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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).

Copyright 2009–2016 by Codility Limited. All Rights Reserved. Unauthorized copying, publication or disclosure prohibited.

 

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");

 

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++; }

        }       

    }

}

 

class Solution { public int solution(int X, int Y, int D) { // write your code in Java SE 8 int distance = Y-X; if(distance % D == 0) return Y/D; else return Y/D+1; } }

 

출처: <https://codility.com/demo/results/training2QR7GB-N88/>