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) = 5Hence, 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 =
- 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++
// 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) |
Javascript
<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)