Leetcode self-record [day 5]

[Data Structure Study Plan] - Day 5

387. First Unique Character in a String

Given a string s, find the first non-repeating character in it and return its index. If it does not exist, return -1.

Solution 1:

class Solution:
    def firstUniqChar(self, s):
        for i in s:
            if s.count(i) == 1:
                return s.index(i)
        return -1

Feedback:

Runtime: 4872 ms, faster than 5.00% of Python3 online submissions for First Unique Character in a String.

Memory Usage: 14.2 MB, less than 85.28% of Python3 online submissions for First Unique Character in a String

Solution 2:

class Solution:
    def firstUniqChar(self, s):
        return min([s.find(c) for c in string.ascii_lowercase if s.count(c)==1] or [-1])
    

Remark: same idea with solution 1, but improved with built-in function string.ascii_lowercase.

Feedback: 

Runtime: 52 ms, faster than 99.19% of Python3 online submissions for First Unique Character in a String.

Memory Usage: 14.2 MB, less than 85.28% of Python3 online submissions for First Unique Character in a String.

383. Ransom Note

Given two strings ransomNote and magazine, return true if ransomNote can be constructed from magazine and false otherwise.

Each letter in magazine can only be used once in ransomNote.

class Solution:
    def canConstruct(self, ransomNote, magazine):
        if set(ransomNote) <= set(magazine):
            for i in set(ransomNote):
                if ransomNote.count(i) <= magazine.count(i):
                    pass
                else:
                    return False
            return True
        return False

Feedback:

Runtime: 61 ms, faster than 75.10% of Python3 online submissions for Ransom Note.

Memory Usage: 14.1 MB, less than 82.56% of Python3 online submissions for Ransom Note.

class Solution:
    def canConstruct(self, ransomNote, magazine):
        for i in set(ransomNote):
            if ransomNote.count(i) > magazine.count(i):
                return False
        return True

Feedback:

Runtime: 37 ms, faster than 96.08% of Python3 online submissions for Ransom Note.

Memory Usage: 14.1 MB, less than 82.56% of Python3 online submissions for Ransom Note.

242. Valid Anagram

Given two strings s and t, return true if t is an anagram of s, and false otherwise.

An Anagram is a word or phrase formed by rearranging the letters of a different word or phrase, typically using all the original letters exactly once.

class Solution:
    def isAnagram(self, s, t):
        m1, m2 = {}, {}
        for i in set(s):
            m1[i] = s.count(i)
        for j in set(t):
            m2[j] = t.count(j)
        return m1 == m2
    

Feedback:

Runtime: 36 ms, faster than 97.63% of Python3 online submissions for Valid Anagram.

Memory Usage: 14.5 MB, less than 77.52% of Python3 online submissions for Valid Anagram.

To be continued... : )