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.
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 ?> |
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