# 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 **i ^{th}** point

**(x**one can move from

_{i}, y_{i})**(a, b)**to

**(a + x**.

_{i}, b + y_{i})**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(N^{2}) approach.

Below is the implementation of the above approach:

`// C++ implementation of the approach ` `#include <bits/stdc++.h> ` `using` `namespace` `std; ` ` ` `// Function to find the maximum possible ` `// distance from origin using given points. ` `void` `Max_Distance(vector<pair<` `int` `, ` `int` `> >& 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<pair<` `int` `, ` `int` `> > vec = { { 1, 1 }, ` ` ` `{ 2, 2 }, ` ` ` `{ 3, 3 }, ` ` ` `{ 4, 4 } }; ` ` ` ` ` `int` `n = vec.size(); ` ` ` ` ` `// Function call ` ` ` `Max_Distance(vec, n); ` ` ` ` ` `return` `0; ` `} ` |

*chevron_right*

*filter_none*

**Output:**

14.14

## Recommended Posts:

- Find K Closest Points to the Origin
- Find points at a given distance on a line of given slope
- Find the integer points (x, y) with Manhattan distance atleast N
- Haversine formula to find distance between two points on a sphere
- Ways to choose three points with distance between the most distant points <= L
- Find maximum and minimum distance between magnets
- Find maximum distance between any city and station
- Find the maximum distance covered using n bikes
- Program to calculate distance between two points in 3 D
- Program to calculate distance between two points
- Hammered distance between N points in a 2-D plane
- Program for distance between two points on earth
- Check whether it is possible to join two points given on circle such that distance between them is k
- Maximum points of intersection n circles
- Maximum points of intersection n lines

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.