def beautifulBinaryString(n,b):
    move=0
    b=list(b)
    for i in range(n-2):
        if b[i]=='0' and b[i+1]=='1' and b[i+2]=='0':
            move+=1
            b[i+2]='1'
    return move

int beautifulBinaryString(string b) {
    int ans = 0;
    for(int i = 0; i < b.size();){
        string sub = b.substr(i, 3);
        if(sub == "010") {
            ans++;
            i+=3;
        }else i++;
    }
    return ans;
}

int beautifulBinaryString(string b) {
    regex re("010");
    string c = regex_replace(b, re, "");
    return (b.size() - c.size()) / 3;
}

 static boolean findPattern(int index, String b){
    int n = b.length();
    return (index < n - 2 && b.substring(index, index + 3).equals("010") );
    }
 
  public static int beautifulBinaryString(String b) {
    int count = 0;
    int n = b.length();
    int i = 0;
    while(i < n){
      if(findPattern(i, b)){
        count++;
        i += 3;
        }
      else
        i++;
       
      }
    return count;
    }
 

int beautifulBinaryString(char* b) {
    int count = 0;
    int length = strlen(b);
    
    for (int i = 0; i <= length - 3; i++) {
        // Check for the substring "010"
        if (b[i] == '0' && b[i + 1] == '1' && b[i + 2] == '0') {
            count++;
            // Skip the next two characters to avoid overlapping
            i += 2; 
        }
    }
    
    return count;
}