def kangaroo(x1, v1, x2, v2):
    counter = 0
    while counter < 10000:
        x1+=v1
        x2+=v2
        counter+=1
        if x1 == x2:
            return "YES"
            break
        elif counter == 10000:
            return "NO"

i limit permutation on python

Typescript: if(v1 <= v2) return "NO"; const hasSameLocation = (x2 - x1)%(v1 - v2) === 0; return hasSameLocation ? "YES" : "NO";

java

    public static String kangaroo(int x1, int v1, int x2, int v2) {
        double time = (double)(x1- x2) / (double)(v2 - v1);
        return time >= 0 && time % 1 == 0 ? "YES" : "NO";
    }

function kangaroo(x1, v1, x2, v2) { // Write your code here

// Debug
// 2081 8403 9107 8400

let answer = "";
let count = 0;

let number  = (x1 + v1 + x2 + v2) * 10

for( let i = 0; i <= number; i++){

  count += x1 + (i * v1) ===  x2 + (i * v2) ? 1 : 0;


}

answer = count > 0 ? "YES" : "NO"


return answer

}

Typescript solution:

We can compose the problem as an algebraic expression: x1 + (j * v1) = x2 + (j * v2) where j is the number of jumps. The result line below is reducing the above formula to solve for j. For j to be valid in our problem, it has to be a positive whole number which we check within our if statement.

let result = (x2 - x1) / (v1 - v2);

if (result > 0 && result % 1 === 0)
    return "YES";
else
    return "NO";