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:
// 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 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 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# 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 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) |
14.14
Time Complexity: O(n^2)
Auxiliary Space: O(1)