Number of ordered points pair satisfying line equation
Given an array of n integers, slope of a line i. e., m and the intercept of the line i.e c, Count the number of ordered pairs(i, j) of points where i ? j, such that point (Ai, Aj) satisfies the line formed with given slope and intercept.
Note : The equation of the line is y = mx + c, where m is the slope of the line and c is the intercept
Examples :
Input : m = 1, c = 1, arr[] = [ 1, 2, 3, 4, 2 ]
Output : 5 ordered pointsExplanation : The equation of the line with given slope and intercept is : y = x + 1. The Number of pairs (i, j), for which (arri, arrj) satisfies the above equation of the line are : (1, 2), (1, 5), (2, 3), (3, 4), (5, 3).
Input : m = 2, c = 1, arr[] = [ 1, 2, 3, 4, 2, 5 ]
Output : 3 ordered points
Method 1 (Brute Force):
Generate all possible pairs (i, j) and check if a particular ordered pair (i, j) is such that, (arri, arrj) satisfies the given equation of the line y = mx + c, and i ? j. If the point is valid(a point is valid if the above condition is satisfied), increment the counter which stores the total number of valid points.
Implementation:
C++
// CPP code to count the number of ordered// pairs satisfying Line Equation#include <bits/stdc++.h>using namespace std;/* Checks if (i, j) is valid, a point (i, j) is valid if point (arr[i], arr[j]) satisfies the equation y = mx + c And i is not equal to j*/bool isValid(int arr[], int i, int j, int m, int c){ // check if i equals to j if (i == j) return false; // Equation LHS = y, and RHS = mx + c int lhs = arr[j]; int rhs = m * arr[i] + c; return (lhs == rhs);}/* Returns the number of ordered pairs (i, j) for which point (arr[i], arr[j]) satisfies the equation of the line y = mx + c */int findOrderedPoints(int arr[], int n, int m, int c){ int counter = 0; // for every possible (i, j) check // if (a[i], a[j]) satisfies the // equation y = mx + c for (int i = 0; i < n; i++) { for (int j = 0; j < n; j++) { // (firstIndex, secondIndex) // is same as (i, j) int firstIndex = i, secondIndex = j; // check if (firstIndex, // secondIndex) is a valid point if (isValid(arr, firstIndex, secondIndex, m, c)) counter++; } } return counter;}// Driver Codeint main(){ int arr[] = { 1, 2, 3, 4, 2 }; int n = sizeof(arr) / sizeof(arr[0]); // equation of line is y = mx + c int m = 1, c = 1; cout << findOrderedPoints(arr, n, m, c); return 0;} |
Java
// Java code to find number of ordered// points satisfying line equationimport java.io.*;public class GFG { // Checks if (i, j) is valid, // a point (i, j) is valid if // point (arr[i], arr[j]) // satisfies the equation // y = mx + c And // i is not equal to j static boolean isValid(int []arr, int i, int j, int m, int c) { // check if i equals to j if (i == j) return false; // Equation LHS = y, // and RHS = mx + c int lhs = arr[j]; int rhs = m * arr[i] + c; return (lhs == rhs); } /* Returns the number of ordered pairs (i, j) for which point (arr[i], arr[j]) satisfies the equation of the line y = mx + c */ static int findOrderedPoints(int []arr, int n, int m, int c) { int counter = 0; // for every possible (i, j) check // if (a[i], a[j]) satisfies the // equation y = mx + c for (int i = 0; i < n; i++) { for (int j = 0; j < n; j++) { // (firstIndex, secondIndex) // is same as (i, j) int firstIndex = i, secondIndex = j; // check if (firstIndex, // secondIndex) is a // valid point if (isValid(arr, firstIndex, secondIndex, m, c)) counter++; } } return counter; } // Driver Code public static void main(String args[]) { int []arr = { 1, 2, 3, 4, 2 }; int n = arr.length; // equation of line is y = mx + c int m = 1, c = 1; System.out.print( findOrderedPoints(arr, n, m, c)); }}// This code is contributed by// Manish Shaw (manishshaw1) |
Python3
