The problem states that x2 is strictly greater than x1.

Thus, if v2 == v1 then the answer is "NO".

Thus, both x2-x1 and v1-v2 are guaranteed positive and non-zero.

Absolute value should take care of the negatives.

Yes there are a number of conditions to check:

int num = x2 - x1;
int den = v1 - v2;
if (den == 0) cout << "NO";
else if (num % den != 0) cout << "NO";
else if (num > 0 && den < 0) cout << "NO";
else cout << "YES";

If x1 < x2 would not be stated, there are just two more conditions to check:

if (num == 0) cout << "YES";
else if (den == 0) cout << "NO";
else if (num % den != 0) cout << "NO";
else if (num < 0 && den > 0) cout << "NO";
else if (num > 0 && den < 0) cout << "NO";
else cout << "YES";

What is the logic (physics) behind this (x2 - x1) * (v2 - v1) < 0 condition?

can anyone explain the logic behind..... (x2-x1)%(v2-v1)==0

What if x2=x1 and v2=v1 ??

Can you please explain how did you do (x2 - x1) * (v2 - v1) < 0?

First you need to solve for y where there is no remainder as discussed above.

Then you need make sure the velocity is correct in that one of the kangaroos will eventually catch up to the other kangaroo.

// javascript solution 
	var remainder = (x1 - x2) % (v2 - v1) == 0;
	var properVelocity = ((x1-x2 > 0) && v2 > v1) || ((x1-x2 < 0) && v1 > v2);
	return  remainder && properVelocity ? 'YES': 'NO';

I found y didn't really need to be calculated if using a loop that performs the x1+v1 and x2+v2 calculation at the same time. That way you can do a simple check at the beginning of each calculation to see if their positions are equal

If the two kangaroos met, then there woulde be an integer n such that x1 + n * v1 = x2 + n * v2. Subtracting x1 from both sides, we get that n * v1 = x2 - x1 + n * v2. Then we subtract n * v2 from both sides and we get n * v1 - n * v2 = x2 - x1. Factoring out n from the left side of the equation, n * (v1 - v2) = (x2 - x1). We can the divide both sides by (v1 - v2) and we get (x2 - x1) / (v1 - v2). If this is an integer, then (x2 - x1) % (v1 - v2) has to equal 0.

I knew something was missing and it took me until here in the comments to actually understand what it was.

Because if one kangaroo takes a jump , the other one would have also taken as both jump at same time . So number of jumps for a paticular period of time remains the same . Hence , y is same for both cases .

No, because the 2nd kangaroo started ahead of the 1st one and will keep getting more ahead. However, this is not because v2 - v1 = 1. This is because x1 - x2 is negative while v2 - v1 is not, so it would take negative steps to get there, which can't happen.

public static void main(String[] args) { Scanner in = new Scanner(System.in); int x1 = in.nextInt(); int v1 = in.nextInt(); int x2 = in.nextInt(); int v2 = in.nextInt();

    if ((v1 == v2) 
        System.out.println("NO");
    else if (((x1 - x2) % (v2 - v1)) == 0 && ((x1 - x2) / (v2 - v1)) > 0) {
       System.out.println("YES");
   } else
       System.out.println("NO");


}

Let me explain how I do it. Let y1 be the total amounts of jumps that kangaroo 1 does so we have:

x1 + y1*v1 = S1, the distance that kangoroo 1 will get to after jumping y1 times.

Similarly, x2 + y2*v2 = S2, the distance that kangoroo 2 will get to after jumping y2 times.

Question: Given the starting locations and movement rates for each kangaroo, can you determine if they'll ever land at the same location at the same time?

so it means they will meet when S1 = S2 => x1 + y1*v1 = x2 + y2*v2, now we assume that both kangaroo did the same jump then met each other at the same location so the equation now is

x1 + y*v1 = x2 + y*v2 <=> y = (x1 - x2)/(v2 - v1) now can you guess what is the condtion for y is base on the above equation?

that is how I understand it but I may be wrong and sorry for my bad explaination, English is not my first lanquage.

Hey Python buddy! You don't need to check whether x2>x1, it's already given in the constraints.

This should work. Cheers!

if v1 > v2 and (x2-x1)%(v1-v2) == 0:
    print "YES
else:
    print "NO"

Let n = number of times each kangaroo hops to eventually land at the same exact distance. To see whether that does happen we need the following expression to be true:

x1 + (v1 * n) = x2 + (v2 * n)

We can check this by looking at the sample input 0, 3, 5, 2 for x1, v1, x2, v2 respectively.

