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 Java (OpenJDK 11.0.6) (judge verdict: Accepted)


import java.io.*;
import java.io.BufferedReader; 
import java.io.IOException; 
import java.io.InputStreamReader; 
import java.util.StringTokenizer; 
import java.util.*;


public  class Main 
{

	   
	   static class FastReader 
	    { 
	        BufferedReader br; 
	        StringTokenizer st; 
	       
	  
	        public FastReader() 
	        { 
	            br = new BufferedReader(new
	                     InputStreamReader(System.in)); 
	        } 
	  
	        String next() 
	        { 
	            while (st == null || !st.hasMoreTokens()) 
	            { 
	                try
	                { 
	                    st = new StringTokenizer(br.readLine()); 
	                } 
	                catch (IOException  e) 
	                { 
	                    e.printStackTrace(); 
	                } 
	            } 
	            return st.nextToken();
	            
	        } 
	  
	        int nextInt() 
	        { 
	            return Integer.parseInt(next()); 
	        } 
	  
	        long nextLong() 
	        { 
	            return Long.parseLong(next()); 
	        } 
	  
	        double nextDouble() 
	        { 
	            return Double.parseDouble(next()); 
	        } 
	  
	        String nextLine() 
	        { 
	            String str = ""; 
	            try
	            { 
	                str = br.readLine(); 
	            } 
	            catch (IOException e) 
	            { 
	                e.printStackTrace(); 
	            } 
	            return str; 
	        }

			public Character charAt(int i) {
				// TODO Auto-generated method stub
				return null;
			} 
	    } 

  

	  // Complete the hurdleRace function below.
//    public static void solve() {
//    	FastReader s=new FastReader();

    public static void main(String[] args) throws IOException {
    	FastReader s=new FastReader();
   			int n = s.nextInt();
   			String s1=s.next();
   			List<Character> l = new ArrayList<>();
   			int c=1;
   			for(int i=0;i<s1.length();i+=c) {
   				c=1;
   				int j=i+1;
   				while( j<s1.length() && s1.charAt(i)==s1.charAt(j)) {
   					c++;
   					j++;
   				}
   				n-=c+1;
   				
   			}
   			
			
			System.out.println(Math.abs(n));
    }	
}