Triangle with no point inside

Given N points in 2-dimensional space, we need to find three points such that triangle made by choosing these points should not contain any other points inside. All given points will not lie on the same line so solution will always exist.
Examples:



In above diagram possible triangle with no point 
inside can be formed by choosing these triplets,
[(0, 0), (2, 0), (1, 1)]
[(0, 0), (1, 1), (0, 2)]
[(1, 1), (2, 0), (2, 2)]
[(1, 1), (0, 2), (2, 2)]

So any of the above triplets can be the final answer.

Recommended: Please try your approach on {IDE} first, before moving on to the solution.

The solution is based on the fact that if there exist triangle(s) with no points inside, then we can form a triangle with any random point among all points.
We can solve this problem by searching all three points one by one. The first point can be chosen randomly. After choosing the first point, we need two points such that their slope should be different and no point should lie inside the triangle of these three points. We can do this by choosing the second point as nearest point to the first and third point as second nearest point with the different slope. For doing this, we first iterate over all points and choose the point which is nearest to the first one and designate that as second point of the required triangle. Then we iterate one more time to find point which has different slope and has the smallest distance, which will be the third point of our triangle.

C++

// C/C++ program to find triangle with no point inside 
#include <bits/stdc++.h> 
using namespace std; 
  
// method to get square of distance between 
// (x1, y1) and (x2, y2) 
int getDistance(int x1, int y1, int x2, int y2) 
{ 
    return (x2 - x1)*(x2 - x1) + 
           (y2 - y1)*(y2 - y1); 
} 
  
// Method prints points which make triangle with no 
// point inside 
void triangleWithNoPointInside(int points[][2], int N) 
{ 
    //    any point can be chosen as first point of triangle 
    int first = 0; 
    int second, third; 
    int minD = INT_MAX; 
  
    // choose nearest point as second point of triangle 
    for (int i = 0; i < N; i++) 
    { 
        if (i == first) 
            continue; 
  
        // Get distance from first point and choose 
        // nearest one 
        int d = getDistance(points[i][0], points[i][1], 
                    points[first][0], points[first][1]); 
        if (minD > d) 
        { 
            minD = d; 
            second = i; 
        } 
    } 
  
    // Pick third point by finding the second closest 
    // point with different slope. 
    minD = INT_MAX; 
    for (int i = 0; i < N; i++) 
    { 
        // if already chosen point then skip them 
        if (i == first || i == second) 
            continue; 
  
        // get distance from first point 
        int d = getDistance(points[i][0], points[i][1], 
                     points[first][0], points[first][1]); 
  
        /*  the slope of the third point with the first 
            point should not be equal to the slope of 
            second point with first point (otherwise 
            they'll be collinear)     and among all such 
            points, we choose point with the smallest 
            distance  */
        // here cross multiplication is compared instead 
        // of division comparison 
        if ((points[i][0] - points[first][0]) * 
            (points[second][1] - points[first][1]) != 
            (points[second][0] - points[first][0]) * 
            (points[i][1] - points[first][1]) && 
            minD > d) 
        { 
            minD = d; 
            third = i; 
        } 
    } 
  
    cout << points[first][0] << ", "
         << points[first][1] << endl; 
    cout << points[second][0] << ", "
         << points[second][1] << endl; 
    cout << points[third][0] << ", "
         << points[third][1] << endl; 
} 
  
// Driver code to test above methods 
int main() 
{ 
    int points[][2] = {{0, 0}, {0, 2}, {2, 0}, 
                       {2, 2}, {1, 1}}; 
    int N = sizeof(points) / sizeof(points[0]); 
    triangleWithNoPointInside(points, N); 
    return 0; 
} 

Java

// Java program to find triangle 
// with no point inside 
import java.io.*; 
  
class GFG  
{ 
    // method to get square of distance between 
    // (x1, y1) and (x2, y2) 
    static int getDistance(int x1, int y1, int x2, int y2) 
    { 
        return (x2 - x1)*(x2 - x1) + 
                  (y2 - y1)*(y2 - y1); 
    } 
      
    // Method prints points which make triangle with no 
    // point inside 
    static void triangleWithNoPointInside(int points[][], int N) 
    { 
        // any point can be chosen as first point of triangle 
        int first = 0; 
        int second =0; 
        int third =0; 
        int minD = Integer.MAX_VALUE; 
      
        // choose nearest point as second point of triangle 
        for (int i = 0; i < N; i++) 
        { 
            if (i == first) 
                continue; 
      
            // Get distance from first point and choose 
            // nearest one 
            int d = getDistance(points[i][0], points[i][1], 
                        points[first][0], points[first][1]); 
            if (minD > d) 
            { 
                minD = d; 
                second = i; 
            } 
        } 
      
        // Pick third point by finding the second closest 
        // point with different slope. 
        minD = Integer.MAX_VALUE; 
        for (int i = 0; i < N; i++) 
        { 
            // if already chosen point then skip them 
            if (i == first || i == second) 
                continue; 
      
            // get distance from first point 
            int d = getDistance(points[i][0], points[i][1], 
                        points[first][0], points[first][1]); 
      
            /* the slope of the third point with the first 
                point should not be equal to the slope of 
                second point with first point (otherwise 
                they'll be collinear) and among all such 
                points, we choose point with the smallest 
                distance */
            // here cross multiplication is compared instead 
            // of division comparison 
            if ((points[i][0] - points[first][0]) * 
                (points[second][1] - points[first][1]) != 
                (points[second][0] - points[first][0]) * 
                (points[i][1] - points[first][1]) && 
                minD > d) 
            { 
                minD = d; 
                third = i; 
            } 
        } 
      
        System.out.println(points[first][0] + ", "
            + points[first][1]); 
        System.out.println(points[second][0]+ ", "
            + points[second][1]) ; 
        System.out.println(points[third][0] +", "
            + points[third][1]); 
    } 
      
    // Driver code  
    public static void main (String[] args)  
    { 
        int points[][] = {{0, 0}, {0, 2}, {2, 0}, 
                         {2, 2}, {1, 1}}; 
        int N = points.length; 
        triangleWithNoPointInside(points, N); 
    } 
} 
  
// This article is contributed by vt_m.  

