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 it’s 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 with 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 point:
- Find the distance of the point from the reference point P using the below formulae:
```
Distance = 
```
- Append the distance in an array
Sort the array of distance and print the points based on the sorted distance.

Below is the implementation of the above approach:

C++

// C++ implementation to sort the 
// array of points by its distance 
// from the given point 
  
#include <bits/stdc++.h> 
using namespace std; 
  
// Function to sort the array of 
// points by its distance from P 
void sortArr(vector<pair<int, int> > arr, 
             int n, pair<int, 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.first - arr[i].first), 2) 
              + pow((p.second - arr[i].second), 2); 
  
        vp.push_back(make_pair( 
            dist, 
            make_pair( 
                arr[i].first, 
                arr[i].second))); 
    } 
  
    // Sorting the array with 
    // respect to its distance 
    sort(vp.begin(), vp.end()); 
  
    // Output 
    for (int i = 0; i < vp.size(); i++) { 
        cout << "("
             << vp[i].second.first << ", "
             << vp[i].second.second << ") "; 
    } 
} 
  
// Driver code 
int main() 
{ 
    // Array of points 
    vector<pair<int, int> > arr 
        = { { 5, 5 }, { 6, 6 }, { 1, 0 }, { 2, 0 }, { 3, 1 }, { 1, -2 } }; 
    int n = 6; 
    pair<int, int> p = { 0, 0 }; 
  
    // Sorting Array 
    sortArr(arr, n, p); 
    return 0; 
} 

Java

// Java implementation to sort the 
// array of points by its distance 
// from the given point 
import java.util.*; 
  
class Pair<K, V>  
{ 
    K first; 
    V second; 
    Pair(K a, V b) 
    { 
        first = a; 
        second = b; 
    } 
} 
  
class GFG{ 
      
// Function to sort the array of 
// points by its distance from P 
static void sortArr(ArrayList<Pair<Integer, 
                                   Integer>> arr,  
              int n, Pair<Integer, Integer> p) 
{ 
  
    // Vector to store the distance 
    // with respective elements 
    ArrayList<Pair<Integer, 
              Pair<Integer,  
                   Integer>>> vp = new ArrayList<>(); 
  
    // Storing the distance with its 
    // distance in the vector array 
    for(int i = 0; i < n; i++) 
    { 
        int dist = (int)Math.pow( 
                   (p.first - arr.get(i).first), 2) + 
                   (int)Math.pow( 
                   (p.second - arr.get(i).second), 2); 
  
        vp.add( 
            new Pair<Integer, Pair<Integer, Integer>>( 
                    dist, new Pair<Integer, Integer>( 
                        arr.get(i).first, 
                        arr.get(i).second))); 
    } 
  
    // Sorting the array with 
    // respect to its distance 
    Collections.sort(vp, (a, b) -> a.first - b.first); 
  
    // Output 
    for(int i = 0; i < vp.size(); i++)  
    { 
        System.out.print("(" +  
        vp.get(i).second.first + ", " +  
        vp.get(i).second.second + ") "); 
    } 
} 
  
// Driver code 
public static void main(String[] args) 
{ 
      
    // Array of points 
    int a[][] = { { 5, 5 }, { 6, 6 }, 
                  { 1, 0 }, { 2, 0 },  
                  { 3, 1 }, { 1, -2 } }; 
                    
    ArrayList<Pair<Integer,  
                   Integer>> arr = new ArrayList<>(); 
                     
    for(int i = 0; i < a.length; i++) 
        arr.add(new Pair<Integer,  
                         Integer>(a[i][0], 
                                  a[i][1])); 
    int n = 6; 
    Pair<Integer,  
         Integer> p = new Pair<Integer,  
                               Integer>(0, 0); 
  
    // Sorting Array 
    sortArr(arr, n, p); 
} 
} 
  
// This code is contributed by jrishabh99 

Python3

# Python3 implementation to sort the 
# array of points by its distance 
# from the given point 
  
# 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 = [] 
      
    # 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.append([dist,[arr[i][0],arr[i][1]]]) 
          
    # Sorting the array with 
    # respect to its distance 
    vp.sort() 
      
    # Output 
    for i in range(len(vp)): 
        print("(",vp[i][1][0],", ",vp[i][1][1], ") ",sep="",end="") 
      
# Driver code 
arr = [[5, 5] , [6, 6] , [ 1, 0] , [2, 0] , [3, 1] , [1, -2]]  
n = 6
p = [0, 0]  
  
# Sorting Array 
sortArr(arr, n, p) 
  
# This code is contributed by shivanisinghss2110 

Output:

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

Performance Analysis:

Time Complexity: As in the above approach, there is sorting of an array of length N, which takes O(N*logN) time in 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 pair. Hence the auxiliary space complexity will be O(N).

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My Personal Notes

C++

Java

Python3

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