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;
}

 public static int beautifulBinaryString(String b) {
        //goal: determine min removals to a string not have 010
        int min = 0;
        for (int i = 0; i < b.length() - 2; i++) {
            if (b.substring(i, i + 3).equals("010")) {
                min++;
                i += 2; // Skip next two characters to avoid overlapping patterns
            }
        }
        return min;
    }

int beautifulBinaryString(string b) {
    int val =0;
    int n = b.size();
    for (int i=2; i< n; i++) {
        if (b.at(i) == '0' and b.at(i-2)=='0' and b.at(i-1)=='1') {
            b.at(i) = '1';
            val++;
        }
    }
    return val;
}

def beautifulBinaryString(b):
    m=0
    while '010' in b:
        for i in range(len(b)-2):
            if b[i]=='0' and b[i+1]=='1' and b[i+2]=='0':
                m+=1
                b=b[i+3:]
                break
    return m

def beautifulBinaryString(b):
    # Write your code here
    return len(b.split('010'))-1