# Find the maximum possible distance from origin using given points

Last Updated : 13 Sep, 2022

Given N 2-Dimensional points. The task is to find the maximum possible distance from the origin using given points. Using the ith point (xi, yi) one can move from (a, b) to (a + xi, b + yi)
Note: N lies between 1 to 1000 and each point can be used at most once.
Examples:

Input: arr[][] = {{1, 1}, {2, 2}, {3, 3}, {4, 4}}
Output: 14.14
The farthest point we can move to is (10, 10).
Input: arr[][] = {{0, 10}, {5, -5}, {-5, -5}}
Output: 10.00

Approach: The key observation is that when the points are ordered by the angles their vectors make with the x-axis, the answer will include vectors in some contiguous range. A proof of this fact can be read from here. Then, the solution is fairly easy to implement. Iterate over all possible ranges and compute the answers for each of them, taking the maximum as the result. When implemented appropriately, this is an O(N2) approach.
Below is the implementation of the above approach:

## CPP

 `// C++ implementation of the approach` `#include ` `using` `namespace` `std;` ` `  `// Function to find the maximum possible` `// distance from origin using given points.` `void` `Max_Distance(vector >& xy, ``int` `n)` `{` `    ``// Sort the points with their tan angle` `    ``sort(xy.begin(), xy.end(), [](``const` `pair<``int``, ``int``>&l,` `                                  ``const` `pair<``int``, ``int``>& r) {` `        ``return` `atan2l(l.second, l.first)` `               ``< atan2l(r.second, r.first);` `    ``});` ` `  `    ``// Push the whole vector` `    ``for` `(``int` `i = 0; i < n; i++)` `        ``xy.push_back(xy[i]);` ` `  `    ``// To store the required answer` `    ``int` `res = 0;` ` `  `    ``// Find the maximum possible answer` `    ``for` `(``int` `i = 0; i< n; i++) {` `        ``int` `x = 0, y = 0;` `        ``for` `(``int` `j = i; j > vec = { { 1, 1 },` `                                    ``{ 2, 2 },` `                                    ``{ 3, 3 },` `                                    ``{ 4, 4 } };` ` `  `    ``int` `n = vec.size();` ` `  `    ``// Function call` `    ``Max_Distance(vec, n);` ` `  `    ``return` `0;` `}`

## Java

 `// Java implementation of the approach`   `import` `java.util.*;`   `class` `GFG {`   `    ``// Function to find the maximum possible` `    ``// distance from origin using given points.` `    ``static` `void` `    ``Max_Distance(ArrayList > xy, ``int` `n)` `    ``{`   `        ``// Sort the points with their tan angle` `        ``Collections.sort(` `            ``xy, ``new` `Comparator >() {` `                ``@Override` `                ``public` `int` `compare(ArrayList x,` `                                   ``ArrayList y)` `                ``{` `                    ``return` `(``int``)((Math.atan2(x.get(``1``),` `                                             ``x.get(``0``)))` `                                 ``- (Math.atan2(y.get(``1``),` `                                               ``y.get(``0``))));` `                ``}` `            ``});`   `        ``// Push the whole vector` `        ``for` `(``int` `i = ``0``; i < n; i++)` `            ``xy.add(xy.get(i));`   `        ``// To store the required answer` `        ``int` `res = ``0``;`   `        ``// Find the maximum possible answer` `        ``for` `(``int` `i = ``0``; i < n; i++) {` `            ``int` `x = ``0``, y = ``0``;` `            ``for` `(``int` `j = i; j < i + n; j++) {` `                ``x += xy.get(j).get(``0``);` `                ``y += xy.get(j).get(``1``);` `                ``res = Math.max(res, x * x + y * y);` `            ``}` `        ``}`   `        ``// Print the required answer` `        ``System.out.println(` `            ``(``double``)Math.round(Math.sqrt(res) * ``100``) / ``100``);` `    ``}`   `    ``// Driver code` `    ``public` `static` `void` `main(String[] args)` `    ``{` `        ``ArrayList > vec` `            ``= ``new` `ArrayList >();`   `        ``ArrayList a1 = ``new` `ArrayList();` `        ``a1.add(``1``);` `        ``a1.add(``1``);` `        ``vec.add(a1);`   `        ``ArrayList a2 = ``new` `ArrayList();` `        ``a2.add(``2``);` `        ``a2.add(``2``);` `        ``vec.add(a2);`   `        ``ArrayList a3 = ``new` `ArrayList();` `        ``a3.add(``3``);` `        ``a3.add(``3``);` `        ``vec.add(a3);`   `        ``ArrayList a4 = ``new` `ArrayList();` `        ``a4.add(``4``);` `        ``a4.add(``4``);` `        ``vec.add(a4);`   `        ``int` `n = ``4``;`   `        ``// Function call` `        ``Max_Distance(vec, n);` `    ``}` `}`   `// This code is contributed by phasing17`

