Something like this should work.

int len = p.size() , curr = 0 ;
	vector<int> counts(26,0) ;
	auto next_char = [](char c){
		if(c == 'z') return 'a' ; 
		char x = c + 1 ; 
		return x ;
	};
	for(int i = 0 ; i < len ; ++i) {
		if(i == 0 || p[i] != next_char(p[i &mdash; 1])) {
			curr = 1 ; 
		}
		else ++curr ;
		counts[p[i] &mdash; 'a'] = max(counts[p[i] &mdash; 'a'] , curr) ; 
	}
	cout << accumulate(counts.begin(),counts.end(),0) << endl ;

Ah dammit, you finding this "straightforward," is pretty dis-heartening. I'm kind of disappointed I couldn't come up with this.

How do you come up with solutions like this? Ie, how did you think of this?