CareerCup

You guys missed the most efficient solution. If the remainder units (k%N) is more than than (N+1)/2, Harry can choose to hop down 1 step of time to a position which is exactly N units, then hop up N units. That is more efficient than directly hop up 1 step at a time.

For example, N is 7 and there are 5 units left. The naive way is one unit up at a time and need 5 more steps. Since 5 is more than (7+1)/2 = 4, the more efficient way is hop down 2 units, make the remaining units 7, then hop up 7 units, total remaining steps is 3.

The answer is (k/n) + (((k%n)>((k+1)/2))? (n+1 - (k%n)) : (k%n))

if(k > n):
return (k/n) + ( k%n > n/2 ? (n - (k%n) + 1) : (k%n) );

if(k == n):
return 1;

if(k < n):
return k;

Simple O(1) solution :)

EDIT: edited based on observation by Anthony Mai's answer.

public int minimumHops(int k, int n) {
return (k/n + k%n);
}

For example : k = 68 and n = 10
Answer = 6 (N hop) + 8 (1 hop) = 14hops
k/n = 6 + k%n = 8 = 14

public int minimumHops(int k, int n) {
	return (k/n + k%n);
}

Example : k = 68 and n = 10
Answer : 6 (n hops : 60) + 8 (1 hops : 8) = 14hops
k/n (68/10 = 6) + k%n (68%10 = 8) = 14hops