FirstUniqueCharacterInAString

Problem

https://leetcode.com/problems/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.

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:sup:`5`

• s consists of only lowercase English letters.

Solution

Use a dictionary to count the number of times each character appears in s. Iterate through the string again. The first character that has a count of 1 is the first unique character.

Pattern

Hash Table, String, Queue, Counting

Code

def firstUniqChar(s: str) -> int:
    """Returns the first character in ``s`` that doens't repeat.
    """
    counts = {}
    for char in s:
        if char in counts:
            counts[char] += 1
        else:
            counts[char] = 1

    for i, char in enumerate(s):
        if counts[char] == 1:
            return i
    return -1

Test

>>> from FirstUniqueCharacterInAString import firstUniqChar
>>> firstUniqChar('leetcode')
0
>>> firstUniqChar('loveleetcode')
2
>>> firstUniqChar('aabb')
-1

FirstUniqueCharacterInAString.firstUniqChar(s: str) → int

Returns the first character in s that doens’t repeat.