Everybody loves a kangaroo right? Except maybe if you are Australian! Anyway in this post we shall talk both about snakes and mammals, or more specific Python and Kangaroos. I have started on learning a bit of Python and using HackerRank to learn my ways around. I have been solving a few puzzles by now. The ones I have solved so far, are no where near the complexity of Project Euler. Apart from the fact that quite a few of the project euler problems are actually represented at HackerRank, but with more test cases, so you have to write the code a little more generically. But more about that later, because I will likely solve a few of the Project Euler problems in Python while learning it.

Anyway, the first problem I solved, which might be worth spending a minute or two about, is the problem called Kangaroo.

The short description of the problem is as follows

There are two kangaroos on a number line ready to jump in the positive direction (i.e, toward positive infinity). The first kangaroo starts at location x

_{1}and moves at a rate of v_{1}meters per jump. The second kangaroo starts at location x_{2}and moves at a rate of v_{2}meters per jump. Given the starting locations and movement rates for each kangaroo, can you determine if they’ll ever landat the same location at the same time?

Input Format

A single line of four space-separated integers denoting the respective values of x_{1}, v_{1}, x_{2}and v_{2}

Note:The two kangaroos must land at the same locationafter making the same number of jumps.

## Simple algebraic solution for Kangaroo

Of course we could make a loop that terminates once the fastest kangaroo is in front. That would definitely work for us. However, we don’t need loops. Since we know that they must make the same number of jumps j. We can solve the equation

This can be rewritten as

If that have a positive solution integer solution then the kangaroos will meet. However, there is one special case we need to check as well. If the kangaroos have the same speed, then the divisor will end up zero, and then we will get an error. However, if they have the same speed, they only meet if they also have the same start location. So that is pretty easy to check for.

#!/bin/python3 import sys def kangaroo(x1, v1, x2, v2): #The case of same speed if v1 == v2 and x1 == x2: return 'YES' elif v1 == v2: return 'NO' x = (x2-x1) / (v1-v2) if (x == round(x) and x > 0): return 'YES' else: return 'NO' x1, v1, x2, v2 = input().strip().split(' ') x1, v1, x2, v2 = [int(x1), int(v1), int(x2), int(v2)] result = kangaroo(x1, v1, x2, v2) print(result)

This passes all the test cases for me. To be quite honest I have yet to figure out how to actually time and thereby compare different solutions. But I guess I will find out eventually.