Related Articles
Find the maximum possible distance from origin using given points
• Last Updated : 26 Sep, 2019

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
```

Attention reader! Don’t stop learning now. Get hold of all the important mathematical concepts for competitive programming with the Essential Maths for CP Course at a student-friendly price. To complete your preparation from learning a language to DS Algo and many more,  please refer Complete Interview Preparation Course.

My Personal Notes arrow_drop_up