[Codility Lessons] FrogJmp(6/n)

[Lesson3] Time Complexity

FrogJmp : Count minimal number of jumps from position X to Y.

---

Java Solution

class Solution {
    public int solution(int X, int Y, int D) {
        // frog가 가야할 거리 계산
        int n = Y - X;
        // frog가 Jump하는 수 계산
        int count = n/D;
        // n/D 나머지가 있을 경우 +1
        if(n%D > 0){
            count = count+1;
        }
        return count;
    }
}

---

PHP Solution

function solution($X, $Y, $D) {
    $n = $Y - $X;
    $count = $n/$D;
    if($n%$D > 0){
        $count++;    
    }
    return (int)$count;

}