Introduction

In this blog post, we will explore how to solve the LeetCode problem "2746" using AI. LeetCode is a popular platform for preparing for coding interviews, and with the help of AI tools like Chatmagic, we can generate solutions quickly and efficiently - helping you pass the interviews and get the job offer without having to study for months.

Problem Statement

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

Explanation

Here's a breakdown of the solution approach, followed by the Python code:

• Dynamic Programming: Use dynamic programming to explore all possible join orders. The state will represent the current string's length and its starting and ending characters.
• State Definition: The DP state dp[i][start][end] stores the minimum length achievable after processing the first i+1 words (from words[0] to words[i]), where start is the first character and end is the last character of the resulting string.

• Transitions: For each word at index i, consider joining it to the left or right of the previously formed string. Update the DP table with the minimum length obtained from both possibilities.

• Runtime Complexity: O(n 26 26) where n is the number of words.

• Storage Complexity: O(n 26 26)

def solve():
    words = ["aa","ab","bc"]
    print(min_length(words))

    words = ["ab","b"]
    print(min_length(words))

    words = ["aaa","c","aba"]
    print(min_length(words))

def min_length(words):
    n = len(words)
    dp = {}

    def calculate_min_length(index, start_char, end_char):
        if index == n:
            return 0

        if (index, start_char, end_char) in dp:
            return dp[(index, start_char, end_char)]

        word = words[index]
        word_start = word[0]
        word_end = word[-1]
        word_len = len(word)

        # Option 1: Join word to the right
        len1 = float('inf')
        join_len = word_len
        if end_char == word_start:
            join_len = word_len - 1

        if index == 0:
            len1 = word_len + calculate_min_length(index + 1, word_start, word_end)

        else:
            len1 = join_len + calculate_min_length(index + 1, start_char, word_end)

        # Option 2: Join word to the left
        len2 = float('inf')
        join_len = word_len
        if word_end == start_char:
            join_len = word_len - 1

        if index == 0:
            len2 = word_len + calculate_min_length(index + 1, word_start, word_end)

        else:
             len2 = join_len + calculate_min_length(index + 1, word_start, end_char)

        dp[(index, start_char, end_char)] = min(len1, len2)
        return dp[(index, start_char, end_char)]

    first_word = words[0]
    result = len(first_word) + calculate_min_length(1, first_word[0], first_word[-1])

    dp = {}
    def calculate_min_length2(index, start_char, end_char):
        if index == n-1:
            return 0

        if (index, start_char, end_char) in dp:
            return dp[(index, start_char, end_char)]

        word = words[index+1]
        word_start = word[0]
        word_end = word[-1]
        word_len = len(word)

        # Option 1: Join word to the right
        len1 = float('inf')
        join_len = word_len
        if end_char == word_start:
            join_len = word_len - 1

        len1 = join_len + calculate_min_length2(index + 1, start_char, word_end)

        # Option 2: Join word to the left
        len2 = float('inf')
        join_len = word_len
        if word_end == start_char:
            join_len = word_len - 1

        len2 = join_len + calculate_min_length2(index + 1, word_start, end_char)

        dp[(index, start_char, end_char)] = min(len1, len2)
        return dp[(index, start_char, end_char)]

    first_word = words[0]
    dp = {}
    result = len(first_word) + calculate_min_length2(0, first_word[0], first_word[-1])

    return result


    # Code
    ```python
    def min_length(words):    n = len(words)
    dp = {}

    def solve(index, first, last):
        if index == n:
            return 0

        if (index, first, last) in dp:
            return dp[(index, first, last)]

        word = words[index]
        word_first = word[0]
        word_last = word[-1]
        word_len = len(word)

        # Join to the right
        right_cost = word_len
        if last == word_first:
            right_cost -= 1
        right_res = right_cost + solve(index + 1, first, word_last)

        # Join to the left
        left_cost = word_len
        if word_last == first:
            left_cost -= 1
        left_res = left_cost + solve(index + 1, word_first, last)

        dp[(index, first, last)] = min(right_res, left_res)
        return dp[(index, first, last)]

    first_word = words[0]
    result = len(first_word) + solve(1, first_word[0], first_word[-1])
    return result