Decremental String Concatenation

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class Solution:
    def minimizeConcatenatedLength(self, words: List[str]) -> int:
        n = len(words)

        @lru_cache(None)
        def dp(i, last1, first2):
            if i == n:
                return 0

            len1 = len(words[i]) - (last1 == words[i][0])
            res1 = len1 + dp(i + 1, words[i][-1], first2)

            len2 = len(words[i]) - (first2 == words[i][-1])
            res2 = len2 + dp(i + 1, last1, words[i][0])

            return min(res1, res2)

        return len(words[0]) + dp(1, words[0][-1], words[0][0])

Identifying Problem Isomorphism

“Decremental String Concatenation” shows some similarities to “Minimum Length of String After Deleting Similar Ends”.

In “Minimum Length of String After Deleting Similar Ends”, you’re given a string and you need to remove characters from the start and end of the string as long as they are the same until you can’t do this anymore. The goal is to minimize the length of the string.

In the “Decremental String Concatenation” problem, you’re given multiple strings and can concatenate them in any order, removing the common character if the end character of one string is the same as the start character of the next string. The goal is also to minimize the final string length.

The two problems share the idea of minimizing the final string length by removing matching characters. However, “Minimum Length of String After Deleting Similar Ends” deals with only one string and the deletions are always at the ends, whereas “Decremental String Concatenation” involves multiple strings and the deletions occur between them.

“Decremental String Concatenation” appears to be the more complex problem, as it involves making decisions about the order in which to concatenate the strings to achieve the smallest length, which adds an extra layer of complexity.

The problem “Decremental String Concatenation” required understanding of string manipulation and some aspects of greedy algorithms. Here are 10 problems of lesser complexity to prepare:

  1. 344. Reverse String: Basic problem to practice string manipulation.

  2. 14. Longest Common Prefix: This problem helps understand how to manipulate and compare strings.

  3. 387. First Unique Character in a String: Helps in understanding how to iterate over a string for certain conditions.

  4. 151. Reverse Words in a String: This problem is about manipulating strings in a more complex way.

  5. 3. Longest Substring Without Repeating Characters: This helps you understand how to deal with substrings.

  6. 28. Implement strStr(): Teaches pattern searching in a string.

  7. 455. Assign Cookies: A simple problem to understand the basics of greedy algorithms.

  8. 392. Is Subsequence: This problem helps understand the concept of subsequences which might be useful.

  9. 122. Best Time to Buy and Sell Stock II: An introductory problem to understand the basic concepts of greedy algorithms.

  10. 860. Lemonade Change: This problem provides practice for making greedy choices.

Each of these problems will give you practice with string manipulations and greedy algorithms, needed for solving more complex problems.

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class Solution:
    def minimizeConcatenatedLength(self, words: List[str]) -> int:
        @lru_cache(maxsize=None)
        def f(first,last,index):
            if len(words)-1 < index:
                return 0
            w = words[index]
            return max((w[-1] == first) + f(w[0], last, index+1), (last == w[0]) + f(first, w[-1], index+1))

        return len(''.join(words)) - f(words[0][0],words[0][-1],1)

Problem Classification

This problem falls under the category of String Manipulation and Dynamic Programming.

What Components:

  1. Array of strings ‘words’.
  2. Definition of a ‘join’ operation between two strings.
  3. A process that involves performing ’n - 1’ join operations.
  4. The goal to minimize the length of the final string after ’n - 1’ join operations.

This problem is asking for the manipulation of an array of strings, following certain rules (the join operation). The objective is to minimize the length of the final string after all join operations are performed.

This problem can be classified as an Optimization problem, a subcategory of Dynamic Programming problems. It involves finding the best (in this case, shortest) possible outcome from all possible outcomes that could be obtained by different sequences of join operations. This requires devising a strategy to make the optimal decision at each step (which string to join next, and in what order), taking into account the impact of each decision on future steps.

Moreover, given the focus on string operations and the specific join operation, this problem also falls under the String Manipulation category. String manipulation tasks involve working with and altering strings, which is a significant aspect of this problem.

In addition, since the problem involves iterative joining operations which may require recursion or looping through the array, this problem can also be seen as a problem of Iteration or Recursion.

In essence, the problem is a combination of Dynamic Programming (for the optimization part), String Manipulation (for the join operation), and Iteration or Recursion (for performing the join operations).

You are given a 0-indexed array words containing n strings.

Let’s define a join operation join(x, y) between two strings x and y as concatenating them into xy. However, if the last character of x is equal to the first character of y, one of them is deleted.

For example join(“ab”, “ba”) = “aba” and join(“ab”, “cde”) = “abcde”.

You are to perform n - 1 join operations. Let str0 = words[0]. Starting from i = 1 up to i = n - 1, for the ith operation, you can do one of the following:

Make stri = join(stri - 1, words[i]) Make stri = join(words[i], stri - 1) Your task is to minimize the length of strn - 1.

Return an integer denoting the minimum possible length of strn - 1.

Example 1:

Input: words = [“aa”,“ab”,“bc”] Output: 4 Explanation: In this example, we can perform join operations in the following order to minimize the length of str2: str0 = “aa” str1 = join(str0, “ab”) = “aab” str2 = join(str1, “bc”) = “aabc” It can be shown that the minimum possible length of str2 is 4. Example 2:

Input: words = [“ab”,“b”] Output: 2 Explanation: In this example, str0 = “ab”, there are two ways to get str1: join(str0, “b”) = “ab” or join(“b”, str0) = “bab”. The first string, “ab”, has the minimum length. Hence, the answer is 2. Example 3:

Input: words = [“aaa”,“c”,“aba”] Output: 6 Explanation: In this example, we can perform join operations in the following order to minimize the length of str2: str0 = “aaa” str1 = join(str0, “c”) = “aaac” str2 = join(“aba”, str1) = “abaaac” It can be shown that the minimum possible length of str2 is 6.

