# -*- coding: utf-8 -*-
"""
Created on Mon Mar  6 14:47:22 2017

@author: Rui


FrogJmp:

Count minimal number of jumps from position X to Y.


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:

    def 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).



"""


#Solução correcta mas com complexidade linear O(n)
def solution_1(X,Y,D):
    conta = 0
    while X < Y:
        conta += 1
        #print('Conta = ',conta)
        X += D
    return conta    

    
#Solução com complexidade constante O(1)

def solution_2(X,Y,D):
    salto = (Y - X) // D
    if (Y - X) % D > 0:
        return salto +1
    else:
        return salto


        
        
     
        

#print(solution_2(3,999111321,7))
#print(solution_2(6,11,3))
