Minimum Cost to Separate Sentence Into Rows

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Dynamic Programming

You are given a string sentence that consists of lowercase English letters and spaces. You want to split the sentence into multiple lines such that each line has at most k characters. You must split the sentence into as few lines as possible.

The length of each line (without spaces between words) must be less than or equal to k. You must break the sentence where there is a space and you cannot break up individual words.

The cost to have a line of length len is (k - len)². The total cost is the sum of the costs for each line. If a line consists of only one word, the length of the line is considered to be the length of the word. Return the minimum total cost to split the sentence into multiple lines.

Example 1:

Input: sentence = "this is a very long sentence", k = 10
Output: 6
Explanation: We can split the sentence into the following lines:
- this is a  (length = 9, cost = 1)
- very long (length = 9, cost = 1)
- sentence    (length = 8, cost = 4)
Thus, the minimum cost to split the sentence is 1 + 1 + 4 = 6.

Example 2:

Input: sentence = "hello world", k = 8
Output: 0
Explanation: We can split the sentence into the following lines:
- hello world (length = 10, cost = 0)
Thus, the minimum cost to split the sentence is 0.

Example 3:

Input: sentence = "a b c d e", k = 4
Output: 12
Explanation: We can split the sentence into the following lines:
- a   (length = 1, cost = 9)
- b   (length = 1, cost = 9)
- c   (length = 1, cost = 9)
- d   (length = 1, cost = 9)
- e   (length = 1, cost = 9)
Thus, the minimum cost to split the sentence is 9 + 9 + 9 + 9 + 9 = 45.

Constraints:

1 <= sentence.length <= 5000
1 <= k <= 5000
sentence consists of lowercase English letters and spaces.
There is at least one word in sentence.
All words in sentence are separated by a single space.

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 constraints on the length of the sentence string and the value of maxRowWidth?
Can the sentence string be empty, and if so, what should I return?
Are there any constraints on the length of individual words in the sentence? Specifically, is it possible for a single word to exceed maxRowWidth?
If it's impossible to separate the sentence into rows such that each row's length is no more than maxRowWidth (e.g., a single word is longer than maxRowWidth), what should I return?
Is the input sentence guaranteed to have words separated by single spaces only, or do I need to handle multiple spaces or leading/trailing spaces?

Brute Force Solution

Approach

The brute force strategy considers every single way the sentence could be broken down into lines. We explore every possible combination of word groupings to find the arrangement with the lowest cost. This involves trying everything and picking the best result.

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

Begin by considering putting only the first word on the first line.
Next, consider putting the first two words on the first line, then the first three words, and so on, until you have considered putting all the words on the first line.
For each possible first line, determine if the number of characters exceeds the maximum allowed length for a line.
If the line is too long, discard that option and move on to the next.
If the line is not too long, calculate the cost of that line based on the spacing rules.
For each valid first line, repeat the process for the remaining words to form the second line, and so on, until all the words have been placed into lines.
Keep track of the total cost for each possible arrangement of words into lines.
Once all possible arrangements have been examined, select the arrangement with the minimum total cost.

Code Implementation

def minimum_cost_to_separate_sentence_into_rows_brute_force(
    sentence, max_row_length, cost_per_extra_space
):
    words = sentence.split()
    number_of_words = len(words)

    def calculate_cost(rows):
        total_cost = 0
        for row in rows:
            row_length = sum(len(word) for word in row) + len(row) - 1
            if row_length > max_row_length:
                return float('inf')
            extra_space = max_row_length - row_length
            total_cost += extra_space * cost_per_extra_space
        return total_cost

    def generate_all_possible_row_arrangements(index, current_rows):
        if index == number_of_words:
            # All words have been placed, calculate the cost.
            return [calculate_cost(current_rows)]

        all_costs = []

        # Explore different lengths for the current row
        for number_of_words_in_row in range(1, number_of_words - index + 1):
            new_row = words[index:index + number_of_words_in_row]

            # Recursive call to handle the rest of the sentence
            costs = generate_all_possible_row_arrangements(
                index + number_of_words_in_row, current_rows + [new_row]
            )

            all_costs.extend(costs)

        return all_costs

    # The exploration of all possible row arrangements
    all_costs = generate_all_possible_row_arrangements(0, [])

    # Find the minimum cost among all arrangements
    return min(all_costs)

Big(O) Analysis

Time Complexity

O(2^n) – The brute force approach explores every possible way to break the sentence into lines. For each word, we have a choice: either include it in the current line or start a new line. This binary choice (include/new line) exists for each of the n words in the sentence. Therefore, the total number of possible arrangements grows exponentially with the number of words. Consequently, the algorithm examines approximately 2^n different arrangements, leading to an exponential time complexity of O(2^n).

Space Complexity

O(N) – The brute force approach explores all possible arrangements of words into lines using recursion. The maximum depth of the recursion is determined by the number of words in the sentence, where N is the number of words. Each recursive call adds a new frame to the call stack. Therefore, the space complexity due to the recursion stack is proportional to N. Thus, the space complexity is O(N).

Optimal Solution

Approach

The best way to solve this problem is using a technique that remembers the best results we've seen so far. Instead of trying all possible combinations, it builds the solution step-by-step, making the best choice at each point based on previous calculations.

