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: 
  • Sort the array of distance and print the points based on the sorted distance.
  • Below is the implementation of the above approach:



    C++

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    // 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;
    }

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    Java

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    // 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

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    Python3

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    # 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

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    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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