Introduction

In this blog post, we will explore how to solve the LeetCode problem "1239" 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 an array of strings arr. A string s is formed by the concatenation of a subsequence of arr that has unique characters. Return the maximum possible length of s. A subsequence is an array that can be derived from another array by deleting some or no elements without changing the order of the remaining elements. Example 1: Input: arr = ["un","iq","ue"] Output: 4 Explanation: All the valid concatenations are: - "" - "un" - "iq" - "ue" - "uniq" ("un" + "iq") - "ique" ("iq" + "ue") Maximum length is 4. Example 2: Input: arr = ["cha","r","act","ers"] Output: 6 Explanation: Possible longest valid concatenations are "chaers" ("cha" + "ers") and "acters" ("act" + "ers"). Example 3: Input: arr = ["abcdefghijklmnopqrstuvwxyz"] Output: 26 Explanation: The only string in arr has all 26 characters. Constraints: 1 <= arr.length <= 16 1 <= arr[i].length <= 26 arr[i] contains only lowercase English letters.

Explanation

Here's the solution:

• High-level approach:

  • Represent each string in arr as a bitmask, where each bit corresponds to a letter of the alphabet. This allows efficient checking for unique characters and concatenation conflicts.
  • Use dynamic programming (or recursion with memoization, which is essentially the same thing in this context) to explore all possible subsequences of arr. The state of the DP will be the current index in arr and the current bitmask representing the characters used so far.
  • At each step, we have two choices: either include the current string in the subsequence (if it's compatible with the current bitmask) or exclude it.

• Complexity:

  • Runtime: O(2ⁿ * m), where n is the length of arr and m is the average length of strings in arr.
  • Storage: O(2ⁿ), to store the DP table (memoization).

Code

    def maxLength(arr: list[str]) -> int:
    """
    Finds the maximum possible length of a string formed by the concatenation of a subsequence of arr that has unique characters.
    """

    masks = []
    for s in arr:
        mask = 0
        valid = True
        for char in s:
            bit = ord(char) - ord('a')
            if (mask >> bit) & 1:
                valid = False
                break
            mask |= (1 << bit)
        if valid:
            masks.append(mask)

    memo = {}

    def solve(index, current_mask):
        if (index, current_mask) in memo:
            return memo[(index, current_mask)]

        if index == len(masks):
            return 0

        # Option 1: Exclude the current string
        max_len = solve(index + 1, current_mask)

        # Option 2: Include the current string (if possible)
        if (masks[index] & current_mask) == 0:
            max_len = max(max_len, bin(masks[index]).count('1') + solve(index + 1, current_mask | masks[index]))

        memo[(index, current_mask)] = max_len
        return max_len

    return solve(0, 0)