Introduction

In this blog post, we will explore how to solve the LeetCode problem "522" 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

Given an array of strings strs, return the length of the longest uncommon subsequence between them. If the longest uncommon subsequence does not exist, return -1. An uncommon subsequence between an array of strings is a string that is a subsequence of one string but not the others. A subsequence of a string s is a string that can be obtained after deleting any number of characters from s. For example, "abc" is a subsequence of "aebdc" because you can delete the underlined characters in "aebdc" to get "abc". Other subsequences of "aebdc" include "aebdc", "aeb", and "" (empty string). Example 1: Input: strs = ["aba","cdc","eae"] Output: 3 Example 2: Input: strs = ["aaa","aaa","aa"] Output: -1 Constraints: 2 <= strs.length <= 50 1 <= strs[i].length <= 10 strs[i] consists of lowercase English letters.

Explanation

Here's the breakdown of the approach, complexity, and the Python code:

• High-Level Approach:

  • Iterate through each string in the input array strs.
  • For each string, check if it's a subsequence of any other string in strs.
  • If a string is not a subsequence of any other string, it's a candidate for the longest uncommon subsequence. Keep track of the maximum length found so far.

• Complexity:

  • Runtime Complexity: O(n^2 * m), where n is the number of strings and m is the average length of the strings.
  • Storage Complexity: O(1)

Code

    def isSubsequence(s, t):
    """
    Checks if string s is a subsequence of string t.
    """
    i = 0
    j = 0
    while i < len(s) and j < len(t):
        if s[i] == t[j]:
            i += 1
        j += 1
    return i == len(s)

def findLUSlength(strs):
    """
    Finds the length of the longest uncommon subsequence.
    """
    max_len = -1
    for i in range(len(strs)):
        is_uncommon = True
        for j in range(len(strs)):
            if i != j and isSubsequence(strs[i], strs[j]):
                is_uncommon = False
                break
        if is_uncommon:
            max_len = max(max_len, len(strs[i]))
    return max_len