LeetCode 387 – Python 3 (Week 06 – 12)

Solution 1

class Solution:
    def firstUniqChar(self, s: str) -> int:
        
        # build hash map : character and how often it appears
        count = collections.Counter(s)
class Solution:
    """
    @param s: a string
    @return: it's index
    """
    def firstUniqChar(self, s):
        # write your code here
       count = collections.Counter(s)
       
       #alp = {}
       # for c in s:
       #     if c not in alp:
       #         alp[c] = 1
       #     else:
       #         alp[c] += 1
       
       
        # find the index
        for i in range(len(s)):
            if count[s[i]] == 1:
                return i     
        return -1

Time complexity is O(n). Space complexity is O(n).

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