }, { 6, 6 }, { 1, 0 },
            { 2, 0 }, { 3, 1 }, { 1, -2 } };
    int n = 6;
    vector<int> p = { 0, 0 };
  
    // Function to perform sorting
    sortArr(arr, n, p);
}

// The code is contributed by Gautam goel (gautamgoel962)

// Java program to implement the approach
import java.util.*;

class GFG
{

  // Function to sort the array of
  // points by its distance from P
  static void sortArr(int[][] arr, int n, int[] p)
  {

    // List to store the distance
    // with respective elements

    ArrayList<ArrayList<Integer>> vp = new ArrayList<ArrayList<Integer>>();
    // Storing the distance with its
    // distance in the List array
    for (int i = 0; i < n; i++) {

      int dist = (int)Math.pow((p[0] - arr[i][0]), 2)
        + (int)Math.pow((p[1] - arr[i][1]), 2);
      ArrayList<Integer> l1 = new ArrayList<Integer> ();
      l1.add(dist);
      l1.add(arr[i][0]);
      l1.add(arr[i][1]);

      vp.add(l1);
    }

    // Sorting the array with
    // respect to its distance

    Collections.sort(vp, new Comparator<ArrayList<Integer>> () {
      @Override
      public int compare(ArrayList<Integer> a, ArrayList<Integer> b) {
        return a.get(0).compareTo(b.get(0));
      }
    });


    // Output
    for (int i = 0; i < n; i++) {
      System.out.print("(" + vp.get(i).get(1) + ", "
                       + vp.get(i).get(2) +") ");
    }
  }

  public static void main(String[] args)
  {
    int[][] arr = { { 5, 5 }, { 6, 6 }, { 1, 0 },
                   { 2, 0 }, { 3, 1 }, { 1, -2 } };
    int n = 6;
    int[] p = { 0, 0 };

    // Function to perform sorting
    sortArr(arr, n, p);
  }
}

// The code is contributed by phasing17

# Python3 program code to implement the approach
import functools

def sortFunction(a, b) :
    if (a[0] == b[0]): 
        return 0;
    
    else :
        if (a[0] < b[0]):
            return -1
        return 1;
    
# Function to sort the array of
# points by its distance from P
def sortArr(arr, n, p):

    # Vector to store the distance
    # with respective elements
    
    vp = [0 for _ in range(n)]
    # Storing the distance with its
    # distance in the vector array
    for i in range(n):
  
        dist = pow((p[0] - arr[i][0]), 2) + pow((p[1] - arr[i][1]), 2);
        vp[i] = [dist, [arr[i][0], arr[i][1]]];
    
    # Sorting the array with
    # respect to its distance
    vp.sort(key=functools.cmp_to_key(sortFunction));
  
    # Output
    for i in range(n):
        print("(", vp[i][1][0], ", ", vp[i][1][1], ") ", sep = "", end = " ");
    
arr = [[ 5, 5 ], [ 6, 6 ], [ 1, 0], [ 2, 0 ], [ 3, 1 ], [ 1, -2 ]];
n = 6;
p = [ 0, 0 ];

# Function to perform sorting
sortArr(arr, n, p);

# This code is contributed by phasing17

// C# program to implement the approach
using System;
using System.Linq;
using System.Collections.Generic;
class GFG
{

  // Function to sort the array of
  // points by its distance from P
  static void sortArr(int[, ] arr, int n, int[] p)
  {

    // List to store the distance
    // with respective elements

    List<List<int>> vp = new List<List<int>>();
    // Storing the distance with its
    // distance in the List array
    for (int i = 0; i < n; i++) {

      int dist = (int)Math.Pow((p[0] - arr[i, 0]), 2)
        + (int)Math.Pow((p[1] - arr[i, 1]), 2);
      List<int> l1 = new List<int>();
      l1.Add(dist);
      l1.Add(arr[i, 0]);
      l1.Add(arr[i, 1]);

      vp.Add(l1);
    }

    // Sorting the array with
    // respect to its distance
    vp = vp.OrderBy(ele => ele[0]).ToList();

    // Output
    for (int i = 0; i < n; i++) {
      Console.Write("(" + vp[i][1] + ", "
                    + vp[i][2] +") ");
    }
  }

  public static void Main(string[] args)
  {
    int[, ] arr = new [,] { { 5, 5 }, { 6, 6 }, { 1, 0 },
                           { 2, 0 }, { 3, 1 }, { 1, -2 } };
    int n = 6;
    int[] p = { 0, 0 };

    // Function to perform sorting
    sortArr(arr, n, p);
  }
}

// The code is contributed by phasing17

<script>
// Javascript program

function sortFunction(a, b) {
    if (a[0] === b[0]) {
        return 0;
    }
    else {
        return (a[0] < b[0]) ? -1 : 1;
    }
}

// Function to sort the array of
// points by its distance from P
function sortArr(arr, n, p)
{
    // Vector to store the distance
    // with respective elements
    
    var vp = new Array(n);
    // Storing the distance with its
    // distance in the vector array
    for (var i = 0; i < n; i++) {
  
        var dist = Math.pow((p[0] - arr[i][0]), 2)
              + Math.pow((p[1] - arr[i][1]), 2);
        vp[i] = [dist, [arr[i][0], arr[i][1]]];
    }
    
    // Sorting the array with
    // respect to its distance
    vp.sort(sortFunction);
  
    // Output
    for (var i = 0; i < n; i++) {
        document.write("(" + vp[i][1][0] + ", " + vp[i][1][1] + ") ");
    }
}

var arr = [[ 5, 5 ], [ 6, 6 ], [ 1, 0], [ 2, 0 ], [ 3, 1 ], [ 1, -2 ]];
var n = 6;
var p = [ 0, 0 ];
// Function to perform sorting
sortArr(arr, n, p);
</script>

Output:

(1, 0) (2, 0) (1, -2) (3, 1) (5, 5) (6, 6) 

spp____

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Article Tags :

• Mathematical
• Geometric
• DSA
• Arrays
• cpp-pair

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Practice Tags :

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