Codility---FrogJmp

Task description

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

 

Solution

 

Programming language used: Java

Total time used: 6 minutes

Code: 11:01:41 UTC, java, final, score:  100

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// 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 res = (Y-X)/D;
        if((Y-X) % D == 0)
            return res;
        else
            return res + 1;
        
    }
}