Python 3

# Python3 program to find triangle  
# with no point inside  
import sys 
  
# method to get square of distance between  
# (x1, y1) and (x2, y2) 
def getDistance(x1, y1, x2, y2): 
    return (x2 - x1) * (x2 - x1) + \ 
           (y2 - y1) * (y2 - y1) 
  
# Method prints points which make triangle  
# with no point inside 
def triangleWithNoPointInside(points, N): 
      
    # any point can be chosen as  
    # first point of triangle 
    first = 0
    second = 0
    third = 0
    minD = sys.maxsize 
  
    # choose nearest point as  
    # second point of triangle 
    for i in range(0, N): 
        if i == first: 
            continue
  
        # Get distance from first point and choose  
        # nearest one 
        d = getDistance(points[i][0], points[i][1], 
                        points[first][0],  
                        points[first][1]) 
        if minD > d: 
            minD = d 
            second = i 
      
    # Pick third point by finding the second closest  
    # point with different slope. 
    minD = sys.maxsize 
    for i in range (0, N): 
          
        # if already chosen point then skip them  
        if i == first or i == second: 
            continue
  
        # get distance from first point 
        d = getDistance(points[i][0], points[i][1], 
                        points[first][0],  
                        points[first][1]) 
  
        """ the slope of the third point with the first  
            point should not be equal to the slope of  
            second point with first point (otherwise  
            they'll be collinear) and among all such  
            points, we choose point with the smallest  
            distance """
              
        # here cross multiplication is compared instead  
        # of division comparison 
        if ((points[i][0] - points[first][0]) *
            (points[second][1] - points[first][1]) !=
            (points[second][0] - points[first][0]) *
            (points[i][1] - points[first][1]) 
            and minD > d) : 
            minD = d 
            third = i 
  
    print(points[first][0], ', ', points[first][1]) 
    print(points[second][0], ', ', points[second][1]) 
    print(points[third][0], ', ', points[third][1]) 
  
# Driver code 
points = [[0, 0], [0, 2],  
          [2, 0], [2, 2], [1, 1]] 
N = len(points) 
triangleWithNoPointInside(points, N) 
  
# This code is contributed by Gowtham Yuvaraj 

C#

using System; 
  
// C# program to find triangle  
// with no point inside  
  
public class GFG 
{ 
    // method to get square of distance between  
    // (x1, y1) and (x2, y2)  
    public static int getDistance(int x1, int y1, int x2, int y2) 
    { 
        return (x2 - x1) * (x2 - x1) + (y2 - y1) * (y2 - y1); 
    } 
  
    // Method prints points which make triangle with no  
    // point inside  
    public static void triangleWithNoPointInside(int[][] points, int N) 
    { 
        // any point can be chosen as first point of triangle  
        int first = 0; 
        int second = 0; 
        int third = 0; 
        int minD = int.MaxValue; 
  
        // choose nearest point as second point of triangle  
        for (int i = 0; i < N; i++) 
        { 
            if (i == first) 
            { 
                continue; 
            } 
  
            // Get distance from first point and choose  
            // nearest one  
            int d = getDistance(points[i][0], points[i][1], points[first][0], points[first][1]); 
            if (minD > d) 
            { 
                minD = d; 
                second = i; 
            } 
        } 
  
        // Pick third point by finding the second closest  
        // point with different slope.  
        minD = int.MaxValue; 
        for (int i = 0; i < N; i++) 
        { 
            // if already chosen point then skip them  
            if (i == first || i == second) 
            { 
                continue; 
            } 
  
            // get distance from first point  
            int d = getDistance(points[i][0], points[i][1], points[first][0], points[first][1]); 
  
            /* the slope of the third point with the first  
                point should not be equal to the slope of  
                second point with first point (otherwise  
                they'll be collinear) and among all such  
                points, we choose point with the smallest  
                distance */
            // here cross multiplication is compared instead  
            // of division comparison  
            if ((points[i][0] - points[first][0]) * (points[second][1] - points[first][1]) != (points[second][0] - points[first][0]) * (points[i][1] - points[first][1]) && minD > d) 
            { 
                minD = d; 
                third = i; 
            } 
        } 
  
        Console.WriteLine(points[first][0] + ", " + points[first][1]); 
        Console.WriteLine(points[second][0] + ", " + points[second][1]); 
        Console.WriteLine(points[third][0] + ", " + points[third][1]); 
    } 
  
    // Driver code   
    public static void Main(string[] args) 
    { 
        int[][] points = new int[][] 
        { 
            new int[] {0, 0}, 
            new int[] {0, 2}, 
            new int[] {2, 0}, 
            new int[] {2, 2}, 
            new int[] {1, 1} 
        }; 
        int N = points.Length; 
        triangleWithNoPointInside(points, N); 
    } 
} 
  
  // This code is contributed by Shrikant13 

Output:

0, 0
1, 1
0, 2

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My Personal Notes

Recommended: Please try your approach on {IDE} first, before moving on to the solution.

C++

Java

Python 3

C#

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