Constraints:

1 <= words.length <= 1000 1 <= words[i].length <= 50 Each character in words[i] is an English lowercase letter

Language Agnostic Coding Drills

  1. Identifying distinct concepts:

The code combines several distinct concepts, including:

a. Function Definition b. Recursion c. Function Decorators d. Memoization e. String Manipulation f. List Operations g. Control Flow with Conditional Statements

  1. List of concepts in order of increasing difficulty:

    a. Function Definition - This involves defining a function that accomplishes a certain task. It’s a basic concept and a fundamental building block of any program.

    b. String Manipulation - This involves working with and altering strings. It’s a more complex task as it requires understanding how strings are indexed and how different string methods work.

    c. List Operations - This involves working with and manipulating lists. It’s similar to string manipulation in terms of difficulty, as it also involves understanding indexing and list methods.

    d. Control Flow with Conditional Statements - This involves directing the flow of the program based on certain conditions. It’s slightly more complex as it requires logical thinking to determine the conditions.

    e. Recursion - This involves a function calling itself in order to solve a problem. It’s a more complex concept as it requires thinking about problems in a recursive way, which can be counter-intuitive.

    f. Function Decorators - This involves using decorators to modify the behavior of a function. This is a more advanced concept, as it requires understanding how decorators work and how they interact with functions.

    g. Memoization - This involves storing the results of expensive function calls and reusing them when the same inputs occur again. It’s the most complex concept here, as it requires understanding how to store and retrieve function results and when it’s beneficial to do so.

  2. Problem-solving approach:

The given code solves the problem by using a recursive function to explore all possible sequences of join operations, while employing memoization to avoid repeated calculations. Here’s how each of the identified coding drills contributes to the solution:

a. Function Definition: The code begins by defining a recursive function ‘f’ that calculates the maximum possible number of characters that can be saved by performing the join operations in an optimal way.

b. String Manipulation: The code uses string manipulation to check if the last character of the current string matches the first character of the next string, and vice versa.

c. List Operations: The code uses list operations to join all the words together and calculate the total length of the words.

d. Control Flow with Conditional Statements: The code uses conditional statements to compare different possibilities at each step and choose the one that results in the maximum saving of characters.

e. Recursion: The code uses recursion to explore all possible sequences of join operations. The function ‘f’ calls itself for the next index, with different parameters depending on the current choice.

f. Function Decorators: The code uses the ‘@lru_cache’ decorator to enable memoization for the function ‘f’.

g. Memoization: The code uses memoization to store the results of function calls for each (first, last, index) tuple, so that they can be reused if the same tuple occurs again.

Each of these concepts plays a crucial role in the solution. By breaking down the problem into these separate coding drills, it becomes easier to understand how each part of the code contributes to the solution.

Targeted Drills in Python

a. Function Definition: def hello_world(): print("Hello, World!") hello_world()

b. String Manipulation: str1 = "Hello" str2 = "World" str3 = str1 + str2 # String concatenation first_char = str1[0] # Accessing the first character of a string last_char = str1[-1] # Accessing the last character of a string

c. List Operations: list1 = ["Hello", "World"] length = len(list1) # Get the number of elements in a list first_item = list1[0] # Access the first item in a list joined_string = ''.join(list1) # Join all strings in a list

d. Control Flow with Conditional Statements: x = 5 if x > 3: print("x is greater than 3") else: print("x is not greater than 3")

e. Recursion: def factorial(n): if n == 0: return 1 else: return n * factorial(n-1) print(factorial(5))

f. Function Decorators: ``` def my_decorator(func): def wrapper(): print(“Before function call”) func() print(“After function call”) return wrapper

  @my_decorator
  def say_hello():
      print("Hello")
  
  say_hello()
  ```

g. Memoization: ``` from functools import lru_cache

  @lru_cache(maxsize=None)
  def fibonacci(n):
      if n < 2:
          return n
      else:
          return fibonacci(n-1) + fibonacci(n-2)
  
  print(fibonacci(10))
  ```

  1. Additional drills for the problem:

    a. Writing a recursive function with multiple base cases and multiple recursive calls.

    def binary_strings(n, prefix=""):
    if n == 0:
        print(prefix)
    else:
        binary_strings(n-1, prefix + "0")
        binary_strings(n-1, prefix + "1")
binary_strings(3)

  2. Merging the drills:

    a. Start by defining the recursive function ‘f’ using the Function Definition drill.

    b. Inside ‘f’, use String Manipulation and Conditional Statements to check the conditions of the join operation and decide which option to take.

    c. Use Recursion to call ‘f’ for the next index with the appropriate parameters based on the chosen option.

    d. Use Function Decorators and Memoization to optimize ‘f’ by storing and reusing the results of function calls.

    e. Outside ‘f’, use List Operations and String Manipulation to calculate the total length of the words and subtract the result of ‘f’ to get the minimum possible length of the final string.

Remember to carefully design the recursive function and the memoization strategy to ensure that all possible sequences of join operations are explored and that no unnecessary calculations are performed.