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

• Difficulty Level : Hard
• Last Updated : 29 Apr, 2021

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), (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:
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-arr) / (2-1) is slope of line made by (1, arr[i]) and (2, arr). Take in consideration of only three points say P0(0, arr), P1(1, arr), and P2(2, arr) 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-arr
• P1 and P2 are on the same line hence their slope will be arr-arr
• P0 and P2 are on the same line hence their slope will be arr-arr/2

Take one out of three cases, say P0 and P1 lie on the same line, in this case, let m=arr-arr is 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 using namespace std; // Find if slope is good with only two interceptbool isSlopeGood(double slope, int arr[], int n){    set 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 existbool checkForParallel(int arr[], int n){    // check the result by processing    // the slope by starting three points    bool slope1 = isSlopeGood(arr - arr, arr, n);    bool slope2 = isSlopeGood(arr - arr, arr, n);    bool slope3 = isSlopeGood((arr - arr) / 2, arr, n);     return (slope1 || slope2 || slope3);} // Driver codeint main(){    int arr[] = { 1, 6, 3, 8, 5 };    int n = sizeof(arr) / sizeof(arr);    cout << (int)checkForParallel(arr, n);     return 0;}

Java

 // Java implementation of the above approachimport java.util.*; class GfG{ // Find if slope is good// with only two interceptstatic boolean isSlopeGood(double slope,                        int arr[], int n){    Set setOfLines = new HashSet ();    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 existstatic boolean checkForParallel(int arr[], int n){    // check the result by processing    // the slope by starting three points    boolean slope1 = isSlopeGood(arr - arr, arr, n);    boolean slope2 = isSlopeGood(arr - arr, arr, n);    boolean slope3 = isSlopeGood((arr - arr) / 2, arr, n);     return (slope1 == true || slope2 == true || slope3 == true);} // Driver codepublic 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 interceptdef 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 existdef checkForParallel(arr, n):         # check the result by processing    # the slope by starting three points    slope1 = isSlopeGood(arr - arr, arr, n)    slope2 = isSlopeGood(arr - arr, arr, n)    slope3 = isSlopeGood((arr - arr) // 2, arr, n)     return (slope1 or slope2 or slope3) # Driver codearr = [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 approachusing System;using System.Collections.Generic; class GfG{ // Find if slope is good// with only two interceptstatic bool isSlopeGood(double slope,                        int []arr, int n){     HashSet setOfLines = new HashSet ();    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 existstatic bool checkForParallel(int []arr, int n){    // check the result by processing    // the slope by starting three points    bool slope1 = isSlopeGood(arr - arr, arr, n);    bool slope2 = isSlopeGood(arr - arr, arr, n);    bool slope3 = isSlopeGood((arr - arr) / 2, arr, n);     return (slope1 == true || slope2 == true || slope3 == true);} // Driver codepublic 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.



Javascript


Output:
1

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