LeetCode Challenge (Python3). Problem#3

Source

Introduction of this challenge:

I recently decided to improve my programming skills, so I started using LeetCode to achieve this goal.

However, since I don't have much programming experience and this is my first time using LeetCode, there may be some difficulties in the process, but I believe I will be able to persevere.

There are currently 2230 problems on LeetCode, and I will be working on the challenge by problem number with the goal of solving all of them.

After solving problem 2, I spent a few days solving problem 3. For me, problem 3 is much easier than problem 2, and one of the main reasons I think is that python is much easier and faster than other programming languages in terms of data processing, so I don't think it's a technical improvement for me, I think it's just a language advantage.

So basically I only spent some time solving the problem3, part of which was spent on understanding python's string processing methods, and the rest on thinking about the logic part.

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Today's problem is "3. Longest Substring Without Repeating Characters".

Link: https://leetcode.com/problems/longest-substring-without-repeating-characters/
Difficulty: Medium
Related Topics: Hash Table String Sliding Window

Problem:

Given a string s, find the length of the longest substring without repeating characters.

Example 1:

Input: s = "abcabcbb"
 Output: 3
 Explanation: The answer is "abc", with the length of 3.

Example 2:

Input: s = "bbbbb"
 Output: 1
 Explanation: The answer is "b", with the length of 1.

Example 3:

Input: s = "pwwkew"
 Output: 3
 Explanation: The answer is "wke", with the length of 3.
 Notice that the answer must be a substring, "pwke" is a subsequence and not a substring.

Constraints:

• 0 <= s.length <= 5 * 10^4
• s consists of English letters, digits, symbols and spaces.

My solution：

class Solution:
    def lengthOfLongestSubstring(self, s: str) -> int:
        ss=''
        i=0  

        for c in range(len(s)):
            if s[c] not in ss: # Determine if a character is a duplicate
                ss=ss+s[c]
                if c==len(s)-1 and len(ss) > i: # Determine if it is the last character and if the length of the substring is the longest 
                    i=len(ss)
            elif len(ss) > i: # Determines if the length of the substring is the longest when the character is repeated
                i=len(ss)
                nc=s.rfind(s[c],0,c) # Find the last occurrence of duplicate characters
                ss=s[nc+1:c+1] # Let the substring be the content between the previous duplicate character and the duplicate character
            else:
                nc=s.rfind(s[c],0,c)
                ss=s[nc+1:c+1]
        return i

After coding, I submitted my solution.

Memory usage seems to be more than most, and I think this will be the focus of my future rethinking of the solution, and of course speed needs to be improved.

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This is the end of the leetcode challenge.

Feel free to give some suggestions based on my code, whether it's a way to improve performance or shorten the code.

Thanks for stopping by here and reading my post.