Whoa.. Bam!! Didn't think that. Still need to take care of things like zero and negative.

Doesn't handle the case where 2 starts ahead and 1 never catches up (e.g., 0 2 5 3). You need to ensure that the sign of the position difference and the sign of the velocity difference are opposite, i.e., the velocity will overcome the position difference rather than increase it:

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

This also short-circuits a divide-by-zero error if the velocity difference is zero (e.g., 43 2 70 2).

can you please explain, how the value of "y" is calculated...

it is too perfect !!!!

Simpler than my algo..Thanks for sharing.. Thumbs Up !!

on theory if (x1 - x2) % (v2 - v1) == 0 is good, if v2-v1 equal to one will fail, and if condition is if (v1 == v2 && x1 != x2) or (x2 > x1 && v2 > v1) or (x1 > x2 && v1 > v2) then don't need waste time calculate it, the answer will be no

Hello umm i solved it in a different way using arrays but it was way more tedious than this mod(%) thing. I just checked it out and it works much much much better than mine. Its just that how did u get to that thing. I mean did u read any formulae or anything coz i really wanna get this. Just a little background i have been coding since 2 months(I am a first year student and have learnt java in my school and thats all) and I am practicing after one whole year . So i would be grateful if someone really guide me on this. Thanks in advance :)

wrong for given case( 0 2 5 3) and many others....

The condition should be (V2 < V1) && (X2 - X1) % (V1 - V2) == 0 :)

not true,,,,,,,

in second case :0 2 5 3 in this case x1=0,x2=5,v1=2,v2=3 so (x1 - x2) % (v2 - v1) == 0 as -5%1==0, so this is not correct

Yes we just need a couple of check cases to deal with negatives and zeros

why y is same for two jumps ? why not y1 y2 ?

What if (v2-v1) == 1? For example, if x1=0, v1 =3 , x2 = 100 and v2= 4, will each other meet?

112 9563 8625 244 this is one of the test case and according to your logic the answer will be NO instead of YES which is the right answer!

Can you provide the name of the equation which you applied and solved? How can I learn about this... Please any link from wikipedia or any other website if possible.

Using this algorithm and additional tweaks, here is the python 2 code:

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

how can you solve these equation? please can you explain it.

Java8 1 line:

int b = in.nextInt();
        int a = in.nextInt();
        int d = in.nextInt();
        int c = in.nextInt();
        
        System.out.println( b == d || (a-c) != 0 && (((d-b) / (a-c))*a)+b > Math.max(b, d) && ((b-d) % (a-c)) == 0 ? "YES" : "NO" );

But do the nummber of jumps need to be the same for both the Kangaroos? I feel that x1+a*v1=x2+b*v2 relates more closely to the problem. Thoughts?

but what is the limit of y? how to identify the limit of jumps from the given problem?

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(); int a = x1 - x2; int b = v1 - v2; String res = "NO"; if(a*b < 0 && a%b == 0){ res = "YES"; } System.out.println(res); }

Please explain a bit more.

i think this will help....

include

include

using namespace std;

int main(){ int x1; int v1; int x2; int v2; cin >> x1 >> v1 >> x2 >> v2; if((x1>x2&&v1v2)){ double remainder; remainder = abs(x1-x2)%abs(v2-v1); if (remainder==0.0){ cout<<"YES"; return 0; }

}
cout<<"NO";   
return 0;

}

Yup. Producing a solution like this...

int num = x1 - x2;
int den = v2 - v1;

System.out.println((den == 0 || num % den != 0 || num/den < 0) ? "NO":"YES");

Yup. cracked that !

The case test 10 is giving a runtime error, else if quite fine, tried different codes... even 'v1 = x2 + y * v2 where "y" is number of jumps... so if (x1 - x2) % (v2 - v1) == 0 ' this

good one man

Your idea helped me solve this. Thank You! :D

really amazing

excellent! this really helped.

woahh thank you! i spent so long on this question

My first attempt was to brute force it with a loop, but JS will timeout. I came to a similar conclusion to fix it, though one or two steps less elegant. Nice deduction!

What do you guys thing about this? This idea came to me on my own lol!

#include <iostream>

using namespace std;
/*Kangaroo functions:
Kangaroo 1 = v1 * n + c1
Kangaroo 2 = v2 * n + c2
No matter how different c is, kangaroo
with biggest speed will always catch up
also v*n is more important than c
cos v*n has higher degree
P.S. I forgot about NOTE ._.*/

int main()
{
    int c1, v1, c2, v2;
    cin >> c1 >> v1 >> c2 >> v2; // below: division by 0 is bad!
    int meet_step = v1-v2 != 0 ? (0 - (c1 - c2))/(v1 - v2) : 0;
    int meet_mod = v1-v2 != 0 ? (0 - (c1 - c2))%(v1 - v2) : 1;
    if(meet_step > 0 && meet_mod == 0)
        cout << "YES";// only if step where our kangaroos
    else   // meet is positive AND is an integer, they meet
        cout << "NO"; //on the ground, not in the air
    
    return 0;
}

this fails for this case

0 3 4 2

the equation states that the kanguru will never match the same point but output is stating that it should collide

for 1: 0 3 6 9

for 2: 4 6

so they will jump for the same x but never at the same time seems like a bug on the problem definition

Great!!!. So simple. Thanks.

