Sort an Array of Points by their distance from a reference Point

Last Updated : 19 May, 2021

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) = 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 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).

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>

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