Given a bracket sequence, print the length of largest prefix that is a regular bracket sequence.
EXPLANATION:
A regular bracket sequence is defined as follows:
S="" is regular.
S="<" + S1 + “>” is regular, if S1 is regular.
S=S1 concat S2 is regular, if S1 and S2 are regular.
If S is regular bracket sequence, for any i, number of closing brackets in S[0,i] should not exceed number of opening brackets. Also, if number of opening brackets is equal to number of closing brackets in S[0,i], S[0,i] is a regular bracket sequence.
def check(s):
t=0,ans=0
for i=0 to N-1:
if s[i]=='<': t++;
else:
t--;
//Now, if t=0, means, the string s[0,i] is valid.
if t==0: ans=max(ans,i+1)
else if t<0: break //string s whole is invalid.
print ans
Actually,
i used to work on Turbo c++ and it used to work on it.
and also i am new to codeshef.
But Yes It seems strange, http://ideone.com/ObhNFA,
its working now but ,
time limit exceed,
but i m trying to develop better logic.