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classGroup(NamedTuple):char:strcount:intdefsubstrCount(n,s):total=0groups=[]group=Group(char=s[0],count=1)# iterate over 's', grouping runs of same charforcharins[1:]:ifgroup.char==char:#incrementcurrentcharcountgroup=Group(char=char,count=1+group.count)else:#registercurrentgroup,restartcountw/newgroupgroups.append(group)total+=(group.count**2+group.count)// 2 # n substringsgroup=Group(char=char,count=1)#restartw/newchar# register final groupgroups.append(group)total+=(group.count**2+group.count)// 2 # n substrings in group# iterate over triplets of groups, counting# odd sized substrings w/ extra middle charforbefore,current,afterinzip(groups,groups[1:],groups[2:]):ifcurrent.count==1andbefore.char==after.char:total+=min(before.count,after.count)returntotal
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Special String Again
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(hopefully) easy-to-read O(n) Python solution