387. First Unique Character in a String

EasyLeetCode

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

Explanation: The character 'l' at index 0 is the first character that does not occur at any other index.

Example 2

Input: s = "loveleetcode"

Output: 2

Example 3

Input: s = "aabb"

Output: -1

Constraints

• 1 <= s.length <= 10^5
• s consists of only lowercase English letters.

How to solve the problem

• Hash Table with Character Counting

Code

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

from collections import Counter

class Solution:
    def firstUniqChar(self, s: str) -> int:
        myHash = Counter(s)
        for i, c in enumerate(s):
            if myHash[c] == 1:
                return i
        return -1

Complexity

• Approach 1: Hash Table with Character Counting

• Time complexity: O(n)

• Space complexity: O(1) We only store character counts, which is bounded by the size of the alphabet (26 for lowercase English letters).