Problem Description

In this problem, we are given an array called words, which contains 'n' strings, each indexed from 0 to n-1. We need to combine these strings using a special 'join' operation which merges two strings together. But there's a twist: if the last character of the first string is the same as the first character of the second string, we have to remove one of these matching characters while joining them. The goal is to perform n - 1 join operations to generate a single string in such a way that the final string's length is as small as possible.

Here's what we are allowed to do during each join operation:

• We can append the next string in the array words to the current string str_i.
• Alternatively, we can prepend the next string in the array words to the current string str_i.

Intuition

To arrive at the solution approach, consider that at every step of joining, we have two choices: either append or prepend the next word. The impact of this choice on the final string length can vary depending on the matching characters at the edges of the strings being joined.

We aim to minimize the overall length each time we make a join. To do this for n - 1 operations, an intuition may lead us to dynamic programming or recursion to check every possibility, remembering the results of sub-problems to avoid redundant calculations.

Since the problem is about making choices at each step and optimizing for the 'smallest length', and considering that each choice might impact the subsequent ones, we can visualize the decision-making process as a tree. This is a common characteristic of problems where backtracking or dynamic programming is a fit.

The solution uses recursion with memoization (notice the use of the @cache decorator) to explore all possible concatenations and record the results for sub-problems, thus avoiding the recalculation of the same state. The dfs function takes into account the current position i, as well as the potential characters at the beginning and end of the current concatenated string (a and b) to make decisions leading towards the minimum length of the final string.

Each call to the dfs function considers the effect of performing a join operation via appending or prepending the current word to the string being built so far and tracks which operation gives the shorter result. The use of a recursive function with memoization allows the efficient solving of this optimization problem by making the complexity feasible for larger input sizes.

Learn more about Dynamic Programming patterns.

Solution Approach

The implementation relies on Depth-First Search (DFS) recursion with memoization to efficiently explore all possible concatenation sequences. The @cache decorator is used to memoize the results. Let's examine how the solution works by breaking down the key components of the implementation:

• The function dfs is the core of this solution. It takes three parameters: the current index i in the words array, a character a representing the beginning character of the string up to this point, and a character b representing the ending character of the string up to this point.

• The base case for the recursion is when i exceeds the number of words we have. At that point, there are no more words to join, and the function returns 0 because there is no extra length to add from additional words.

• When inside dfs, we consider the next word s to be joined with the current string. We have two choices resembling a binary tree's branches:

  • Append: Where we consider adding s to the end of the current string, so we call dfs(i + 1, a, s[-1]), passing s[-1] as the new ending character since we're appending. We subtract 1 from the length if the current ending character b matches the starting character of s.
  • Prepend: Where we consider adding s at the beginning, so we call dfs(i + 1, s[0], b), passing s[0] as the new starting character. We subtract 1 from the length if the ending character s[-1] matches the current beginning character a.

• The recursive calls return the sum of the length of the current word s and the minimum length from the two choices (append or prepend). The subtraction of 1 accounts for the removed character when there is a match.

• The final return statement return len(words[0]) + dfs(1, words[0][0], words[0][-1]) initiates the process with the first word, taking its length and appending the result of the recursive call starting from the second word with the first word's beginning and ending characters.

The @cache decorator on dfs ensures that for each unique set of inputs to the function (combinations of i, a, and b), the result is stored. If the same inputs are ever called again, the stored result is returned immediately, which significantly reduces the number of recursive calls, especially for large arrays with many possible combinations.

Overall, this solution showcases a classic recursive problem-solving strategy enhanced with memoization, a powerful way to optimize recursive depth-first search algorithms.

Example Walkthrough

Let's take an example array of words: ["abc", "cda", "dac"]. We will walk through the process to understand how we might join these strings to minimize the overall length.

Firstly, starting with the first word "abc", we initiate the recursive process. There are no characters on either side, so a and b are 'a' and 'c' respectively, which represent the first and last characters of the current string.

Now, we look at the next word "cda". We have two choices:

1. Append: If we append "cda" to "abc", the resulting string would be "abccda" because the last character of "abc" is the same as the first of "cda", we can combine them by removing one 'c'. So, the string becomes "abcdac". We then call dfs(2, 'a', 'a') since we have moved on to the next word and the end character has been updated to 'a' from "cda".

2. Prepend: If we prepend "cda" to "abc", the resulting string would be "cdabc". There is no matching character in this case, so no character is removed. We then call dfs(2, 'c', 'c'), updating the start character to 'c'.

The return value for each of these choices would be the length of "cda" (which is 3) plus the minimum of the lengths obtained from appending or prepending the next word "dac".

Continuing with the recursion, we now have the third word "dac" and the choices described above for the second word:

1. If we chose to append during the previous operation, we now have "abcdac":

   • Append: Append "dac" leading to "abcdacdac", and after removing the duplicate 'a', we have "abcdacdc". The recursive call would be dfs(3, 'a', 'c').
   • Prepend: Prepend "dac" leading to "dacabcdac", and after removing the duplicate 'd', it remains the same as prepending will not benefit us this time. The recursive call would be dfs(3, 'd', 'c').

2. If we chose to prepend in the previous operation, our string would be "cdabc":

   • Append: Append "dac" leading to "cdabcdac", with no character removed. The recursive call would be dfs(3, 'c', 'c').
   • Prepend: Prepend "dac" to "cdabc" leading to "daccdabc" and after removing the duplicate 'c', we have "dacdabc". The recursive call would be dfs(3, 'd', 'c').

For each of these calls, we consider the removal of a duplicate character if there's a match at any end.

When i becomes equal to n, the recursion has considered all words, and we return 0 since there are no more words to add length from.

The memoization stores the results of each recursive call like dfs(2, 'a', 'a'), dfs(2, 'c', 'c'), dfs(3, 'a', 'c'), etc., so if the function with the same set of parameters is called again, it doesn't recalculate but retrieves the value from the cache, saving time.

In our example, assuming optimal choices were made, the solution might deduce that calling dfs(3, 'a', 'c') results in the smallest final string length. Ending with an efficient concatenated string "abcdacdc" or "dacdabc", depending on the sequence of operations chosen by the DFS algorithm.

Solution Implementation

Time and Space Complexity

The given Python code implements a solution to minimize the concatenated length of a list of words by exploiting memoization (using the @cache decorator, which caches the results of the dfs function calls). Analyzing the provided code, we can deduce the complexity as follows:

Time Complexity

The function dfs(i: int, a: str, b: str) is recursively called with different parameters, which correspond to the current index (i), the first character of the last added word to string a (a), and the last character of the last added word to string b (b). Since memoization is used, each state will be computed at most once. There are n possible indices (n being the size of words), and each word can provide 26 possible characters for a and b (assuming the input is lower-case English letters). Therefore, we have:

• Number of states: n * 26 * 26

For each state, operations are performed in constant time (O(1)) considering there is no loop within dfs, just two recursive calls and a few constant-time conditions and assignments:

• Operation per state: O(1)

Therefore, the total time complexity is the number of states times the operations per state, which is:

• Total Time Complexity: O(n * 26^2)

Space Complexity

The space complexity is mainly affected by the call stack during the recursive calls, and the space used for memoization. In the worst case, the recursion depth can go up to n:

• Recursion stack: O(n)

The memoization will store results for every possible state:

• Memoization storage: O(n * 26^2)

Hence, the total space complexity can be seen as the maximum of the above two concerns. Since n * 26^2 is typically greater than just n:

• Total Space Complexity: O(n * 26^2)

Note: The exact constants for the number of characters would depend on the character set allowed in the input.

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