Solution Explanation: Minimum Cost to Separate Sentence Into Rows

This problem involves finding the minimum cost to split a sentence into rows, with each row having a maximum length of k characters. The cost of a row is calculated as (k - n)^2, where n is the length of the row. The total cost excludes the last row.

Approach: Dynamic Programming with Memoization

The most efficient approach utilizes dynamic programming with memoization to avoid redundant calculations. Here's a breakdown:

1. Preprocessing:

   • The input sentence is split into an array of words.
   • A prefix sum array s is created. s[i] stores the total length of words from index 0 to i-1 (inclusive), including spaces between them. This allows for efficient calculation of the length of any substring of words.

2. Recursive Function dfs(i):

   • This function takes an index i as input, representing the starting word index for a row.
   • Base Case: If placing all remaining words (from i onwards) on a single row doesn't exceed the length k, the cost is 0 (no more rows needed).
   • Recursive Step: Otherwise, the function iterates through possible splitting points j. For each j, it calculates:
     • The length m of the row from i to j-1.
     • The cost of this row: (k - m)^2.
     • The minimum cost for the remaining part of the sentence: dfs(j).
     • The total cost for this split is the sum of the row cost and the remaining cost: (k - m)^2 + dfs(j).
   • The function returns the minimum total cost among all possible splits.

3. Memoization:

   • The dfs function uses memoization (a dictionary or array f) to store the results of subproblems. If the result for a given i is already computed, it's retrieved from the memoization table, avoiding recalculation. This significantly improves efficiency.

4. Main Function:

   • The main function initializes the prefix sum array s and calls dfs(0) to get the minimum cost for the entire sentence.

Time and Space Complexity

• Time Complexity: O(n^2), where n is the number of words. The dfs function has a nested loop (iterating through splitting points). However, due to memoization, each subproblem is solved only once.

• Space Complexity: O(n). This is due to the space used by the prefix sum array s and the memoization table f.

Code Examples (Python3, Java, C++, Go, TypeScript)

The code examples in the previous response demonstrate this approach in various programming languages. The key elements—prefix sum calculation, recursive dfs function with memoization, and the main function—are common across all the implementations. The memoization technique drastically reduces runtime compared to a purely recursive solution.