문제

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:

public func solution(_ X : Int, _ Y : Int, _ D : Int) -> Int

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

Write an **efficient** algorithm for the following assumptions:

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

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답안

public func solution(_ X : Int, _ Y : Int, _ D : Int) -> Int {
    let remainder = (Y - X) % D
    return ((Y - X) / D) + (0 < remainder ? 1 : 0)
}

1. 총 이동한 거리(Y-X)에서 개구리가 뛸 수 있는 거리(D)를 나눈다.
2. 총 이동한 거리(Y-X)에서 개구리가 뛸 수 있는 거리(D)를 나눈 것의 나머지가 구한다. 없으면 0 처리 한다.
3. 1과 2를 더한다.

Tasks Details

문제 출처

• https://app.codility.com/programmers/lessons/3-time_complexity/frog_jmp/