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 codeint 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 approachimport 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 approachfrom 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 codevec = [[1, 1], [2, 2], [3, 3], [4, 4]];n = len(vec)# Function callMax_Distance(vec, n);# The code is contributed by phasing17 |
C#
// C# implementation of the approachusing 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 codelet vec = [[1, 1], [2, 2], [3, 3], [4, 4]];let n = vec.length;// Function callMax_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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