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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 complexit

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

class Solution {

public int solution(int X, int Y, int D) {

// write your code in Java SE 8

int ret = 0;

while(true) {

if(Y < X) return ret;

else { Y-=D; ret++; }

}

Detected time complexity:

O(Y-X)

출처: <https://codility.com/demo/results/trainingF7SUG2-BAA/>

// you can also use imports, for example:

// import java.util.*;

// you can write to stdout for debugging purposes, e.g.

// System.out.println("this is a debug message");