22232425262728

public class Main {
    public static void main(String[] args) {
        // Binary string input
        String s = "1100110";
        int n = s.length();

        // Pad leading zeros to make length a multiple of 3
        if (n % 3 == 1) {
            s = "00" + s;
        } else if (n % 3 == 2) {
            s = "0" + s;
        }

        // Update length after padding
        n = s.length();
        StringBuilder ans = new StringBuilder();

        // Process every group of 3 bits
        for (int i = 0; i < n; i += 3) {
            int temp = (s.charAt(i) - '0') * 4 + (s.charAt(i + 1) - '0') * 2 + (s.charAt(i + 2) - '0') * 1;
            ans.append((char) (temp + '0'));
        }

        // Output the octal representation
        System.out.println(ans.toString());
    }
}

1234567891011121314151617181920212223

def binary_to_octal(s):
    n = len(s)

    # Pad leading zeros to make length a multiple of 3
    if n % 3 == 1:
        s = "00" + s
    elif n % 3 == 2:
        s = "0" + s

    n = len(s)
    ans = ""

    # Process every group of 3 bits
    for i in range(0, n, 3):
        temp = int(s[i]) * 4 + int(s[i + 1]) * 2 + int(s[i + 2]) * 1
        ans += str(temp)

    return ans

# Driver code
s = "1100110"
print(binary_to_octal(s))

1234567891011121314151617181920212223242526

function binaryToOctal(s) {
    let n = s.length;

    // Pad leading zeros to make length a multiple of 3
    if (n % 3 === 1) {
        s = "00" + s;
    } else if (n % 3 === 2) {
        s = "0" + s;
    }

    n = s.length;
    let ans = "";

    // Process every group of 3 bits
    for (let i = 0; i < n; i += 3) {
        let temp = (parseInt(s[i]) * 4) + (parseInt(s[i + 1]) * 2) + (parseInt(s[i + 2]) * 1);
        ans += temp.toString();
    }

    return ans;
}

// Driver code
const s = "1100110";
console.log(binaryToOctal(s));

Complexity Analysis

Time Complexity: O(n), where n is the length of the binary string. Each bit is visited exactly once during the grouping and conversion process.

Space