# Python code to count the number of ordered# pairs satisfying Line Equation# Checks if (i, j) is valid, a point (i, j)# is valid if point (arr[i], arr[j])# satisfies the equation y = mx + c And # i is not equal to jdef isValid(arr, i, j, m, c) : # check if i equals to j if (i == j) : return False # Equation LHS = y, and RHS = mx + c lhs = arr[j]; rhs = m * arr[i] + c return (lhs == rhs)# Returns the number of ordered pairs# (i, j) for which point (arr[i], arr[j])# satisfies the equation of the line # y = mx + c */def findOrderedPoints(arr, n, m, c) : counter = 0 # for every possible (i, j) check # if (a[i], a[j]) satisfies the # equation y = mx + c for i in range(0, n) : for j in range(0, n) : # (firstIndex, secondIndex) # is same as (i, j) firstIndex = i secondIndex = j # check if (firstIndex, # secondIndex) is a valid point if (isValid(arr, firstIndex, secondIndex, m, c)) : counter = counter + 1 return counter# Driver Codearr = [ 1, 2, 3, 4, 2 ]n = len(arr)# equation of line is y = mx + cm = 1c = 1print (findOrderedPoints(arr, n, m, c))# This code is contributed by Manish Shaw# (manishshaw1) |
C#
// C# code to find number of ordered// points satisfying line equationusing System;class GFG { // Checks if (i, j) is valid, // a point (i, j) is valid if // point (arr[i], arr[j]) // satisfies the equation // y = mx + c And // i is not equal to j static bool isValid(int []arr, int i, int j, int m, int c) { // check if i equals to j if (i == j) return false; // Equation LHS = y, // and RHS = mx + c int lhs = arr[j]; int rhs = m * arr[i] + c; return (lhs == rhs); } /* Returns the number of ordered pairs (i, j) for which point (arr[i], arr[j]) satisfies the equation of the line y = mx + c */ static int findOrderedPoints(int []arr, int n, int m, int c) { int counter = 0; // for every possible (i, j) check // if (a[i], a[j]) satisfies the // equation y = mx + c for (int i = 0; i < n; i++) { for (int j = 0; j < n; j++) { // (firstIndex, secondIndex) // is same as (i, j) int firstIndex = i, secondIndex = j; // check if (firstIndex, // secondIndex) is a valid point if (isValid(arr, firstIndex, secondIndex, m, c)) counter++; } } return counter; } // Driver Code public static void Main() { int []arr = { 1, 2, 3, 4, 2 }; int n = arr.Length; // equation of line is y = mx + c int m = 1, c = 1; Console.Write(findOrderedPoints(arr, n, m, c)); }}// This code is contributed by// Manish Shaw (manishshaw1) |
PHP
<?php// PHP code to count the // number of ordered pairs// satisfying Line Equation/* Checks if (i, j) is valid,a point (i, j) is valid if point (arr[i], arr[j]) satisfies the equation y = mx + c And i is not equal to j*/function isValid($arr, $i, $j, $m, $c){ // check if i equals to j if ($i == $j) return false; // Equation LHS = y, and // RHS = mx + c $lhs = $arr[$j]; $rhs = $m * $arr[$i] + $c; return ($lhs == $rhs);}/* Returns the number of ordered pairs (i, j) for which point (arr[i], arr[j]) satisfies the equation of the line y = mx + c */function findOrderedPoints($arr, $n, $m, $c){ $counter = 0; // for every possible (i, j) // check if (a[i], a[j]) // satisfies the equation // y = mx + c for ($i = 0; $i < $n; $i++) { for ($j = 0; $j < $n; $j++) { // (firstIndex, secondIndex) // is same as (i, j) $firstIndex = $i; $secondIndex = $j; // check if (firstIndex, // secondIndex) is a valid point if (isValid($arr, $firstIndex, $secondIndex, $m, $c)) $counter++; } } return $counter;}// Driver Code$arr = array( 1, 2, 3, 4, 2 );$n = count($arr);// equation of line // is y = mx + c$m = 1; $c = 1;echo (findOrderedPoints($arr, $n, $m, $c));// This code is contributed by // Manish Shaw (manishshaw1)?> |
Javascript
