TheAlgorithms/Java · error · IllegalArgumentException

Encoded text contains invalid characters: {}

Error message

Encoded text contains invalid characters: {}

What it means

In the multi-character decode path, HuffmanCoding walks left/right based on each bit, so every character in the encoded string must be exactly '0' or '1'. Any other character (a space, letter, '2', newline) makes the traversal direction undefined, and the library rejects it.

Source

Thrown at src/main/java/com/thealgorithms/compression/HuffmanCoding.java:226

            throw new IllegalStateException("Huffman tree is empty.");
        }

        StringBuilder sb = new StringBuilder();

        if (root.isLeaf()) {
            for (char bit : encodedText.toCharArray()) {
                if (bit != '0') {
                    throw new IllegalArgumentException("Invalid binary sequence for single-character tree.");
                }
                sb.append(root.ch);
            }
            return sb.toString();
        }

        Node current = root;
        for (char bit : encodedText.toCharArray()) {
            if (bit != '0' && bit != '1') {
                throw new IllegalArgumentException("Encoded text contains invalid characters: " + bit);
            }

            current = (bit == '0') ? current.left : current.right;

            if (current.isLeaf()) {
                sb.append(current.ch);
                current = root;
            }
        }

        if (current != root) {
            throw new IllegalArgumentException("Malformed encoded string: incomplete sequence ending.");
        }

        return sb.toString();
    }

    /**

View on GitHub (pinned to fdfb9a395b)

Solutions

1. Ensure the encoded string consists solely of '0' and '1' characters before decoding.
2. Sanitize the payload by stripping non-binary characters, or reject it if any are found.
3. Verify the storage/transport layer preserved the exact bit string.

Example fix

// before
String dec = hc.decode(payload); // payload has stray spaces/newlines

// after
if (!payload.matches("[01]*")) {
    throw new IllegalArgumentException("Encoded text must be binary (0/1 only)");
}
String dec = hc.decode(payload);

if (!encodedText.matches("[01]*")) {
    throw new IllegalArgumentException("Encoded text must contain only 0 and 1");
}
String dec = hc.decode(encodedText);

When it happens

Trigger: Calling decode("0 1 0") (spaces); decode("abc"); decode("012") (contains '2'); decode on a string that includes a newline or whitespace.

Common situations: The encoded payload was stored/transmitted and picked up extra characters; a copy-paste introduced whitespace; the payload was never validated as binary before decoding.

Understand the failure class

• Parsing and encoding errors: unexpected token, malformed input — why parsers reject input and how to find the real culprit.

Related errors

• Character '%c' (U+%04X) not found in Huffman dictionary.
• Invalid binary sequence for single-character tree.
• Malformed encoded string: incomplete sequence ending.
• Input string cannot be null or empty.
• Original index must be between 0 and {}, got: {}

AI-assisted analysis of TheAlgorithms/Java@fdfb9a395b (2026-08-13). Data as JSON: /api/errors/2aa73f9fba257ac8. Report an issue: GitHub.