🧠 Problem: Beautiful Binary String

This function counts the minimum number of operations required to make a binary string "beautiful" by replacing occurrences of '010' with '011'.

def beautifulBinaryString(b):
    count = 0
    ans = 0
    while count != 1:
        a = b.find('010')
        if a >= 0:
            b = b[:a+2] + '1' + b[a+3:]
            ans += 1
            count += 0
        else:
            count += 1
    return ans

🧪 Example Usage

s = '0101010'
print(beautifulBinaryString(s))  # Output: 2

✅ Notes:

• The find('010') method locates the first occurrence of the pattern.
• The string is modified to break the pattern by changing the last '0' to '1'.
• The loop continues until no '010' is found.

// c#

public static int beautifulBinaryString(string b)
{       
    string newB = b.Replace("010", "xxx");
    int occur = 0;

    for(int i=0; i<newB.Length; i++)
    {
        if (newB.Substring(i,1) == "x")
        {
            i += 2;
            occur++;
        }
    }

    return occur;
}

Basic solution

int beautifulBinaryString(string b) { int result = 0; while (b.find("010")!=string::npos) { ++b[b.find("010")+2]; result++; } return result; }