main idea:

  1. for nth tx,ty, ( assuming tx > ty), the nth tx comes from the (n-1)th tx + the (n-1)th ty ==> in order to make ty, tx increase, there has to be a difference between these two number
  2. for any sx > tx or sy > ty, there is no chance will change
  3. there is a case when tx == ts, which means all the time only ty is changed ==> ty = sy + k* sx, in this case, just to check if (ty-sy) % sx == 0 
    class Solution:
# time O(lg(max(ty,tx)))  space O(1)
 def reachingPoints(self, sx: int, sy: int, tx: int, ty: int) -> bool:
     # big assumption: ty >= sy and tx >= sx 
     while sx <= tx and sy <= ty:
         # case 1  ty == tx 
         if tx == ty:
             break
         # case2  ty > tx 
         # case 2.1 ty > sy  
         # case 2.2 ty <= sy
         elif ty > tx:
             if tx == sx:
                 return (ty-sy)%sx == 0
             ty = ty % tx
         # case3  ty <= tx 
         # case 3.1 tx > sx
         # case 3.2 tx <= sy 
         elif ty < tx:
             if ty == sy:
                 return (tx-sx)%sy == 0 
             tx = tx%ty
     return sx == tx and sy == ty