here is my C solution: int c=0,i=0; int beautifulBinaryString(char* b) { if(b[i]!='\0') { if(b[i]=='0' && b[i+1]=='1' && b[i+2]=='0') { c++; i=i+3; } else{i++;} beautifulBinaryString(b); } return c; } *

Here are my C++ solution: int beautifulBinaryString(string b) { int i = 0, d = 0; while (i < b.length() - 2) { if (b[i] == '0' && b[i + 1] == '1' && b[i + 2] == '0') { d++; i += 3; // Skip to the character after "010" } else { i++; } } return d; }

Here are my c++ solution, explanation here : https://youtu.be/xf0JsPRH5rs

Solution 1 :

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;
}

Solution 2 :

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