FrogJmp

Dec 26, 2016


FrogJmp

A small frog wants to get to the other side of the road. The frog is currently located at position X and wants to get to a position greater than or equal to Y. The small frog always jumps a fixed distance, D.

Count the minimal number of jumps that the small frog must perform to reach its target.

Write a function:

class Solution { public int solution(int X, int Y, int D); }

that, given three integers X, Y and D, returns the minimal number of jumps from position X to a position equal to or greater than Y.

For example, given:

  X = 10
  Y = 85
  D = 30

the function should return 3, because the frog will be positioned as follows:

after the first jump, at position 10 + 30 = 40
after the second jump, at position 10 + 30 + 30 = 70
after the third jump, at position 10 + 30 + 30 + 30 = 100

Assume that:

X, Y and D are integers within the range [1..1,000,000,000];
X ≤ Y.

Complexity:

expected worst-case time complexity is O(1);
expected worst-case space complexity is O(1).

Solution:

class Solution {
    public int solution(int X, int Y, int D) {
        if(X==Y)
			return 0;
        return (int) Math.ceil((double)(Y-X)/D);
    }
}

FrogJmp | Java Tutorial