0 + 3n = 5 + 2n n = 5

This means it takes the kangaroos jumping 5 times to land at the same distance:

0 > 3 > 6 > 9 > 12 > 15

5 > 7 > 9 > 11 > 13 > 15

Now that we know the equation is right, let's move n to one side:

(v1 * n) - (v2 * n) = x2 - x1

n(v1 - v2) = x2 - x1

n = (x2 - x1) / (v1 - v2)

We don't know what the value of n will be, but we know that it must be an integer. If the division solves for a whole integer, we know there is no remainder value. So, we use modular division

(x2 - x1) % (v1 - v2) == 0 (meaning no remainder value, and therefore the division solves for an integer)

1 line?

4 lines for input reading and the lenght of the println is higher than the maximun allowed by the standards so this final line needs to be splitted into two.

4 + 2 is 6 lines not 1.

I agree - It would be a cooler question. The question does have this note though:

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

You need to watch out for when v1 == v2 because then you will try modulus of 0. In this case or the more general case where v2 >= v1, the answer is always NO because the kangaroo just can't catch up.

The iteration is infinite and btw it meets at 12.

If the two kangaroos met, then there woulde be an integer n such that x1 + n * v1 = x2 + n * v2. Subtracting x1 from both sides, we get that n * v1 = x2 - x1 + n * v2. Then we subtract n * v2 from both sides and we get n * v1 - n * v2 = x2 - x1. Factoring out n from the left side of the equation, n * (v1 - v2) = (x2 - x1). We can the divide both sides by (v1 - v2) and we get (x2 - x1) / (v1 - v2). If this is an integer, then (x2 - x1) % (v1 - v2) has to equal 0.

There's a really nice explanation here: https://www.hackerrank.com/challenges/kangaroo/forum/comments/327371

There's a really nice explanation here

The problem asks whether they will meet at a particular point after same number of jumps. Distance moved after 'y' jumps kangaroo1 d1 = x1 + v1 * y. kangaroo2 d2 = x2 + v2 * y. 'y' must be same for both kangaroos. d1 = d2. x1 + v1 * y = x2 + v2 * y. rearanging the above equation we get y = (x2 - x1)/(v1 - v2). y is number of jumps which must be a positive integer. Therefore which implies that x2-x1 is divisible by v1-v2. This is how the equation has evolved.

Intersection is not always in the form of y = mx + b, this is a linear equation involving one variable (the x, or in our case the number of jumps). We could also have a linear equation with 2 variables, we could also have a mixture of linear and non-linear equations (such as an exponential equation and a linear equation and finding their point of intersection).

Honestly the only shorcut is improving upon math, otherwise you'll be limited by trying to remember certain solutions to equations and when to apply them, but you'll rarely know when NOT to apply them, or which ones to apply for a problem that's a bit more complex :(

A bit more explanation for this one, the Kangaroos jump at a linear rate (their veolcity/jump movement never changes, it's always v1 or v2). Therefore we have something along the lines of:

y = velocity * jumps + startLocation

We are already given velocity AND start location for each kangaroo. let's abbreviate using the actual equation variables (x1,x2,v1,v2), we get two separate equations for y (the total distance travelled after J jumps)

• y = v1 * J + x1
• y = v2 * J + x2

We want to set these equal to each other, this would give us a distance that they both share when their J variable's are equal to one another:

v1 * J + x1 = v2 * J + x2

And using our algebra we get the EQ described above in the other posters solution:

v1 * J - v2 * J + x1 - x2 = 0 Which can be written as: J(v1-v2) = x2 - x1 or (x2-x1) / (v1-v2) = J

So given 2 velocity values and 2 starting positions, like 4 = v1 and 2 = v2 and 0 = x1 and 4 = x2, we can evaluate:

• (4 - 0) / (4-2) = J
• 4 / 2 = J
• J = 2 (so at 2 jumps we should intersect)

Showing the values below of the jumps this is right (the far left value is the starting, arrows point to the next jump location):

• 0 -> 4 -> 8
• 4 -> 6 -> 8

And then of course, as the above states, we need to make sure the jump total is a whole number, otherwise we're intersecting in the air. Which is fine, but this problem doesn't really allow that. After all, given any 2 kangaroos if the "BACK" kangaroo is moving FASTER than the one in front of it, they will eventually collide. It's like 2 cars and the one in back is moving faster, if it doesn't slow down it will hit the one in front of it!

Hi. I just added an explanation to my original post to help answer your question.

HackerRank solutions.