TheAlgorithms/Java · error · IllegalArgumentException

Input is not a valid binary number.

Error message

Input is not a valid binary number.

What it means

Thrown by BinaryToOctal.convertBinaryToOctal(int) when the input's decimal string representation does not match the regex [01]+, meaning it contains a digit other than 0 or 1. This is a whole-input regex check, stricter than per-digit checks in sibling classes. The special case binary == 0 returns "0" before the regex runs.

Source

Thrown at src/main/java/com/thealgorithms/conversions/BinaryToOctal.java:24

    private static final int DECIMAL_BASE = 10;

    private BinaryToOctal() {
    }

    /**
     * This method converts a binary number to an octal number.
     *
     * @param binary The binary number
     * @return The octal number
     * @throws IllegalArgumentException if the input is not a valid binary number
     */
    public static String convertBinaryToOctal(int binary) {
        if (binary == 0) {
            return "0";
        }

        if (!String.valueOf(binary).matches("[01]+")) {
            throw new IllegalArgumentException("Input is not a valid binary number.");
        }

        StringBuilder octal = new StringBuilder();
        int currentBit;
        int bitValueMultiplier = 1;

        while (binary != 0) {
            int octalDigit = 0;
            for (int i = 0; i < BITS_PER_OCTAL_DIGIT && binary != 0; i++) {
                currentBit = binary % DECIMAL_BASE;
                binary /= DECIMAL_BASE;
                octalDigit += currentBit * bitValueMultiplier;
                bitValueMultiplier *= BINARY_BASE;
            }
            octal.insert(0, octalDigit);
            bitValueMultiplier = 1; // Reset multiplier for the next group
        }

View on GitHub (pinned to fdfb9a395b)

Solutions

1. Validate the input int with String.valueOf(binary).matches("[01]+") before calling.
2. For string-based binary input with leading zeros or signs, parse manually: Integer.toString(Integer.parseInt(s, 2), 8).
3. Confirm the input is genuinely a binary-coded decimal integer, not a computed value.

Example fix

// before
String oct = BinaryToOctal.convertBinaryToOctal(102); // '2' fails regex

// after
String oct = BinaryToOctal.convertBinaryToOctal(1010); // valid binary digits

if (!String.valueOf(binary).matches("[01]+")) {
    throw new IllegalArgumentException("not a valid binary number: " + binary);
}
String oct = BinaryToOctal.convertBinaryToOctal(binary);

static boolean isBinaryNumberInt(int n) {
    return String.valueOf(n).matches("[01]+");
}

try {
    String oct = BinaryToOctal.convertBinaryToOctal(n);
} catch (IllegalArgumentException e) {
    throw new DomainException("Invalid binary input: " + n, e);
}

When it happens

Trigger: Passing 102, 123, or any int with a digit 2-9. Passing a negative number (the minus sign fails the regex). Passing a number whose string form has non-binary digits.

Common situations: Confusing decimal int with binary representation. Negative binary inputs (not supported here). Leading zeros lost when stored as int.

Related errors

• Incorrect binary digit: {}
• Incorrect binary digit: {}
• Slope and intercept must be valid numbers.
• For input string: {}
• Input cannot be null

AI-assisted analysis of TheAlgorithms/Java@fdfb9a395b (2026-08-13). Data as JSON: /api/errors/8f746b7d3edc8394. Report an issue: GitHub.