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## Hackerrank - Kangaroo Solution

Beeze Aal

You are choreographing a circus show with various animals. For one act, you are given two kangaroos on a number line ready to jump in the positive direction (i.e, toward positive infinity).

• The first kangaroo starts at location  and moves at a rate of  meters per jump.
• The second kangaroo starts at location  and moves at a rate of  meters per jump.

You have to figure out a way to get both kangaroos at the same location at the same time as part of the show. If it is possible, return YES, otherwise return NO.

For example, kangaroo  starts at  with a jump distance  and kangaroo  starts at  with a jump distance of . After one jump, they are both at , (, ), so our answer is YES.

Function Description

Complete the function kangaroo in the editor below. It should return YES if they reach the same position at the same time, or NO if they don't.

kangaroo has the following parameter(s):

• x1, v1: integers, starting position and jump distance for kangaroo 1
• x2, v2: integers, starting position and jump distance for kangaroo 2

Input Format

A single line of four space-separated integers denoting the respective values of , , , and .

Constraints

Output Format

Print YES if they can land on the same location at the same time; otherwise, print NO.

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

Sample Input 0

0 2 5 3

Sample Output 0

NO

Explanation 0

The two kangaroos jump through the following sequence of locations:

From the image, it is clear that the kangaroos meet at the same location (number  on the number line) after same number of jumps ( jumps), and we print YES.

Sample Input 1

0 3 4 2

Sample Output 1

NO

Explanation 1

The second kangaroo has a starting location that is ahead (further to the right) of the first kangaroo's starting location (i.e., ). Because the second kangaroo moves at a faster rate (meaning ) and is already ahead of the first kangaroo, the first kangaroo will never be able to catch up. Thus, we print NO.

### Solution in Python

def kangaroo(x1, v1, x2, v2):
return "YES" if (v1 > v2) and not (x2 - x1) % (v2 - v1) else "NO"

x1,v1,x2,v2 = map(int,input().split())
result = kangaroo(x1, v1, x2, v2)
print(result)


Let us assume both kangaroos make y jumps.

The end position of kangaroo 1 after making y jumps will be

Initial position(k1) + Distance covered in y jumps (v1*y)

Which becomes

k1+v1*y

The end position of kangaroo 2 after making y jumps will be

k2+v2*y

For kangaroos to meet at same point. End position of both kangaroos must be equal.

k1+v1*y = k2+v2*y

Solving the above equation we get

-> k1+v1*y == k2+v2*y
-> k1-k2 == v2*y-v1*y
-> k1-k2 == (v2-v1)*y

Since k1-k2 is the product of (v2-v1) and some value y

Therefore k1-k2 must be divisible by v2-v1 for the two kangaroos to meet at same point.

(k1-k2) % (v2-v1) == 0

Another condition which needs to be satisfied is

v1>v2

It is given that k1<k2 which means kangaroo 1 lack behind kangaroo 2 therefor for the both kangaroos to meet at a same point. kangaroo 1 must be faster than kangaroo 2. i.e. v1>v2

Alternative solution using while loop (Not recommended)

def kangaroo(x1, v1, x2, v2):
if v1>v2:
while x1<x2:
x1+=v1
x2+=v2
return "YES" if x1==x2 else "NO"

x1,v1,x2,v2 = map(int,input().split())
result = kangaroo(x1, v1, x2, v2)
print(result)