Check whether a straight line can be formed using N co-ordinate points
Given an array arr[] of N co-ordinate points, the task is to check whether a straight line can be formed using these co-ordinate points.
Input: arr[] = {{0, 0}, {1, 1}, {2, 2}}
Output: Yes
Explanation:
Slope of every two points is same. That is 1.
Therefore, a straight line can be formed using these points.Input: arr[] = {{0, 1}, {2, 0}}
Output: Yes
Explanation:
Two points in co-ordinate system always forms a straight line.
Approach: The idea is to find the slope of line between every pair of points in the array and if the slope of every pair of point is same, then these points together forms a straight line.
// Slope of line formed by // two points (y2, y1), (x2, x1) Slope of Line = y2 - y1 --------- x2 - x1
Below is the implementation of the above approach:
C++
// C++ implementation to check // if a straight line // can be formed using N points #include <bits/stdc++.h> using namespace std; // Function to check if a straight line // can be formed using N points bool isStraightLinePossible( vector<pair< int , int > > arr, int n) { // First pair of point (x0, y0) int x0 = arr[0].first; int y0 = arr[0].second; // Second pair of point (x1, y1) int x1 = arr[1].first; int y1 = arr[1].second; int dx = x1 - x0, dy = y1 - y0; // Loop to iterate over the points for ( int i = 0; i < n; i++) { int x = arr[i].first, y = arr[i].second; if (dx * (y - y1) != dy * (x - x1)){ cout << "NO" ; return false ; } } cout << "YES" ; return true ; } // Driver Code int main() { // Array of points vector<pair< int , int > > arr = { { 0, 0 }, { 1, 1 }, { 3, 3 }, { 2, 2 } }; int n = 4; // Function Call isStraightLinePossible(arr, n); return 0; } |
Java
// Java implementation to check // if a straight line can be // formed using N points import java.util.*; class GFG{ static class pair { int first, second; pair( int first, int second) { this .first = first; this .second = second; } } // Function to check if a straight line // can be formed using N points static void isStraightLinePossible( ArrayList<pair> arr, int n) { // First pair of point (x0, y0) int x0 = arr.get( 0 ).first; int y0 = arr.get( 0 ).second; // Second pair of point (x1, y1) int x1 = arr.get( 1 ).first; int y1 = arr.get( 1 ).second; int dx = x1 - x0, dy = y1 - y0; // Loop to iterate over the points for ( int i = 0 ; i < n; i++) { int x = arr.get(i).first; int y = arr.get(i).second; if (dx * (y - y1) != dy * (x - x1)) { System.out.println( "NO" ); } } System.out.println( "YES" ); } // Driver code public static void main(String[] args) { // Array of points ArrayList<pair> arr = new ArrayList<>(); arr.add( new pair( 0 , 0 )); arr.add( new pair( 1 , 1 )); arr.add( new pair( 3 , 3 )); arr.add( new pair( 2 , 2 )); int n = 4 ; // Function Call isStraightLinePossible(arr, n); } } // This code is contributed by offbeat |
Python3
# Python3 implementation to check # if a straight line can be formed # using N points # Function to check if a straight line # can be formed using N points def isStraightLinePossible(arr, n): # First pair of point (x0, y0) x0 = arr[ 0 ][ 0 ] y0 = arr[ 0 ][ 1 ] # Second pair of point (x1, y1) x1 = arr[ 1 ][ 0 ] y1 = arr[ 1 ][ 1 ] dx = x1 - x0 dy = y1 - y0 # Loop to iterate over the points for i in range (n): x = arr[i][ 0 ] y = arr[i][ 1 ] if (dx * (y - y1) ! = dy * (x - x1)): print ( "NO" , end = "") return False print ( "YES" , end = "") return True # Driver code # Array of points arr = [ [ 0 , 0 ], [ 1 , 1 ], [ 3 , 3 ], [ 2 , 2 ] ] n = 4 # Function Call isStraightLinePossible(arr, n) # This code is contributed by divyeshrabadiya07 |
C#
// C# implementation to check // if a straight line can be // formed using N points using System; using System.Collections.Generic; class GFG{ public class pair { public int first, second; public pair( int first, int second) { this .first = first; this .second = second; } } // Function to check if a straight line // can be formed using N points static void isStraightLinePossible( List<pair> arr, int n) { // First pair of point (x0, y0) int x0 = arr[0].first; int y0 = arr[0].second; // Second pair of point (x1, y1) int x1 = arr[1].first; int y1 = arr[1].second; int dx = x1 - x0, dy = y1 - y0; // Loop to iterate over the points for ( int i = 0; i < n; i++) { int x = arr[i].first; int y = arr[i].second; if (dx * (y - y1) != dy * (x - x1)) { Console.WriteLine( "NO" ); } } Console.WriteLine( "YES" ); } // Driver code public static void Main(String[] args) { // Array of points List<pair> arr = new List<pair>(); arr.Add( new pair(0, 0)); arr.Add( new pair(1, 1)); arr.Add( new pair(3, 3)); arr.Add( new pair(2, 2)); int n = 4; // Function call isStraightLinePossible(arr, n); } } // This code is contributed by Princi Singh |
Javascript
<script> // Javascript implementation of the above approach // Function to check if a straight line // can be formed using N points function isStraightLinePossible( arr, n) { // First pair of point (x0, y0) var x0 = arr[0][0]; var y0 = arr[0][1]; // Second pair of point (x1, y1) var x1 = arr[1][0]; var y1 = arr[1][1]; var dx = x1 - x0; var dy = y1 - y0; // Loop to iterate over the points for ( var i = 0; i < n; i++) { var x = arr[i][0], y = arr[i][1]; if (dx * (y - y1) != dy * (x - x1)){ document.write( "NO" ); return false ; } } document.write( "YES" ); return true ; } // Driver Code // Array of points var arr = [[ 0, 0 ], [ 1, 1 ], [ 3, 3 ], [ 2, 2 ]]; var n = 4; // Function Call isStraightLinePossible(arr, n); </script> |
Output:
YES
Time Complexity: O(N)
Auxiliary Space: O(1)
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