Decode the Message

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Strings

You are given the strings key and message, which represent a cipher key and a secret message, respectively. The steps to decode message are as follows:

Use the first appearance of all 26 lowercase English letters in key as the order of the substitution table.
Align the substitution table with the regular English alphabet.
Each letter in message is then substituted using the table.
Spaces ' ' are transformed to themselves.

For example, given key = "happy boy" (actual key would have at least one instance of each letter in the alphabet), we have the partial substitution table of ('h' -> 'a', 'a' -> 'b', 'p' -> 'c', 'y' -> 'd', 'b' -> 'e', 'o' -> 'f').

Return the decoded message.

Example 1:

Input: key = "the quick brown fox jumps over the lazy dog", message = "vkbs bs t suepuv"
Output: "this is a secret"
Explanation: The diagram above shows the substitution table.
It is obtained by taking the first appearance of each letter in "the quick brown fox jumps over the lazy dog".

Example 2:

Input: key = "eljuxhpwnyrdgtqkviszcfmabo", message = "zwx hnfx lqantp mnoeius ycgk vcnjrdb"
Output: "the five boxing wizards jump quickly"
Explanation: The diagram above shows the substitution table.
It is obtained by taking the first appearance of each letter in "eljuxhpwnyrdgtqkviszcfmabo".

Constraints:

26 <= key.length <= 2000
key consists of lowercase English letters and ' '.
key contains every letter in the English alphabet ('a' to 'z') at least once.
1 <= message.length <= 2000
message consists of lowercase English letters and ' '.

Decode the Message

Easy

Asked by:

3 views

Topics:

Strings

You are given the strings key and message, which represent a cipher key and a secret message, respectively. The steps to decode message are as follows:

Use the first appearance of all 26 lowercase English letters in key as the order of the substitution table.
Align the substitution table with the regular English alphabet.
Each letter in message is then substituted using the table.
Spaces ' ' are transformed to themselves.

For example, given key = "happy boy" (actual key would have at least one instance of each letter in the alphabet), we have the partial substitution table of ('h' -> 'a', 'a' -> 'b', 'p' -> 'c', 'y' -> 'd', 'b' -> 'e', 'o' -> 'f').

Return the decoded message.

Example 1:

Input: key = "the quick brown fox jumps over the lazy dog", message = "vkbs bs t suepuv"
Output: "this is a secret"
Explanation: The diagram above shows the substitution table.
It is obtained by taking the first appearance of each letter in "the quick brown fox jumps over the lazy dog".

Example 2:

Input: key = "eljuxhpwnyrdgtqkviszcfmabo", message = "zwx hnfx lqantp mnoeius ycgk vcnjrdb"
Output: "the five boxing wizards jump quickly"
Explanation: The diagram above shows the substitution table.
It is obtained by taking the first appearance of each letter in "eljuxhpwnyrdgtqkviszcfmabo".

Constraints:

26 <= key.length <= 2000
key consists of lowercase English letters and ' '.
key contains every letter in the English alphabet ('a' to 'z') at least once.
1 <= message.length <= 2000
message consists of lowercase English letters and ' '.

Solution

Clarifying Questions

When you get asked this question in a real-life environment, it will often be ambiguous (especially at FAANG). Make sure to ask these questions in that case:

What characters are allowed in the encoded message? Are they all alphanumeric or can they include special characters?
If the message cannot be decoded, what should I return? Should I return an empty string, null, or throw an exception?
Is the encoding guaranteed to be valid (i.e., will there always be a corresponding character for each encoded sequence)?
What is the maximum length of the encoded message?
Are there any specific limitations on memory usage for decoding very long messages?

Brute Force Solution

Approach

The brute force approach for decoding a message tries out every single possible way to interpret it. This involves exploring all combinations of possible decodings until a valid and complete decoding is found. It's like trying every key on a keyring until one opens the lock.

Here's how the algorithm would work step-by-step:

Start by considering the first part of the message and treat it as a potential encoded character.
Check if that part of the message matches any known code or rule for a character.
If it does, remember this possible decoding and move on to decoding the rest of the message after that first part.
If it doesn't, try considering the first two parts of the message together, and check if that matches any known codes.
Continue this process, trying all possible lengths for the first encoded character.
For each possible valid decoding of the first character, repeat the entire process for the rest of the message after that character.
Keep track of all the complete decodings that you find.
Finally, from all the complete decodings, choose the correct one, based on the problem's definition of a 'correct' or 'valid' decoding.

Code Implementation

def decode_the_message_brute_force(encoded_message):
    possible_decodings = []

    def decode_recursive(index, current_decoding):
        # If we've reached the end, we've found a valid decoding
        if index == len(encoded_message):
            possible_decodings.append(current_decoding[:])
            return

        for length in range(1, len(encoded_message) - index + 1):
            encoded_character = encoded_message[index:index + length]

            # Check if the encoded character is a valid character
            if is_valid_character(encoded_character):

                # Add the decoded character to the current decoding
                current_decoding.append(decode_character(encoded_character))

                # Recursively decode the rest of the message
                decode_recursive(index + length, current_decoding)

                # Backtrack: Remove the last added character
                current_decoding.pop()

    decode_recursive(0, [])

    # Find the first valid decoding if there are any
    if possible_decodings:
        return possible_decodings[0]
    else:
        return None

def is_valid_character(encoded_character):
    if encoded_character == '1':
        return True
    elif encoded_character == '2':
        return True
    elif encoded_character == '3':
        return True
    elif encoded_character == '12':
        return True
    elif encoded_character == '23':
        return True
    else:
        return False

def decode_character(encoded_character):
    if encoded_character == '1':
        return 'A'
    elif encoded_character == '2':
        return 'B'
    elif encoded_character == '3':
        return 'C'
    elif encoded_character == '12':
        return 'L'
    elif encoded_character == '23':
        return 'W'
    else:
        return ''

Try it yourself on LeetCode

Big(O) Analysis

Time Complexity

O(2^n) – The brute force approach explores all possible segmentations of the input string of length n. At each position, we have a choice: either decode the current digit as a single character, or combine it with the next digit if valid. This branching effectively creates a binary tree of possibilities where, in the worst case, we explore paths roughly equivalent to the number of subsets of a set with n elements. Consequently, the number of operations grows exponentially with the input size n, leading to a time complexity of O(2^n).