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

Imagine the sentence as a series of words, and we're trying to find the cheapest way to break it into lines.
Start by considering the first word. What's the cost if it starts the first line?
Then, figure out the cost if the first two words are on the first line, and so on, until you can't fit any more words on that first line.
For each of these possibilities, calculate the cost of that first line (based on how much space is wasted).
Now, for each of those first-line arrangements, we need to find the best way to arrange the *rest* of the sentence.
This is where the smart part comes in: we remember the *best* way to arrange the rest of the sentence starting from each word. That is, if we have already calculated that placing words 3 through 7 on the first line costs X, and then calculate the best way to separate the rest of the sentence into rows, we will know that the *total* cost is X + the cost to separate the rest of the sentence.
So, when we are trying to decide how to best separate the sentence into rows, we never have to calculate the cost of separating the rest of the sentence into rows more than once.
Repeat this process until you've figured out the best cost for every possible starting point for the lines.
The overall minimum cost will be the best cost starting from the very first word.
The overall strategy is to calculate the cost for each part of the sentence only once and use these costs to find the final answer. This avoids repeating work and makes the solution much faster.

Code Implementation

def min_cost_to_separate_sentence(sentence, max_width):
    words = sentence.split()
    number_of_words = len(words)
    cost = [0] * (number_of_words + 1)

    for i in range(number_of_words - 1, -1, -1):
        cost[i] = float('inf')
        line_length = -1
        for j in range(i, number_of_words):
            # Calculate the length of line if we include the word at index j
            line_length += len(words[j]) + 1

            if line_length > max_width:
                break

            # Calculate cost if we break the line after the word at index j
            if j == number_of_words - 1:
                current_cost = 0
            else:
                extra_spaces = max_width - line_length

                # Penalize uneven lines using cost function.
                current_cost = extra_spaces ** 3

            cost[i] = min(cost[i], current_cost + cost[j + 1])

    #The overall cost starts at beginning.
    return cost[0]

Big(O) Analysis

Time Complexity

O(n²) – The algorithm iterates through each word in the sentence (n words). For each word, it explores all possible line breaks up to the end of the sentence. In the worst case, for each starting word, the algorithm might consider all subsequent words to form a line. Thus, the number of operations is proportional to the sum of integers from 1 to n, where n is the number of words. This sum is approximately n * (n + 1) / 2. Therefore, the overall time complexity is O(n²).

Space Complexity

O(N) – The algorithm described in the plain English explanation remembers the best way to arrange the *rest* of the sentence starting from each word. This implies the use of a data structure, most likely an array or map, to store the minimum cost for each possible starting position within the sentence. Since there are N words in the sentence, where N is the input size, this data structure would store at most N values. Therefore, the auxiliary space used is proportional to the number of words in the sentence, resulting in O(N) space complexity.

Edge Cases

Case	How to Handle
Null or empty sentence string	Return 0, as there are no rows to calculate cost for, indicating an empty arrangement.
maxRowWidth is zero or negative	Return a large value (e.g., infinity) to indicate an invalid input and prevent infinite loops or division by zero.
Sentence contains a single word longer than maxRowWidth	Return a large value (e.g., infinity) indicating that the sentence cannot be separated into rows according to the constraints.
Sentence contains multiple words, but each word's length is greater than maxRowWidth/2	Dynamic programming approach would need to determine which words to place on each row while keeping the rows less than maxRowWidth.
Very long sentence causing integer overflow when calculating cost (sum of squares)	Use a data type with a larger range (e.g., long long in C++, long in Java) to store the cost, or consider modular arithmetic if appropriate.
maxRowWidth is very large compared to the sentence length	The first row could potentially contain the entire sentence, which may lead to a cost of (maxRowWidth - sentenceLength)^2.
Sentence contains leading or trailing spaces	Trim the sentence to remove leading and trailing spaces before processing it to avoid incorrect length calculations.
Sentence contains multiple consecutive spaces	Normalize spaces to single spaces between words to ensure correct word length calculations.

Null or empty sentence string

How to Handle:

Return 0, as there are no rows to calculate cost for, indicating an empty arrangement.

maxRowWidth is zero or negative

How to Handle:

Return a large value (e.g., infinity) to indicate an invalid input and prevent infinite loops or division by zero.

Sentence contains a single word longer than maxRowWidth

How to Handle:

Return a large value (e.g., infinity) indicating that the sentence cannot be separated into rows according to the constraints.

Sentence contains multiple words, but each word's length is greater than maxRowWidth/2

How to Handle:

Dynamic programming approach would need to determine which words to place on each row while keeping the rows less than maxRowWidth.

Very long sentence causing integer overflow when calculating cost (sum of squares)

How to Handle:

Use a data type with a larger range (e.g., long long in C++, long in Java) to store the cost, or consider modular arithmetic if appropriate.

maxRowWidth is very large compared to the sentence length

How to Handle:

The first row could potentially contain the entire sentence, which may lead to a cost of (maxRowWidth - sentenceLength)^2.

Sentence contains leading or trailing spaces

How to Handle:

Trim the sentence to remove leading and trailing spaces before processing it to avoid incorrect length calculations.

Sentence contains multiple consecutive spaces

How to Handle:

Normalize spaces to single spaces between words to ensure correct word length calculations.