PHP version that worked for me if((v1 && x2)||(x2 && v2)) return "NO"; elseif(x2 || v2) return "NO"; elseif((x2) % (v1) == 0) return "YES"; return "NO";

Ah so if the difference in origin is divisible by the difference in velocity, then at some point they will meet. Awesome. Had to sketch that one out

The condition is working but I don't understand the principle behind the (x1 - x2) % (v2 - v1) == 0 equation. Can anyone help me?

Actually that didn't strike :-)

Awesome

"(x1 - x2) % (v2 - v1) == 0" ??. Can you please explain this??

how these maths logics are finding not a good student in maths lol.

I had to add in a line of "return "NO" if v2 == v1" to avoid runtime divide by 0 errors, but other than that, this worked great.

What equation is this? How did you know to use that equation? Thanks!

Nice way to think about it

can you explain how x1 + y * v1 = x2 + y * v2 is converted into (x1 - x2) % (v2 - v1) == 0

Wow thatz really brilliant logic!!!

by this formula i got wrong result

How you can define using this formula? x1 + y * v1 Why x1 must be add by y and multiply by v1?

Thank you

that a good aproach that i have tried but the problem is that it does not have the notion of time. the problem says that the kangoroos meet at the same time (or after the same number of jumps) and not just meet. im still working on that part if anyone has any ideas :)

Its like Arithmatic Progression. Like we have to find where two Ap meet But please also justify How u take the number of jumps equal?? Take case as 2,6,8..... and 0,3,6,9.....

Also Take the case when number of jumps are not equal

thank you...you just made it easy!

How did mathematically solve x1 + y * v1 = x2 + y * v2 to (x1-x2)%(v2-v1) == 0?

static String kangaroo(int x1, int v1, int x2, int v2) {
        for(int i = 1; i > 0; i++){
        if(x1+i*v1 == x2+i*v2){
            return "YES";
        }else if(x1+i*v1 > x2+i*v2) {
            return "NO";
        }
    }
        return null;
    }

This will perform infinite loops until kangarooA > kangarooB, the result will be NO. Cuz at the beginning, kangarooA's starting location always less than kangarooB's location. Else, if kangarooA equals kangarooB (as Mily94 formula), the result we be YES. But this isn't optimal code, it take over 1s to show result from execution.

bool diff;
string out = "NO";
for(int x = 0; x < MAX; x++){
    diff = ((v1*x+x1)-(v2*x+x2)) == 0;
    //cout << ((v1*x+x1)-(v2*x+x2)) << endl;
    if(diff){
        out = "YES";
        break;
    }
}
return out;

    here is another way to do it, MAX is 10000 since it will never go above that.

def kangaroo(x1, v1, x2, v2):
    if (v2 >= v1 or (x2-x1)%(v1-v2) != 0):
        return 'NO'
    return 'YES'

We first check if v2 (second kangaroo's jump) is bigger or equal to v1 (first kangaroo's jump) because there's a constraint that x2 (starting position of the second kangaroo) has to be bigger than x1 (starting position of the first kangaroo). It means that for first kangaroo to reach the second one it has to have bigger jump than the second kangaroo. If the second kangaroo has bigger or equal jump, the first will never catch it, therefore the statement is satisfied and we return a 'NO'.

If the first condition is not satisfied, we check if the combination of starting positions and steps makes it impossible for the first kangaroo to reach the second. If it indeed is impossible (if its modulus doesn't equal zero) we again return a 'NO'.

If none of these happen, we return a 'YES'.

I could only think of this bit but thanks for helping brothah... Appreciate it...

if((x2+v2)%v1 == 0)
            cout<<"YES";

Not quite. If y is negative your solution can fail, depending on how the '%' is implemented.

Hi, I think your equation is wrong ... Actually you take "y" number of jump as a common factor for both kangaroos, but number of jumps will be different for kangaroo1 and kangaroo2 so we should have y1, y2 number of jump for each kangaroo, am I right ?

This is a great writeup, thank you!

Really great explanantion! Thanks!

That's a very clear & helpful explanation. Thank you.

clever

very good explanation.Thank you :)

this has got no where near enough love.

where can I get better at recognising these sorts of patterns? or is it just a matter of knowing that when something intersects you have to use y = mx + b ?

eitherway, nice one! (I just cheated:)

        if(x1 < x2 && v1 < v2) return "NO";
        
        while(x1 <= x2){
            x1 += v1;
            x2 += v2;
            if(x1 == x2)return "YES";
        }
        return "NO";

Fabulous write-up. Thanks!

wooohooo needed this thank ya!

This is a brilliant write up, thanks for this.

Thank you!