<script> // JavaScript code to find number of ordered // points satisfying line equation // Checks if (i, j) is valid, // a point (i, j) is valid if // point (arr[i], arr[j]) // satisfies the equation // y = mx + c And // i is not equal to j function isValid(arr, i, j, m, c) { // check if i equals to j if (i == j) return false; // Equation LHS = y, // and RHS = mx + c var lhs = arr[j]; var rhs = m * arr[i] + c; return lhs == rhs; } /* Returns the number of ordered pairs (i, j) for which point (arr[i], arr[j]) satisfies the equation of the line y = mx + c */ function findOrderedPoints(arr, n, m, c) { var counter = 0; // for every possible (i, j) check // if (a[i], a[j]) satisfies the // equation y = mx + c for (var i = 0; i < n; i++) { for (var j = 0; j < n; j++) { // (firstIndex, secondIndex) // is same as (i, j) var firstIndex = i, secondIndex = j; // check if (firstIndex, // secondIndex) is a valid point if (isValid(arr, firstIndex, secondIndex, m, c)) counter++; } } return counter; } // Driver Code var arr = [1, 2, 3, 4, 2]; var n = arr.length; // equation of line is y = mx + c var m = 1, c = 1; document.write(findOrderedPoints(arr, n, m, c)); // This code is contributed by rdtank. </script> |
Output
5
Complexity Analysis:
- Time Complexity : O(n2)
- Auxiliary Space: O(1)
Method 2 (Efficient) :
Given a x coordinate of a point, for each x there is a unique value of y and the value of y is nothing but m * x + c. So, for each possible x coordinate of the array arr, calculate how many times the unique value of y which satisfies the equation of the line occurs in that array. Store count of all the integers of array, arr in a map. Now, for each value, arri, add to the answer, the number of occurrences of m * arri + c. For a given i, m * a[i] + c occurs x times in the array, then, add x to our counter for total valid points, but need to check that if a[i] = m * a[i] + c then, it is obvious that since this occurs x times in the array then one occurrence is at the ith index and rest (x – 1) occurrences are the valid y coordinates so add (x – 1) to our points counter.
Implementation:
C++
// CPP code to find number of ordered// points satisfying line equation#include <bits/stdc++.h>using namespace std;/* Returns the number of ordered pairs (i, j) for which point (arr[i], arr[j]) satisfies the equation of the line y = mx + c */int findOrderedPoints(int arr[], int n, int m, int c){ int counter = 0; // map stores the frequency // of arr[i] for all i unordered_map<int, int> frequency; for (int i = 0; i < n; i++) frequency[arr[i]]++; for (int i = 0; i < n; i++) { int xCoordinate = arr[i]; int yCoordinate = (m * arr[i] + c); // if for a[i](xCoordinate), // a yCoordinate exists in the map // add the frequency of yCoordinate // to the counter if (frequency.find(yCoordinate) != frequency.end()) counter += frequency[yCoordinate]; // check if xCoordinate = yCoordinate, // if this is true then since we only // want (i, j) such that i != j, decrement // the counter by one to avoid points // of type (arr[i], arr[i]) if (xCoordinate == yCoordinate) counter--; } return counter;}// Driver Codeint main(){ int arr[] = { 1, 2, 3, 4, 2 }; int n = sizeof(arr) / sizeof(arr[0]); int m = 1, c = 1; cout << findOrderedPoints(arr, n, m, c); return 0;} |
Java
// Java program to find number of ordered// points satisfying line equationimport java.io.*;import java.util.*;class GFG { /* Returns the number of ordered pairs (i, j) for which point (arr[i], arr[j]) satisfies the equation of the line y = mx + c */ static int findOrderedPoints(int arr[], int n, int m, int c){ int counter = 0; // map stores the frequency // of arr[i] for all i HashMap<Integer,Integer> frequency = new HashMap<>(); for(int i=0;i<n;i++){ if(frequency.get(arr[i])==null){ frequency.put(arr[i],1); }else{ frequency.put(arr[i],frequency.get(arr[i])+1); } } for (int i = 0; i < n; i++) { int xCoordinate = arr[i]; int yCoordinate = (m * arr[i] + c); // if for a[i](xCoordinate), // a yCoordinate exists in the map // add the frequency of yCoordinate // to the counter if (frequency.get(yCoordinate) !=null) counter += frequency.get(yCoordinate); // check if xCoordinate = yCoordinate, // if this is true then since we only // want (i, j) such that i != j, decrement // the counter by one to avoid points // of type (arr[i], arr[i]) if (xCoordinate == yCoordinate) counter--; } return counter; } //Driver Code public static void main (String[] args) { int arr[] = { 1, 2, 3, 4, 2 }; int n=5,m=1,c=1; System.out.println(findOrderedPoints(arr,n,m,c)); }}//This code is contributed by shruti456rawal |
