# Find the maximum possible distance from origin using given points

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

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

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:

 `// 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 < i + n; j++) { ` `            ``x += xy[j].first; ` `            ``y += xy[j].second; ` `            ``res = max(res, x * x + y * y); ` `        ``} ` `    ``} ` ` `  `    ``// Print the required answer ` `    ``cout << fixed << setprecision(2) << sqrtl(res); ` `} ` ` `  `// Driver code ` `int` `main() ` `{ ` `    ``vector > vec = { { 1, 1 }, ` `                                    ``{ 2, 2 }, ` `                                    ``{ 3, 3 }, ` `                                    ``{ 4, 4 } }; ` ` `  `    ``int` `n = vec.size(); ` ` `  `    ``// Function call ` `    ``Max_Distance(vec, n); ` ` `  `    ``return` `0; ` `} `

Output:

```14.14
```

My Personal Notes arrow_drop_up Check out this Author's contributed articles.

If you like GeeksforGeeks and would like to contribute, you can also write an article using contribute.geeksforgeeks.org or mail your article to contribute@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.

Please Improve this article if you find anything incorrect by clicking on the "Improve Article" button below.

Article Tags :
Practice Tags :

Be the First to upvote.

Please write to us at contribute@geeksforgeeks.org to report any issue with the above content.