Find whether only two parallel lines contain all coordinates points or not
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++
#include <bits/stdc++.h>
using namespace std;
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));
return setOfLines.size() == 2;
}
bool checkForParallel( int arr[], int n)
{
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);
}
int main()
{
int arr[] = {1, 6, 3, 8, 5 };
int n = sizeof (arr) / sizeof (arr[0]);
cout << ( int )checkForParallel(arr, n);
return 0;
}
|
Java
import java.util.*;
class GfG
{
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));
return setOfLines.size() == 2 ;
}
static boolean checkForParallel( int arr[], int n)
{
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 );
}
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" );
}
}
|
Python3
def isSlopeGood(slope, arr, n):
setOfLines = dict ()
for i in range (n):
setOfLines[arr[i] - slope * (i)] = 1
return len (setOfLines) = = 2
def checkForParallel(arr, n):
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)
arr = [ 1 , 6 , 3 , 8 , 5 ]
n = len (arr)
if checkForParallel(arr, n):
print ( 1 )
else :
print ( 0 )
|
C#
using System;
using System.Collections.Generic;
class GfG
{
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));
return setOfLines.Count == 2;
}
static bool checkForParallel( int []arr, int n)
{
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 );
}
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" );
}
}
|
PHP
<?php
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 );
return ( count ( $setOfLines ) == 2);
}
function checkForParallel( $arr , $n )
{
$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 );
}
$arr = array ( 1, 6, 3, 8, 5 );
$n = count ( $arr );
echo (int)checkForParallel( $arr , $n ) . "\n" ;
?>
|
Javascript
<script>
function isSlopeGood(slope, arr, n)
{
var setOfLines = new Set();
for ( var i = 0; i < n; i++)
setOfLines.add(arr[i] - slope * (i));
return setOfLines.size == 2;
}
function checkForParallel(arr, n)
{
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;
}
var arr = [ 1, 6, 3, 8, 5 ];
var n = arr.length;
document.write( checkForParallel(arr, n));
</script>
|
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.
Last Updated :
22 Jun, 2022
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