2052. Minimum Cost to Separate Sentence Into Rows

Problem Description

In this problem, you are given a sentence which is a string of words separated by spaces. Additionally, you are given an integer k that represents the maximum number of characters allowed in each row. The task is to break the sentence into rows where each row has at most k characters.

The constraints you need to consider are:

• A word cannot be split across two rows.
• Each word must be used exactly once and in the same order as it appears in the given sentence.
• Rows should not begin or end with spaces.
• Between two words in a row, there must be exactly one space.
• The 'cost' of a row is calculated as (k - n)^2, where n is the length of the row. The total cost is the sum of the individual costs for all rows, except the last one, which isn't included in the total cost calculation.

Your goal is to find the arrangement of words into rows that results in the minimum possible total cost.

Intuition

The intuition behind solving this problem is to use dynamic programming to explore all possible ways to break the sentence into rows. This process can be visualized as a decision tree where each node represents a state in which certain words have been grouped into a row, and each branch is a decision to start a new row with the next word. The solution involves recursively computing the cost of each decision and then combining those costs to find the minimum cost arrangement.

Dynamic programming is appropriate here because:

1. The problem has an optimal substructure - the solution to the problem can be composed of the solutions to its subproblems. For example, the minimum cost for the entire sentence depends on the minimum cost for the remainder of the sentence after forming each possible row.
2. It has overlapping subproblems - as we compute costs for different partitions of the sentence, we will encounter the same remaining sentence multiple times.

The recursive function dfs starts at the beginning of the sentence and makes decisions to split the sentence at various points such that the length of the row is at most k. It keeps track of the minimum cost found so far for a given partition. Through memoization (using the @cache decorator), we remember the results of subproblems, which saves us from recalculating the same state multiple times, thus reducing the time complexity of the algorithm.

The solution approach is as follows:

• First, convert the sentence into a list of word lengths (t) and use accumulate to create an array s that holds the accumulated sum of word lengths at each index, which helps in quickly checking the length of a range of words.
• The dfs function then recursively explores different points where the sentence can be split, calculating the cost of creating a new row.
• The base case of the recursive function is when there are enough characters left to fit the rest of the sentence into one row.
• If the base case condition is not met, the function iteratively tries fitting more words into the current row and calls itself to compute the minimum cost for the rest of the sentence from that point onwards.
• The aim is to find the minimum cost across all possible splits.
• The result of the DFS will be the minimum total cost for separating the sentence into rows, excluding the cost of the last row (as per the problem statement).

The provided solution uses the depth-first search (DFS) algorithm and is implemented efficiently thanks to Python's memoization capabilities.

Learn more about Dynamic Programming patterns.

Solution Approach

The provided Python solution employs a top-down dynamic programming approach, also known as memoization, to solve the problem. Let's go through the different components of the solution:

Data Structures:

• An array t is used to store the length of each word from the sentence.
• An array s is created using the accumulate function from the itertools module, which represents the accumulated sum of lengths of words up to a particular index. This helps in efficiently computing the length of any sequence of words in constant time.

Dynamic Programming Function dfs (Depth First Search):

• The dfs function is the heart of the solution. It takes an index i as its argument, which indicates the starting point for the current row within the list of words.
• It returns the minimum cost of fitting the remaining words into rows, starting from index i (inclusive).
• The function is decorated with @cache, which automatically remembers the result of each call with particular arguments (i in this case) so that it doesn't need to be computed again.

Base Case:

• The base case for the recursive dfs function is when all remaining words from index i can fit into one row, in which case the cost is 0 because the last row's cost is not included in the total.

Recursive Case:

• For other cases, we initialize a variable, ans to infinity (inf), representing the minimum cost found so far.
• We iterate through the words starting from index i+1 with a variable j. In this loop, we do the following:
  • Compute the current length t of the sequence of words from i to j-1 (inclusive) as s[j] - s[i] + j - i - 1. The + j - i - 1 accounts for the spaces between the words.
  • If this length t does not exceed k, we compute the cost of this row as (k - t) ** 2 and add it to the result of the recursive call to dfs(j) (which represents the minimum cost for the rest of the sentence starting from j).
  • Update ans with the minimum of its current value and the cost calculated above.
  • Increment j to include the next word in the current row or to start a new row from the next word.

Once the loop finishes, ans will hold the minimum cost found for breaking the sentence into rows from index i.

Triggering the Dynamic Programming:

• We call dfs(0) to start the process from the beginning of the sentence. The return value is the minimum total cost for breaking the sentence into rows as per the problem's constraint.

In summary, the algorithm cleverly combines memoization and the intuitive approach of trying every possible break point to ensure all scenarios are covered, yet avoids redundant computation by caching the costs of subproblems already solved.

Example Walkthrough

Let's illustrate the solution approach using a small example. Consider the sentence "LeetCode is great" and let's say the maximum number of characters allowed per row, k, is 10.

First, we prepare our data structures:

• The word lengths are t = [8, 2, 5].
• The accumulated sum array is s = [8, 11, 17] using accumulate.

Now, we invoke dfs(0) to begin our dynamic programming search:

• At i = 0, the sentence's first word "LeetCode" is 8 characters, which fits within the 10 character limit.
• We can't add the next word "is" to this row because it would exceed our limit (8 (LeetCode) + 1 (space) + 2 (is) = 11 > k).
• Therefore, we split after "LeetCode", yielding a cost of (10 - 8)^2 = 4 for this row, and recursively invoke dfs(1) for the remaining sentence "is great".

Now, dfs(1) takes over:

• Starting at i = 1, we have a remaining length of 7 characters for "is great", which fits into a row, so we consider the next word.
• Since "great" is the last word, we terminate here. The cost for this penultimate row is (10 - 7)^2 = 9.
• However, since "great" forms the last row, we do not include this cost in our total, and the function returns 0 for the cost at this index.

For our initial call dfs(0), the minimum cost would be equal to the cost of the first row plus the cost returned from the call to dfs(1): 4 + 0 = 4.

Therefore, our rows are "LeetCode" and "is great" with a minimum total cost of 4, with the split chosen such that the cost for the penultimate row is minimal, and the algorithm efficiently finds this by memoization to avoid repetitive calculations.

Solution Implementation

Time and Space Complexity

Time Complexity: O(n * k)

The solution uses dynamic programming with memoization (indicated by the use of the @cache decorator). The function dfs(i) is recursively called while iterating through each word from the index i to the end of the sentence. In the worst case, for every word, it checks all other words up to the maximum allowed length k. Therefore, the time complexity is O(n * k) where n is the number of words.

Space Complexity: O(n)

The space complexity comes from two main contributions: the memoization of the dfs function and the creation of the prefix sum array s. Since the dfs function will compute a result for each starting index i, the cache will at most store n results. Additionally, the prefix sum array s is of size n+1. Thus, the overall space complexity is O(n) due to these factors.

Learn more about how to find time and space complexity quickly using problem constraints.