Minimum Cost Good Caption

Hard

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Strings

You are given a string caption of length n. A good caption is a string where every character appears in groups of at least 3 consecutive occurrences.

For example:

"aaabbb" and "aaaaccc" are good captions.
"aabbb" and "ccccd" are not good captions.

You can perform the following operation any number of times:

Choose an index i (where 0 <= i < n) and change the character at that index to either:

The character immediately before it in the alphabet (if caption[i] != 'a').
The character immediately after it in the alphabet (if caption[i] != 'z').

Your task is to convert the given caption into a good caption using the minimum number of operations, and return it. If there are multiple possible good captions, return the lexicographically smallest one among them. If it is impossible to create a good caption, return an empty string "".

Example 1:

Input: caption = "cdcd"

Output: "cccc"

Explanation:

It can be shown that the given caption cannot be transformed into a good caption with fewer than 2 operations. The possible good captions that can be created using exactly 2 operations are:

"dddd": Change caption[0] and caption[2] to their next character 'd'.
"cccc": Change caption[1] and caption[3] to their previous character 'c'.

Since "cccc" is lexicographically smaller than "dddd", return "cccc".

Example 2:

Input: caption = "aca"

Output: "aaa"

Explanation:

It can be proven that the given caption requires at least 2 operations to be transformed into a good caption. The only good caption that can be obtained with exactly 2 operations is as follows:

Operation 1: Change caption[1] to 'b'. caption = "aba".
Operation 2: Change caption[1] to 'a'. caption = "aaa".

Thus, return "aaa".

Example 3:

Input: caption = "bc"

Output: ""

Explanation:

It can be shown that the given caption cannot be converted to a good caption by using any number of operations.

Constraints:

1 <= caption.length <= 5 * 10⁴
caption consists only of lowercase English letters.

Try coding it in LeetCode

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 are the maximum lengths of the `caption` string and the words in the `words` array?
Can the `words` array contain empty strings, and if so, should they be considered as substrings that always exist?
Are the words in the `words` array guaranteed to be unique, or could there be duplicates?
If it's impossible to transform the caption into a good caption, what value should I return (e.g., -1, Integer.MAX_VALUE)?
Are there any specific character restrictions for the caption and the words (e.g., only ASCII characters, alphanumeric only)?

Brute Force Solution

Approach

The brute force strategy involves trying every conceivable way to break the caption into lines and then picking the best one. We exhaustively examine all possible arrangements of words on each line to ensure optimal space usage.

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

Consider every possible number of words for the first line, starting with just the first word and extending to all words.
For each possible first line, check if the length is within the limit.
If the first line is valid, then consider all possible combinations of words for the second line, again checking if each line length is valid.
Continue this process, creating different line combinations until all the words are assigned to lines.
For each complete caption created (where all words are assigned to lines), calculate its cost based on the spacing rules.
Compare the costs of all possible complete captions.
Finally, choose the caption with the lowest cost as the best caption.

Code Implementation

def minimum_cost_good_caption_brute_force(
    words, ideal_length, cost_per_char
):
    number_of_words = len(words)
    minimum_total_cost = float('inf')

    def calculate_cost(line):
        line_length = sum(len(word) for word in line) + len(line) - 1
        if len(line) == 0:
            line_length = 0
        return abs(line_length - ideal_length) * cost_per_char

    def solve(index, current_caption):
        nonlocal minimum_total_cost

        # If we've used all words, calculate and update the minimum cost
        if index == number_of_words:
            total_cost = 0
            for line in current_caption:
                total_cost += calculate_cost(line)
            minimum_total_cost = min(minimum_total_cost, total_cost)
            return

        # Try adding words to the last line or starting a new line.
        if current_caption:
            # Option 1: Add the current word to the last line
            new_caption_with_added_word = current_caption[:-1] + [
                current_caption[-1] + [words[index]]
            ]
            solve(index + 1, new_caption_with_added_word)

        # Option 2: Start a new line with the current word
        new_caption_with_new_line = current_caption + [[words[index]]]
        solve(index + 1, new_caption_with_new_line)

    # Start with an empty caption.
    solve(0, [])

    return minimum_total_cost

Big(O) Analysis

Time Complexity

O(2^n) – The provided brute force approach explores all possible ways to break the caption into lines. For each word, we have a choice: either it starts a new line or it continues the current line. Since there are n words, this leads to 2^n possible caption arrangements. Calculating the cost for each arrangement takes O(n) time in the worst case to iterate through all the words. Therefore, the overall time complexity is O(n * 2^n), which is dominated by the exponential term, making the final complexity O(2^n).

Space Complexity

O(N^2) – The brute-force approach explores all possible ways to break the caption into lines. In the worst-case scenario, the recursion depth could be proportional to the number of words, N. For each recursive call, the algorithm potentially stores a partial caption, which could involve storing up to N words. Therefore, the space complexity is potentially O(N * N), where each of N recursive calls stores a caption of up to N length. After simplifying, we get O(N^2).

