Problem:
Score : 300 points

Problem Statement

There are N slimes lining up from left to right. The colors of these slimes will be given as a string S of length N consisting of lowercase English letters. The i-th slime from the left has the color that corresponds to the i-th character of S.

Adjacent slimes with the same color will fuse into one larger slime without changing the color. If there were a slime adjacent to this group of slimes before fusion, that slime is now adjacent to the new larger slime.

Ultimately, how many slimes will be there?

Constraints

1 \leq N \leq 10^5

|S| = N

S consists of lowercase English letters.

Input

Input is given from Standard Input in the following format:

N
S

Output

Print the final number of slimes.

Sample Input 1

10
aabbbbaaca

Sample Output 1

5

Ultimately, these slimes will fuse into abaca.

Sample Input 2

5
aaaaa

Sample Output 2

1

All the slimes will fuse into one.

Sample Input 3

20
xxzaffeeeeddfkkkkllq

Sample Output 3

10

Solution in C++14 (GCC 5.4.1) (judge verdict: Wrong Answer)

#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
int main(){
    int n;
    cin >> n;
    vector<int> l(n);
    for(int i = 0;i < n;i++)cin >> l[i];
    //l[n] = -10;
    //l[n+1] = 100000;
    sort(l.begin(),l.end());
    ll ans = 0;
    for(int i = 0;i < n;i++){
        for(int j = i+1;j < n;j++){
            int a = l[i];
            int b = l[j];
            int m = upper_bound(l.begin(),l.end(),b-a)-l.begin();
            int M = lower_bound(l.begin(),l.end(),a+b)-l.begin()-1;
            //cout << a+b <<" " << l[M] << " "  << M <<" " << l[m] << " "  << m << " "<< b-a << endl;
            int un = 0;
            if(m <= i && i <= M)un--;
            if(m <= j && j <= M)un--;
            //cout << un << endl;
            ans += max(0,M-m+1+un);
            //cout << ans << endl;
        }
    }
    cout << ans/3 << endl;
    //cin >> n;
}