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  1. Prepare
  2. Algorithms
  3. Strings
  4. Beautiful Binary String
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Beautiful Binary String

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  • yashdeora98294
     + 0 comments

    Here is solution in python, java, c++ and c programming - https://programmingoneonone.com/hackerrank-beautiful-binary-string-problem-solution.html

  • jaykayudo
     + 0 comments
    function beautifulBinaryString(b: string): number {
    // Write your code here
    let number_of_steps = 0;
    
    function finder(str: string){
        let substring_index = str.indexOf('010');
        if(substring_index === -1){
            return
        }
        number_of_steps++;
        if(substring_index + 3 >= str.length){
            return
        }
        finder(str.slice(substring_index + 3));
    }
    finder(b);
    return number_of_steps;
}

  • anarod_val25
     + 0 comments
    int beautifulBinaryString(string b) {
    int ptrn = 0;
    
    for (int i = 0; i<b.size(); i++){
        string sub = b.substr(i, 3);
        
        if(sub=="010"){
            ptrn++;
            i+=2;
        }
    }
    
    return ptrn;
}

  • greivin_lopez
     + 0 comments
    # Beautiful Binary String
def beautiful_binary_string(b):
    return b.count('101')

  • bibash_tandon
     + 0 comments

    🧠 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.