Space Complexity

O(N) – The brute force approach explores all possible decodings using recursion. In the worst-case scenario, where each digit can be decoded individually, the recursion depth could be N, where N is the length of the message. Each recursive call creates a new stack frame, consuming memory. Thus, the maximum space used by the recursion stack is proportional to N, leading to O(N) auxiliary space.

Optimal Solution

Approach

The best way to decode this message is to work line by line, making the most efficient choices for each line. Instead of exploring every possible arrangement, we'll focus on filling each line optimally before moving to the next.

Here's how the algorithm would work step-by-step:

Begin at the start of the message and consider how many encoded parts you can fit on the first line before reaching the limit.
Select the largest possible group of encoded parts to place on the current line. This uses up the most space immediately.
After placing the encoded parts, figure out the best way to distribute the blank spaces between them so the line fills as much as possible.
If there are extra spaces that can't be divided evenly, place them strategically to make sure the decoded message looks good.
Move on to the next line of the decoded message and repeat the same process of selecting and spacing encoded parts.
Handle the very last line of the decoded message differently. Just place the remaining encoded parts normally, without trying to perfectly space them out. Add any remaining spaces at the end of the line.
By making these smart choices line by line, you will efficiently create the decoded message, without having to check every single possibility.

Code Implementation

def decode_the_message(encoded_parts, line_length):
    decoded_message = []
    current_index = 0

    while current_index < len(encoded_parts):
        current_line = []
        current_line_length = 0

        # Find the maximum number of parts that fit on the line
        number_of_parts_on_line = 0
        while current_index + number_of_parts_on_line < len(encoded_parts) and \
              current_line_length + encoded_parts[current_index + number_of_parts_on_line] + number_of_parts_on_line <= line_length:
            number_of_parts_on_line += 1

        # Add the encoded parts to the current line
        for i in range(number_of_parts_on_line):
            current_line.append(encoded_parts[current_index + i])

        current_index += number_of_parts_on_line
        remaining_space = line_length - sum(current_line)

        # Distribute spaces evenly, except for the last line
        if current_index < len(encoded_parts):
            number_of_gaps = len(current_line) - 1

            # Determine spaces between encoded parts
            if number_of_gaps > 0:
                base_space = remaining_space // number_of_gaps
                extra_spaces = remaining_space % number_of_gaps
            else:
                base_space = remaining_space
                extra_spaces = 0

            spaced_line = []
            for i in range(len(current_line)):
                spaced_line.append('*' * current_line[i])
                if i < len(current_line) - 1:
                    # Distribute spaces to create the correct line length.
                    spaces_to_add = base_space
                    if extra_spaces > 0:
                        spaces_to_add += 1
                        extra_spaces -= 1
                    spaced_line.append(' ' * spaces_to_add)

            decoded_message.append(''.join(spaced_line))
        else:
            # Last line: add remaining spaces at the end
            last_line = ['*' * part for part in current_line]
            last_line_string = ''.join(last_line)
            last_line_string += ' ' * (line_length - len(last_line_string))

            # Last line must not perfectly spaced.
            decoded_message.append(last_line_string)

    return '
'.join(decoded_message)

Try it yourself on LeetCode

Big(O) Analysis

Time Complexity

O(n*m) – The algorithm iterates through the encoded message with n parts line by line, aiming to fill each line optimally before proceeding to the next. Within each line, it considers up to m encoded parts (where m is a characteristic of each line, specifically the number of encoded parts considered for placement on a given line before moving to the next), selecting the largest possible group and then distributing spaces. Since the encoded parts are consumed one by one to make lines, n is the total number of parts and m is the maximum number of parts that can fit on a single line. This leads to O(n*m) runtime since the placement of each of the n parts might need the examination of m encoded parts on average.

Space Complexity

O(L) – The algorithm builds the decoded message line by line. Each line constructed requires temporary storage, primarily to hold the encoded parts and spaces before being added to the final result. The largest possible length of any line is dictated by the line limit (L), which needs to be stored temporarily. Therefore, the auxiliary space used is proportional to the line length, giving a space complexity of O(L).

Edge Cases

Case	How to Handle
Null or empty input string	Return an empty string or throw an IllegalArgumentException, depending on requirements
Input string containing non-alphanumeric characters	Define whether to ignore, remove, or throw an error for invalid characters.
Input string is very long (potentially exceeding memory limits)	Consider streaming the input or using a more memory-efficient data structure if string length is unbounded.
Decoding results in an extremely long or computationally expensive output	Implement safeguards to prevent excessive resource consumption and consider returning a partial or truncated result with appropriate notification.
Decoding instructions are self-referential, leading to infinite loops	Implement cycle detection mechanisms within the decoding algorithm and gracefully exit after a specific number of iterations or recursion depth.
Decoding instructions contain ambiguous or conflicting mappings	Define a precedence rule for conflicting mappings and apply it consistently or throw an exception.
Input string contains only one unique encoded character	Handle this edge case gracefully, either returning the decoded value corresponding to the character, or returning an empty/default string depending on requirements.
Decoding algorithm relies on external resources that become unavailable during execution	Implement proper error handling and retry mechanisms with appropriate fallback strategies.