Find whether only two parallel lines contain all coordinates points or not

Given an array which represents y coordinates of a set of points on a coordinate plane, where (i, arr[i]) represent a single point. Find whether it is possible to draw a pair of parallel lines which includes all the coordinate points given and also both lines must contain a point. Print 1 for possible and 0 if not possible.

Examples:

Input: arr[] = {1, 4, 3, 6, 5};
Output: 1
(1, 1), (3, 3) and (5, 5) lie on one line
where as (2, 4) and (4, 6) lie on another line.

Input: arr[] = {2, 4, 3, 6, 5};
Output: 0
Minimum 3 lines needed to cover all points.

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

Approach: Slope of a line made by ponits (x1, y1) and (x2, y2) is y2-y2/x2-x1. As the given array consist of coordinates of points as (i, arr[i]). So, (arr[2]-arr[1]) / (2-1) is slope of line made by (1, arr[i]) and (2, arr[2]). Take in consideration of only three points say P0(0, arr[0]), P1(1, arr[1]) and P2(2, arr[2]) as the requirement is of only two parallel line this is mandatory that two of these three points lie on the same line. So, three possible cases are:

P0 and P1 are on same line hence their slope will be arr[1]-arr[0]
P1 and P2 are on same line hence their slope will be arr[2]-arr[1]
P0 and P2 are on same line hence their slope will be arr[2]-arr[0]/2

Take one out of three cases say P0 and P1 lies on same line, in this case let m=arr[1]-arr[0] is our slope. For the a general point in array (i, arr[i]) the equation of line is:

=> (y-y1) = m (x-x1)
=> y-arr[i] = m (x-i)
=> y-mx = arr[i] - mi

Now, as y-mx=c is general equation of straight line here c = arr[i] -mi. Now, if the solution will be possible for given arraythen we must have exact two intercept (c).
So, if two distinct intercept exists for any of above mentioned three possible, the required solution is possible and print 1 else 0.

Below is the implementation of the above approach:

C++

// C++ implementation of the above approach 
#include <bits/stdc++.h> 
using namespace std; 
  
// Find if slope is good with only two intercept 
bool isSlopeGood(double slope, int arr[], int n) 
{ 
    set<double> setOfLines; 
    for (int i = 0; i < n; i++) 
        setOfLines.insert(arr[i] - slope * (i)); 
  
    // if set of lines have only two distinct intercept 
    return setOfLines.size() == 2; 
} 
  
// Function to check if required solution exist 
bool checkForParallel(int arr[], int n) 
{ 
    // check the result by processing 
    // the slope by starting three points 
    bool slope1 = isSlopeGood(arr[1] - arr[0], arr, n); 
    bool slope2 = isSlopeGood(arr[2] - arr[1], arr, n); 
    bool slope3 = isSlopeGood((arr[2] - arr[0]) / 2, arr, n); 
  
    return (slope1 || slope2 || slope3); 
} 
  
// Driver code 
int main() 
{ 
    int arr[] = { 1, 6, 3, 8, 5 }; 
    int n = sizeof(arr) / sizeof(arr[0]); 
    cout << (int)checkForParallel(arr, n); 
  
    return 0; 
} 

Java

// Java implementation of the above approach  
import java.util.*; 
  
class GfG  
{ 
  
// Find if slope is good  
// with only two intercept  
static boolean isSlopeGood(double slope, 
                        int arr[], int n)  
{  
    Set<Double> setOfLines = new HashSet<Double> ();  
    for (int i = 0; i < n; i++)  
        setOfLines.add(arr[i] - slope * (i));  
  
    // if set of lines have only two distinct intercept  
    return setOfLines.size() == 2;  
}  
  
// Function to check if required solution exist  
static boolean checkForParallel(int arr[], int n)  
{  
    // check the result by processing  
    // the slope by starting three points  
    boolean slope1 = isSlopeGood(arr[1] - arr[0], arr, n);  
    boolean slope2 = isSlopeGood(arr[2] - arr[1], arr, n);  
    boolean slope3 = isSlopeGood((arr[2] - arr[0]) / 2, arr, n);  
  
    return (slope1 == true || slope2 == true || slope3 == true);  
}  
  
// Driver code  
public static void main(String[] args)  
{  
    int arr[] = { 1, 6, 3, 8, 5 };  
    int n = arr.length;  
    if(checkForParallel(arr, n) == true) 
    System.out.println("1"); 
    else
    System.out.println("0"); 
} 
}  
  
// This code is contributed by Prerna Saini. 

