LeetCode Blog for course "Algorithms" -- Problem 3 & 5

Problem 3. Longest Substring Without Repeating Characters

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

Examples:
Given "abcabcbb", the answer is "abc", which the length is 3.
Given "bbbbb", the answer is "b", with the length of 1.
Given "pwwkew", the answer is "wke", with the length of 3. Note that the answer must be a substring, "pwke" is a subsequence and not a substring.


My solution in Python:

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class Solution(object):    def lengthOfLongestSubstring(self, s):        """        :type s: str        :rtype: int        """        answer = 0;        left = 0;        last = {};        for i in range(len(s)):            if s[i] in last and last[s[i]] >= left:                left = last[s[i]] +1;            last[s[i]] = i;            answer = max(answer, i - left + 1);        return answer;

Review:

We make good use of the “dictionary” in Python here. Starting with the first character in the string, we go through the string characters one by one, and store one item in the dictionary, with the character being the key, and the place it appears in the string being the value. When we encounter a character that has already appeared previously, we immediately know because this particular key (the character) is already in the dictionary. Thus we move the starting character of the substring to the next position of which the repeated character first appeared in the original string. We then compare the length of the current non-repeated-character-substring with the longest substring we already know. When “i” reaches the last character of the original string, the “answer” should be the length of the longest substring without repeated character.





Problem 5. Longest Palindromic Substring



Given a string s, find the longest palindromic substring in s. You may assume that the maximum length of s is 1000.



Example 1:

Input: “babad”
Output: “bab”
Note: “aba” is also a valid answer.





Example 2:

Input: “cbbd”
Output: “bb”



My solution in Python: 

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class Solution(object):    def longestPalindrome(self, s):        """        :type s: str        :rtype: str        """        ansl, ansr, maxx = 0, 0, 0        length = len(s)        for i in range(1, length * 2):            if i & 1 :                left = i / 2                right = left            else :                left = i / 2 - 1                right = left + 1            while (left >= 0) and (right < length) and (s[left] == s[right]):                left -= 1                right += 1            left += 1            right -= 1            if right - left > maxx:                maxx = right - left                ansl = left                ansr = right        return s[ansl: ansr + 1]

Review:

We observe that a palindrome mirrors around its center. Therefore, a palindrome can be expanded from its center, and there are only 2n-1 such centers. By treating odd “i” and even “i” differently, we are able to examine both palindromes with even length and with odd length. This question is actually very interesting and has many (at least 5) different solutions, each with different time and space efficiency. The most advanced one, the Manacher’s algorithm, can be found here.