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Find the maximum possible distance from origin using given points

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  • Difficulty Level : Basic
  • Last Updated : 13 Sep, 2022
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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 <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;
}

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<ArrayList<Integer> > xy, int n)
    {
 
        // Sort the points with their tan angle
        Collections.sort(
            xy, new Comparator<ArrayList<Integer> >() {
                @Override
                public int compare(ArrayList<Integer> x,
                                   ArrayList<Integer> 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<ArrayList<Integer> > vec
            = new ArrayList<ArrayList<Integer> >();
 
        ArrayList<Integer> a1 = new ArrayList<Integer>();
        a1.add(1);
        a1.add(1);
        vec.add(a1);
 
        ArrayList<Integer> a2 = new ArrayList<Integer>();
        a2.add(2);
        a2.add(2);
        vec.add(a2);
 
        ArrayList<Integer> a3 = new ArrayList<Integer>();
        a3.add(3);
        a3.add(3);
        vec.add(a3);
 
        ArrayList<Integer> a4 = new ArrayList<Integer>();
        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 <i + n; j++) {
            x += xy[j][0];
            y += xy[j][1];
            res = Math.max(res, x * x + y * y);
        }
    }
 
    // Print the required answer
    console.log(Math.sqrt(res).toFixed(2));
 
}
 
// Driver code
let vec = [[1, 1],
           [2, 2],
           [3, 3],
           [4, 4]];
 
let n = vec.length;
 
// Function call
Max_Distance(vec, n);
 
// The code is contributed by Gautam goel (gautmgoel962)

Output: 

14.14

 

Time Complexity: O(n^2)

Auxiliary Space: O(1)


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