We use cookies to ensure you have the best browsing experience on our website. Please read our cookie policy for more information about how we use cookies.
Number Line Jumps
Number Line Jumps
+ 0 comments JavaScript with O(n) complexity
function kangaroo(x1, v1, x2, v2) { for(let i=0;i<=10000;i++){ if(x1 === x2){ return 'YES' } x1+=v1 x2+=v2 } return 'NO' }
+ 0 comments PHP: wanted to cover all possibilities :P
function kangaroo($x1, $v1, $x2, $v2) { // Write your code here $pk1 = $x1; //position kangaroo No1 $pk2 = $x2; //position kangaroo No2 if ($x1 == $x2 && $v1 != $v2 or $x1 != $x2 && $v1 == $v2 or $x1 > $x2 && $v1 > $v2 or $x1 < $x2 && $v1 < $v2) { return "NO"; }else if ($x1 < $x2) { for ($t=1; $pk1<=$pk2; $t++) { if (($pk1 +($v1*$t)) == ($pk2 + ($v2*$t))) { return "YES"; } else if (($pk1 +($v1*$t)) > ($pk2 + ($v2*$t))) { return "NO"; } } }else if ($x2 < $x1) { for ($t=1; $pk2<=$pk1; $t++) { if (($pk1 +($v1*$t)) == ($pk2 + ($v2*$t))) { return "YES"; } else if (($pk1 +($v1*$t)) < ($pk2 + ($v2*$t))) { return "NO"; } } } else return "NO"; }
+ 0 comments def kangaroo(x1, v1, x2, v2): if ((x1 > x2 and v1 >= v2) or (x2 > x1 and v2 >= v1)) or ((x1 - x2) % (v2-v1) != 0) : return 'NO' return 'YES'
+ 0 comments def kangaroo(x1, v1, x2, v2): # Image the time in an x axis # and position in a y axis # kangaroo movements can be represented by two straight lines if v1 == v2: if x1 == x2: # They are always moving # in the same locations # at the same time return "YES" else: # Same velocity but different initial positions return "NO" num = x2 - x1 den = v1 - v2 if num / den < 0: # Intersection of the two lines happens before x=0 (initial time) return "NO" if num % den == 0: # Intersection happens in a specific location return "YES" # Intersection happens, but not in a quantized world made by jumps return "NO"
+ 0 comments My solution in Java:
public static String kangaroo(int x1, int v1, int x2, int v2) { return ((v1-v2<=0) || (x2-x1)%(v1-v2)!=0) ? "NO" : "YES"; }
Load more conversations
Sort 3392 Discussions, By:
Please Login in order to post a comment