문제출처

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:

function solution(X, Y, 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).

풀이과정

① X가 처음으로 Y보다 커지려면 D를 몇 번 더해야 하는 지 알아내야 하는 것이기 때문에, Y와 X의 차이를 D로 나눈 결과를 활용해 볼 수 있을 것이다. ② (Y - X)를 D로 나눈 나머지가 0이라면 그 자체로 답이 되고, 나머지가 있다면 몫에 1을 더한 값을 리턴 해줘야 한다. ③ 만약 X와 Y의 값이 같다면 결과는 0이다.(예외처리 해야 하는 부분)

작성코드

function solution(X, Y, D) {
    if (X === Y) {
        return 0;
    }
    return Math.ceil((Y - X) / D);
}