# 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)}

## Recommended: Please try your approach on {IDE} first, before moving on to the solution.

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

## 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 - arr[i]), 2)+ pow((p - arr[i]), 2)                     vp.append([dist,[arr[i],arr[i]]])                 # Sorting the array with      # respect to its distance      vp.sort()             # Output      for i in range(len(vp)):          print("(",vp[i],", ",vp[i], ") ",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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Improved By : shivanisinghss2110