Evil solution



In [1]:

    
evil = lambda d,p=[0,0],o=0:sum(abs(a) for a in p) if len(d)==0 else evil(d[1:],p=[a+int(d[0][1:])*b for a,b in zip(p,[(1,0),(0,-1),(-1,0),(0,1)][[[1,3],[2,0],[3,1],[0,2]][o][0 if d[0][0]=='L' else 1]])],o=[[1,3],[2,0],[3,1],[0,2]][o][0 if d[0][0]=='L' else 1])

Test examples



In [2]:

    
evil(["R2", "L3"])









    Out[2]:





5



In [3]:

    
evil(["R2", "R2", "R2"])









    Out[3]:





2



In [4]:

    
evil(["R5", "L5", "R5", "R3"])









    Out[4]:





12

Process input



In [5]:

    
with open("inputs/day1a.txt") as fd:
    print("Easter Bunny HQ is {} blocks away".format(evil([i.strip() for i in fd.read().split(',')])))









    



Easter Bunny HQ is 291 blocks away



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