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
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
#include <bits/stdc++.h>
using namespace std;
void Max_Distance(vector<pair< int , int > >& xy, int n)
{
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);
});
for ( int i = 0; i < n; i++)
xy.push_back(xy[i]);
int res = 0;
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);
}
}
cout << fixed << setprecision(2) << sqrtl(res);
}
int main()
{
vector<pair< int , int >> vec = { { 1, 1 },
{ 2, 2 },
{ 3, 3 },
{ 4, 4 } };
int n = vec.size();
Max_Distance(vec, n);
return 0;
}
|
Java
import java.util.*;
class GFG {
static void
Max_Distance(ArrayList<ArrayList<Integer> > xy, int n)
{
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 ))));
}
});
for ( int i = 0 ; i < n; i++)
xy.add(xy.get(i));
int res = 0 ;
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);
}
}
System.out.println(
( double )Math.round(Math.sqrt(res) * 100 ) / 100 );
}
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 ;
Max_Distance(vec, n);
}
}
|
Python3
from math import *
def myCustomSort(l):
return atan2(l[ 1 ], l[ 0 ]);
def Max_Distance(xy, n):
xy.sort(key = myCustomSort);
xy + = xy
res = 0 ;
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 ( round (res * * 0.5 , 2 ))
vec = [[ 1 , 1 ], [ 2 , 2 ], [ 3 , 3 ], [ 4 , 4 ]];
n = len (vec)
Max_Distance(vec, n);
|
C#
using System;
using System.Linq;
using System.Collections.Generic;
class GFG
{
static void Max_Distance(List< int []> xy, int n)
{
xy = xy.OrderBy(x => Math.Atan2(x[1], x[0]))
.ToList();
for ( int i = 0; i < n; i++)
xy.Add(xy[i]);
int res = 0;
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);
}
}
Console.WriteLine(Math.Round(Math.Sqrt(res), 2));
}
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;
Max_Distance(vec, n);
}
}
|
Javascript
function myCustomSort(l, r){
return Math.atan2(l[1], l[0]) < Math.atan2(r[1], r[0]);
}
function Max_Distance(xy, n)
{
xy.sort(myCustomSort);
for (let i = 0; i < n; i++)
xy.push(xy[i]);
let res = 0;
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);
}
}
console.log(Math.sqrt(res).toFixed(2));
}
let vec = [[1, 1],
[2, 2],
[3, 3],
[4, 4]];
let n = vec.length;
Max_Distance(vec, n);
|
Time Complexity: O(n^2)
Auxiliary Space: O(1)
Last Updated :
13 Sep, 2022
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