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.

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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# 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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