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

0 0 6 minutes read

Given an array that represents the 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 that 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.

Approach: The slope of a line made by points (x1, y1) and (x2, y2) is y2-y2/x2-x1. As the given array consists 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 into consideration 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 lines this is mandatory that two of these three points lie on the same line. So, three possible cases are:

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

Take one out of three cases, say P0 and P1 lie on the same line, in this case, let m=arr[1]-arr[0] be our slope. For the general point in the array (i, arr[i]) the equation of the 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 a given array then we must have exactly two intercepts (c).
So, if two distinct intercepts exist for any of the 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+1));
 
    // 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]) / 2.0, 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
?>

Javascript

<script>
 
// Javascript implementation of the above approach
 
// Find if slope is good with only two intercept
function isSlopeGood(slope, arr, n)
{
    var setOfLines = new Set();
    for (var 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
function checkForParallel(arr, n)
{
    // check the result by processing
    // the slope by starting three points
    var slope1 = isSlopeGood(arr[1] - arr[0], arr, n);
    var slope2 = isSlopeGood(arr[2] - arr[1], arr, n);
    var slope3 = isSlopeGood(parseInt((arr[2] - arr[0]) / 2), arr, n);
 
    if(slope1 || slope2 || slope3)
    {
        return 1;
    }
    return 0;
}
 
// Driver code
var arr = [ 1, 6, 3, 8, 5 ];
var n = arr.length;
document.write( checkForParallel(arr, n));
 
</script>

Output:

Time Complexity: O(N), since there runs a loop from 0 to (n – 1).

Auxiliary Space: O(N), since N extra space has been taken.

Feeling lost in the world of random DSA topics, wasting time without progress? It’s time for a change! Join our DSA course, where we’ll guide you on an exciting journey to master DSA efficiently and on schedule.
Ready to dive in? Explore our Free Demo Content and join our DSA course, trusted by over 100,000 zambiatek!