Python3

# Python3 implementation of the  
# above approach 
  
# Find if slope is good with only  
# two intercept 
def isSlopeGood(slope, arr, n): 
  
    setOfLines = dict() 
    for i in range(n): 
        setOfLines[arr[i] - slope * (i)] = 1
  
    # if set of lines have only  
    # two distinct intercept 
    return len(setOfLines) == 2
  
# Function to check if required solution exist 
def checkForParallel(arr, n): 
      
    # check the result by processing 
    # the slope by starting three points 
    slope1 = isSlopeGood(arr[1] - arr[0], arr, n) 
    slope2 = isSlopeGood(arr[2] - arr[1], arr, n) 
    slope3 = isSlopeGood((arr[2] - arr[0]) // 2, arr, n) 
  
    return (slope1 or slope2 or slope3) 
  
# Driver code 
arr = [1, 6, 3, 8, 5 ] 
n = len(arr) 
if checkForParallel(arr, n): 
    print(1) 
else: 
    print(0) 
      
# This code is contributed by Mohit Kumar     

C#

// C# implementation of the above approach  
using System; 
using System.Collections.Generic; 
  
class GfG  
{  
  
// Find if slope is good  
// with only two intercept  
static bool isSlopeGood(double slope,  
                        int []arr, int n)  
{  
  
    HashSet<Double> setOfLines = new HashSet<Double> ();  
    for (int i = 0; i < n; i++)  
        setOfLines.Add(arr[i] - slope * (i));  
  
    // if set of lines have only two distinct intercept  
    return setOfLines.Count == 2;  
}  
  
// Function to check if required solution exist  
static bool checkForParallel(int []arr, int n)  
{  
    // check the result by processing  
    // the slope by starting three points  
    bool slope1 = isSlopeGood(arr[1] - arr[0], arr, n);  
    bool slope2 = isSlopeGood(arr[2] - arr[1], arr, n);  
    bool slope3 = isSlopeGood((arr[2] - arr[0]) / 2, arr, n);  
  
    return (slope1 == true || slope2 == true || slope3 == true);  
}  
  
// Driver code  
public static void Main()  
{  
    int []arr = { 1, 6, 3, 8, 5 };  
    int n = arr.Length;  
    if(checkForParallel(arr, n) == true)  
        Console.WriteLine("1");  
    else
        Console.WriteLine("0");  
}  
}  
  
// This code is contributed by Ryuga.  
  

PHP

<?php 
// PHP implementation of the above approach 
  
// Find if slope is good with only 
// two intercept 
function isSlopeGood($slope, $arr, $n) 
{ 
    $setOfLines = array_fill(0, max($arr) * $n, 0); 
    for ($i = 0; $i < $n; $i++) 
        $setOfLines[$arr[$i] - $slope * $i] = 1; 
    $setOfLines = array_unique($setOfLines); 
      
    // if set of lines have only two 
    // distinct intercept 
    return (count($setOfLines) == 2); 
} 
  
// Function to check if required  
// solution exist 
function checkForParallel($arr, $n) 
{ 
    // check the result by processing 
    // the slope by starting three points 
    $slope1 = isSlopeGood($arr[1] - $arr[0],  
                                    $arr, $n); 
    $slope2 = isSlopeGood($arr[2] - $arr[1],  
                                    $arr, $n); 
    $slope3 = isSlopeGood((int)(($arr[2] -  
                                 $arr[0]) / 2), 
                                 $arr, $n); 
  
    return ($slope1 || $slope2 || $slope3); 
} 
  
// Driver code 
$arr = array( 1, 6, 3, 8, 5 ); 
$n = count($arr); 
echo (int)checkForParallel($arr, $n) . "\n"; 
  
// This code is contributed by mits 
?> 

Output:

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My Personal Notes

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

C++

Java

Python3

C#

PHP

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