Python3
# Python code to find number of ordered# points satisfying line equation""" Returns the number of ordered pairs (i, j) for which point (arr[i], arr[j]) satisfies the equation of the line y = mx + c """def findOrderedPoints(arr, n, m, c): counter = 0 # map stores the frequency # of arr[i] for all i frequency = dict() for i in range(n): if(arr[i] in frequency): frequency[arr[i]] += 1 else: frequency[arr[i]] = 1 for i in range(n): xCoordinate = arr[i] yCoordinate = (m * arr[i] + c) # if for a[i](xCoordinate), # a yCoordinate exists in the map # add the frequency of yCoordinate # to the counter # console.log(counter); if (yCoordinate in frequency): counter += frequency[yCoordinate] # check if xCoordinate = yCoordinate, # if this is true then since we only # want (i, j) such that i != j, decrement # the counter by one to avoid points # of type (arr[i], arr[i]) # console.log(counter); if (xCoordinate == yCoordinate): counter -= 1 return counter# Driver Codearr = [1, 2, 3, 4, 2]n = len(arr)m = 1c = 1print(findOrderedPoints(arr, n, m, c))# The code is contributed by Saurabh Jaiswal |
Javascript
// JavaScript code to find number of ordered// points satisfying line equation/* Returns the number of ordered pairs (i, j) for which point (arr[i], arr[j]) satisfies the equation of the line y = mx + c */function findOrderedPoints(arr, n, m, c){ let counter = 0; // map stores the frequency // of arr[i] for all i let frequency = new Map(); for (let i = 0; i < n; i++){ if(frequency.has(arr[i])){ frequency.set(arr[i], frequency.get(arr[i]) + 1); } else{ frequency.set(arr[i], 1); } } for (let i = 0; i < n; i++) { let xCoordinate = arr[i]; let yCoordinate = (m * arr[i] + c); // if for a[i](xCoordinate), // a yCoordinate exists in the map // add the frequency of yCoordinate // to the counter // console.log(counter); if (frequency.has(yCoordinate)) counter += frequency.get(yCoordinate); // check if xCoordinate = yCoordinate, // if this is true then since we only // want (i, j) such that i != j, decrement // the counter by one to avoid points // of type (arr[i], arr[i]) // console.log(counter); if (xCoordinate == yCoordinate) counter--; } return counter;}// Driver Codelet arr = [1, 2, 3, 4, 2];let n = arr.length;let m = 1, c = 1;console.log(findOrderedPoints(arr, n, m, c));// The code is contributed by Nidhi goel |
C#
using System;using System.Collections.Generic;public static class GFG { // C# code to find number of ordered // points satisfying line equation /* Returns the number of ordered pairs (i, j) for which point (arr[i], arr[j]) satisfies the equation of the line y = mx + c */ public static int findOrderedPoints(int[] arr, int n, int m, int c) { int counter = 0; // map stores the frequency // of arr[i] for all i Dictionary<int, int> frequency = new Dictionary<int, int>(); for (int i = 0; i < n; i++) { if (frequency.ContainsKey(arr[i])) { var val = frequency[arr[i]]; frequency.Remove(arr[i]); frequency.Add(arr[i], val + 1); } else { frequency.Add(arr[i], 1); } } for (int i = 0; i < n; i++) { int xCoordinate = arr[i]; int yCoordinate = (m * arr[i] + c); // if for a[i](xCoordinate), // a yCoordinate exists in the map // add the frequency of yCoordinate // to the counter if (frequency.ContainsKey(yCoordinate)) { counter += frequency[yCoordinate]; } // check if xCoordinate = yCoordinate, // if this is true then since we only // want (i, j) such that i != j, decrement // the counter by one to avoid points // of type (arr[i], arr[i]) if (xCoordinate == yCoordinate) { counter--; } } return counter; } // Driver Code public static void Main() { int[] arr = { 1, 2, 3, 4, 2 }; int n = arr.Length; int m = 1; int c = 1; Console.Write(findOrderedPoints(arr, n, m, c)); }}// The code is contributed by Aarti_Rathi |
Output
5
Time Complexity: O(n), where n is the size of the given array.
Auxiliary Space: O(n) because it is using extra space for unoredered_map frequency.
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