LeetCode 387 solution

problem

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

Example 1:

Input: s = "leetcode"
Output: 0

Example 2:

Input: s = "loveleetcode"
Output: 2

Example 3:

Input: s = "aabb"
Output: -1

Constraints:

• 1 <= s.length <= 105
• s consists of only lowercase English letters.

python

class Solution:
    def firstUniqChar(self, s: str) -> int:
        hash_t = {}
        for i in s:
            hash_t[i] = hash_t.get(i,0) + 1
        for idx,j in enumerate(s):
            if hash_t[j] == 1:
                return idx
        return -1

Its not that difficult problem. Just check the frequency of each char in s with hash table. And if the frequency is 1 -> return it's index, else return -1.

The important point of this code is hash_[t] = hash_t.get(i,0) + 1 this part. No need to conditionally make is there is no pre existence of char in hash table. This is python dict data structure's good point to get.