Pattern matching - c++

Question

I was just wondering if any other logic is possible for this problem:
Question : Find the number of pairs in a given string and output the sum of all the pairs and the unpaired elements. PS: The input is case sensitive.
Example O/P:
eeqe 3
aaaa
2
rwertr
5
I figured out the solution to this problem by first sorting the input string and then comparing adjacent elements as shown in the code below:
int main()
{
int t,count=0,pos=0;
char swap;
char a[201];
cin>>a;
int len=strlen(a);
//cout<<a<<endl;
for (int c = 0 ; c < ( len - 1 ); c++)
{
for (int d = 0 ; d < len - c - 1; d++)
{
if (a[d] > a[d+1]) /* For decreasing order use < */
{
swap = a[d];
a[d] = a[d+1];
a[d+1] = swap;
}
}
}
//cout<<a<<endl;
count=0;
for(int i=0;i<len;){
if(a[i]==a[i+1]){
count++;
i+=2;
//if(i== len-2)i++;
}
else{ count++; i++;}
}
//if(a[len-1]!=a[len-2])count++;
cout<<count<<endl;
return 0;
}
This code works fine. But, I was just wondering if there is any other efficient solution to this problem that doesn't involve sorting the entire input array.

Answer 1

It basically avoids sorting based on the idea that there are only 256 possible chars, so it's sufficent to count them.This is my solution:
int main()
{
std::string s; std::cin >> s;
int cnt[256] = {};
for (std::size_t i = 0; i < s.size(); ++i)
++cnt[static_cast<unsigned char>(s[i])];
int sum = 0;
for (std::size_t i = 0; i < 256; ++i)
sum += cnt[i]/2 + cnt[i]%2;
std::cout << sum << std::endl;
}
If, for example, the string contains 5 times an 'a', this allows 5/2 pairs (integer division) and 1 remains unpaired (because 5 is odd => 5%2 is 1)
Edit: Because we are here in SO:
int main()
{
std::array<int, 256> cnt{-1}; // last char will be '\0', ignore this.
std::for_each(std::istreambuf_iterator<char>(std::cin.rdbuf()),
std::istreambuf_iterator<char>{},
[&](unsigned char c){++cnt[c];});
std::cout << std::accumulate(cnt.begin(), cnt.end(), 0,
[](int i, int c)->int{return i+(c/2)+(c%2);}) << '\n';
}