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Which test case my code is failing? (TYPING|python3.6)

Question code: TYPING

My code:

s=set() l=["d","f"] r=["j","k"] for i in range (int(input())): a=[] for j in range (int(input())): str1=input() t=0 li=list(str1) b=len(str1) if str1 not in s: s.add(str1) for x in range(0,b-1): if (li[x] in l and li[x+1] in l) or (li[x] in r and li[x+1] in r): t=t+1 ans =(b+t)*0.2 #this is actually (b-t)*0.2 +t*0.4 a.append(ans) else: for x in range(0,b-1): if (li[x] in l and li[x+1] in l) or (li[x] in r and li[x+1] in r): t=t+1 ans =(b+t)*0.1 a.append(ans) fans=sum(a)*10 print (int(fans))

asked 08 Nov, 15:58

v_shah's gravatar image

2★v_shah
93
accept rate: 0%

edited 08 Nov, 16:03


Your system for adding a time penalty per same-hand letter pair seems accurate.

You would do better to work in integers throughout, basically adjusting so that your units of time are deciseconds (tenths of a second). I don't think that it is causing a problem here, with low numbers, but you could avoid the issue completely.

However your actual main problem is that you don't reset your set s of known words between test cases. So encountering a word for the first time in a test case, you may score it like it is not the first time.

link

answered 09 Nov, 04:41

joffan's gravatar image

5★joffan
9148
accept rate: 12%

Thanks a lot!

(09 Nov, 17:13) v_shah2★

I tried to understand what's happening when "str1 in s" and "not in s" but it's too late here and I'm sleepy. Here check my solution, everything is clear there: solution link

link

answered 09 Nov, 03:57

tieros's gravatar image

4★tieros
734
accept rate: 12%

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question asked: 08 Nov, 15:58

question was seen: 110 times

last updated: 09 Nov, 17:13