## Python3

 `# Python3 implementation of the approach` `from` `math ``import` `*`   `# Function to implement the custom sort ` `def` `myCustomSort(l):` `    ``return` `atan2(l[``1``], l[``0``]);`   `# Function to find the maximum possible` `# distance from origin using given points.` `def` `Max_Distance(xy, n):`   `    ``# Sort the points with their tan angle` `    ``xy.sort(key ``=` `myCustomSort);`   `    ``# Push the whole vector` `    ``xy ``+``=` `xy`   `    ``# To store the required answer` `    ``res ``=` `0``;`   `    ``# Find the maximum possible answer` `    ``for` `i ``in` `range``(n):` `        ``x ``=` `0` `        ``y ``=` `0` `        ``for` `j ``in` `range``(i, i ``+` `n):` `            ``x ``+``=` `xy[j][``0``];` `            ``y ``+``=` `xy[j][``1``];` `            ``res ``=` `max``(res, x ``*` `x ``+` `y ``*` `y);` `        `  `    ``# Print the required answer` `    ``print``(``round``(res ``*``*` `0.5``, ``2``)) `   `# Driver code` `vec ``=` `[[``1``, ``1``], [``2``, ``2``], [``3``, ``3``], [``4``, ``4``]];` `n ``=` `len``(vec)`   `# Function call` `Max_Distance(vec, n);`   `# The code is contributed by phasing17`

## C#

 `// C# implementation of the approach` `using` `System;` `using` `System.Linq;` `using` `System.Collections.Generic;`   `class` `GFG` `{`   `  ``// Function to find the maximum possible` `  ``// distance from origin using given points.` `  ``static` `void` `Max_Distance(List<``int``[]> xy, ``int` `n)` `  ``{`   `    ``// Sort the points with their tan angle` `    ``xy = xy.OrderBy(x => Math.Atan2(x[1], x[0]))` `      ``.ToList();`   `    ``// Push the whole vector` `    ``for` `(``int` `i = 0; i < n; i++)` `      ``xy.Add(xy[i]);`   `    ``// To store the required answer` `    ``int` `res = 0;`   `    ``// Find the maximum possible answer` `    ``for` `(``int` `i = 0; i < n; i++) {` `      ``int` `x = 0, y = 0;` `      ``for` `(``int` `j = i; j < i + n; j++) {` `        ``x += xy[j][0];` `        ``y += xy[j][1];` `        ``res = Math.Max(res, x * x + y * y);` `      ``}` `    ``}`   `    ``// Print the required answer` `    ``Console.WriteLine(Math.Round(Math.Sqrt(res), 2));` `  ``}`   `  ``// Driver code` `  ``public` `static` `void` `Main(``string``[] args)` `  ``{` `    ``List<``int``[]> vec = ``new` `List<``int``[]>();` `    ``vec.Add(``new``[] { 1, 1 });` `    ``vec.Add(``new``[] { 2, 2 });` `    ``vec.Add(``new``[] { 3, 3 });` `    ``vec.Add(``new``[] { 4, 4 });`   `    ``int` `n = vec.Count;`   `    ``// Function call` `    ``Max_Distance(vec, n);` `  ``}` `}`   `// This code is contributed by phasing17`

## Javascript

 `// JavaScript implementation of the approach`   `// Function to implement the custom sort ` `function` `myCustomSort(l, r){` `    ``return` `Math.atan2(l[1], l[0]) < Math.atan2(r[1], r[0]);` `}`   `// Function to find the maximum possible` `// distance from origin using given points.` `function` `Max_Distance(xy, n)` `{` `    ``// Sort the points with their tan angle` `    ``xy.sort(myCustomSort);`   `    ``// Push the whole vector` `    ``for` `(let i = 0; i < n; i++)` `        ``xy.push(xy[i]);`   `    ``// To store the required answer` `    ``let res = 0;`   `    ``// Find the maximum possible answer` `    ``for` `(let i = 0; i< n; i++) {` `        ``let x = 0, y = 0;` `        ``for` `(let j = i; j

Output:

`14.14`

Time Complexity: O(n^2)

Auxiliary Space: O(1)