Optimal Solution

Approach

To minimize the caption cost, we need to find the optimal way to break the caption into lines. The best approach is to use a technique that figures out the cheapest arrangement line by line, rather than trying every single possible combination.

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

Start at the beginning of the caption and look at each word in sequence.
Figure out, for each starting point, what is the minimum cost of arranging the remaining words into lines.
Consider putting just the first word on the first line, then consider putting the first two words, then the first three, and so on, up to the maximum number of words that can fit on the line.
For each of those options, calculate the cost of the line (based on the number of spaces needed) and then add the minimum cost of arranging the remaining words, which you've already calculated.
Pick the option for the first line that results in the overall lowest cost, and record that cost.
Repeat this process, considering each word as the possible start of a new line until you have calculated the minimum cost for the entire caption.
To actually build the caption, trace back through your calculations, following the path that led to the lowest cost decisions, and arrange the words according to that path.

Code Implementation

def minimum_cost_good_caption(words, max_line_length):    total_cost = 0
    current_word_index = 0

    while current_word_index < len(words):
        # Find the maximum number of words that fit on the current line.
        number_of_words_on_line = 0
        current_line_length = 0
        while current_word_index + number_of_words_on_line < len(words) and \
              current_line_length + len(words[current_word_index + number_of_words_on_line]) + number_of_words_on_line <= max_line_length:
            number_of_words_on_line += 1

        # Calculate the empty space on the line.
        line_length = sum(len(words[current_word_index + i]) for i in range(number_of_words_on_line))
        number_of_spaces = number_of_words_on_line - 1
        empty_space = max_line_length - line_length - number_of_spaces

        # Empty space determines the line cost.
        line_cost = empty_space * empty_space
        total_cost += line_cost

        # Move the word index to the next line.
        current_word_index += number_of_words_on_line

    # Sum of line costs gives the minimum caption cost.
    return total_cost

Big(O) Analysis

Time Complexity

O(n²) – The algorithm iterates through each word in the caption (n words) to consider it as the start of a new line. For each starting word, it evaluates all possible line lengths, up to the maximum words that can fit on a line, resulting in another loop of at most n iterations in the worst case. This nested loop structure, where an outer loop iterates n times and the inner loop iterates up to n times for each outer loop iteration, gives a time complexity proportional to n * n, or O(n²).

Space Complexity

O(N) – The primary space complexity arises from storing the minimum cost of arranging the remaining words for each starting point. This is done to avoid redundant calculations. We need to store these minimum costs in an auxiliary data structure, such as an array, where each element represents the minimum cost starting from a specific word. Consequently, this array's size is directly proportional to the number of words, N, in the caption, resulting in O(N) space complexity.

Edge Cases

Case	How to Handle
Empty caption string	If the caption is empty, the cost is the minimum number of characters needed to concatenate all words; calculate this and return.
Empty words array	If the words array is empty, the caption is already 'good', so return 0.
Words array contains an empty string	An empty word is always a substring; remove empty words or ignore them to avoid infinite loops/incorrect results.
One or more words are not present in any possible transformation	If a word is longer than the caption plus the length of all other words, it can never be a substring, return -1 or handle the no-solution case appropriately.
Extremely long caption string or words array leading to potential memory issues or timeout	Optimize the search for substrings, potentially using efficient string matching algorithms (e.g., Knuth-Morris-Pratt) or consider dynamic programming techniques.
Duplicate words in the words array	The solution must account for needing to include duplicates; treat each duplicate as a unique word to be found.
Overlapping words in the words array	The solution should handle overlapping words correctly, ensuring each is considered, even if they share characters with other words within the transformed caption.
All words are already present in the caption.	The solution should immediately return 0 if the caption already contains all words as substrings.

Empty caption string

How to Handle:

If the caption is empty, the cost is the minimum number of characters needed to concatenate all words; calculate this and return.

Empty words array

How to Handle:

If the words array is empty, the caption is already 'good', so return 0.

Words array contains an empty string

How to Handle:

An empty word is always a substring; remove empty words or ignore them to avoid infinite loops/incorrect results.

One or more words are not present in any possible transformation

How to Handle:

If a word is longer than the caption plus the length of all other words, it can never be a substring, return -1 or handle the no-solution case appropriately.

Extremely long caption string or words array leading to potential memory issues or timeout

How to Handle:

Optimize the search for substrings, potentially using efficient string matching algorithms (e.g., Knuth-Morris-Pratt) or consider dynamic programming techniques.

Duplicate words in the words array

How to Handle:

The solution must account for needing to include duplicates; treat each duplicate as a unique word to be found.

Overlapping words in the words array

How to Handle:

The solution should handle overlapping words correctly, ensuring each is considered, even if they share characters with other words within the transformed caption.

All words are already present in the caption.

How to Handle:

The solution should immediately return 0 if the caption already contains all words as substrings.