Algorithm
Problem Name: Mathematics -
https://www.hackerrank.com/challenges/find-point/problem?isFullScreen=true
In this HackerRank in Mathematics -
Consider two points, p = (px,py) and q = (qx,qy) . We consider the inversion or point reflection, r = (rx,ry), of point p across point q to be a 180 degree rotation of point p around q.
Given n sets of points p and q , find r for each pair of points and print two space-separated integers denoting the respective values of rx and ry on a new line.
Function Description
Complete the findPoint function in the editor below.
findPoint has the following parameters:
- int px, py, qx, qy: x and y coordinates for points p and q.
Returns
- int[2]: x and y coordinates of the reflected point r
Input Format
The first line contains an integer, n, denoting the number of sets of points.
Each of the n subsequent lines contains four space-separated integers that describe the respective values of px ,py , qx and qy defining points p = (px,py) and q = (qx,qy)
Constraints
- 1 <= n <=15
- -100 <= px ,py , qx , qy <= 100
Sample Input
2
0 0 1 1
1 1 2 2
Sample Output
2 2
3 3
Explanation
The graphs below depict points p,q and r for the n = 2 points given as Sample Input:
Code Examples
#1 Code Example with C Programming
Code -
C Programming
#include <stdio.h>
int main()
{
int t;
scanf("%d",&t);
while(t--)
{
int px,py,qx,qy,rx,ry;
scanf("%d%d%d%d",&px,&py,&qx,&qy);
rx=2*qx-px;
ry=2*qy-py;
printf("%d %d\n",rx,ry);
}
return 0;
}
#2 Code Example with C++ Programming
Code -
C++ Programming
#include <bits/stdc++.h>
using namespace std;
int main()
{
int T, a, b, c, d;
cin >> T;
for(int i = 1; i < = T; i++){
cin >> a >> b >> c >> d;
cout << (2 * c) - a << " " << (2 * d) - b << endl;
}
return 0;
}
#3 Code Example with Java Programming
Code -
Java Programming
import java.io.*;
import java.util.*;
import java.text.*;
import java.math.*;
import java.util.regex.*;
public class Solution {
public static void main(String[] args) {
int n, px, py, qx, qy;
Scanner sc = new Scanner(System.in);
n = sc.nextInt();
for (int i = 0; i < n; i++) {
px = sc.nextInt();
py = sc.nextInt();
qx = sc.nextInt();
qy = sc.nextInt();
System.out.println(((qx-px)+qx) + " " + ((qy-py)+qy));
}
}
}
#4 Code Example with Javascript Programming
Code -
Javascript Programming
'use strict';
const fs = require('fs');
process.stdin.resume();
process.stdin.setEncoding('utf-8');
let inputString = '';
let currentLine = 0;
process.stdin.on('data', function(inputStdin) {
inputString += inputStdin;
});
process.stdin.on('end', function() {
inputString = inputString.split('\n');
main();
});
function readLine() {
return inputString[currentLine++];
}
/*
* Complete the 'findPoint' function below.
*
* The function is expected to return an INTEGER_ARRAY.
* The function accepts following parameters:
* 1. INTEGER px
* 2. INTEGER py
* 3. INTEGER qx
* 4. INTEGER qy
*/
function findPoint(px, py, qx, qy) {
let first = qx - px + qx,
last = qy - py + qy;
return [first, last];
}
function main() {
const ws = fs.createWriteStream(process.env.OUTPUT_PATH);
const n = parseInt(readLine().trim(), 10);
for (let nItr = 0; nItr < n; nItr++) {
const firstMultipleInput = readLine().replace(/\s+$/g, '').split(' ');
const px = parseInt(firstMultipleInput[0], 10);
const py = parseInt(firstMultipleInput[1], 10);
const qx = parseInt(firstMultipleInput[2], 10);
const qy = parseInt(firstMultipleInput[3], 10);
const result = findPoint(px, py, qx, qy);
ws.write(result.join(' ') + '\n');
}
ws.end();
}
#5 Code Example with C# Programming
Code -
C# Programming
using System.CodeDom.Compiler;
using System.Collections.Generic;
using System.Collections;
using System.ComponentModel;
using System.Diagnostics.CodeAnalysis;
using System.Globalization;
using System.IO;
using System.Linq;
using System.Reflection;
using System.Runtime.Serialization;
using System.Text.RegularExpressions;
using System.Text;
using System;
class Result
{
/*
* Complete the 'findPoint' function below.
*
* The function is expected to return an INTEGER_ARRAY.
* The function accepts following parameters:
* 1. INTEGER px
* 2. INTEGER py
* 3. INTEGER qx
* 4. INTEGER qy
*/
public static List < int> findPoint(int px, int py, int qx, int qy) =>
new List<int>() { -(px - qx) + qx, -(py - qy) + qy };
}
class Solution
{
public static void Main(string[] args)
{
TextWriter textWriter = new StreamWriter(@System.Environment.GetEnvironmentVariable("OUTPUT_PATH"), true);
int n = Convert.ToInt32(Console.ReadLine().Trim());
for (int nItr = 0; nItr < n; nItr++)
{
string[] firstMultipleInput = Console.ReadLine().TrimEnd().Split(' ');
int px = Convert.ToInt32(firstMultipleInput[0]);
int py = Convert.ToInt32(firstMultipleInput[1]);
int qx = Convert.ToInt32(firstMultipleInput[2]);
int qy = Convert.ToInt32(firstMultipleInput[3]);
List < int> result = Result.findPoint(px, py, qx, qy);
textWriter.WriteLine(String.Join(" ", result));
}
textWriter.Flush();
textWriter.Close();
}
}
#6 Code Example with Python Programming
Code -
Python Programming
#!/bin/python3
import math
import os
import random
import re
import sys
#
# Complete the 'findPoint' function below.
#
# The function is expected to return an INTEGER_ARRAY.
# The function accepts following parameters:
# 1. INTEGER px
# 2. INTEGER py
# 3. INTEGER qx
# 4. INTEGER qy
#
def findPoint(px, py, qx, qy):
p2q_x = qx - px
p2q_y = qy - py
r_pt = [qx + p2q_x, qy + p2q_y]
return r_pt
if __name__ == '__main__':
fptr = open(os.environ['OUTPUT_PATH'], 'w')
n = int(input().strip())
for n_itr in range(n):
first_multiple_input = input().rstrip().split()
px = int(first_multiple_input[0])
py = int(first_multiple_input[1])
qx = int(first_multiple_input[2])
qy = int(first_multiple_input[3])
result = findPoint(px, py, qx, qy)
fptr.write(' '.join(map(str, result)))
fptr.write('\n')
fptr.close()