> For the complete documentation index, see [llms.txt](https://lei-d.gitbook.io/leetcode/llms.txt). Markdown versions of documentation pages are available by appending `.md` to page URLs; this page is available as [Markdown](https://lei-d.gitbook.io/leetcode/string/387first-unique-character-in-a-string.md).

# 387\_First Unique Character in a String

\[easy] \[string, hash table]

Given a string, find the first non-repeating character in it and return it's index. If it doesn't exist, return -1.

**Note:**You may assume the string contain only lowercase letters.

**Examples:**

```
s = "leetcode"
return 0.

s = "loveleetcode",
return 2.
```

## Solution 1: brute force

**Idea**:

* Loop over each character from left to right, count the number of its occurrence. If it only occur once, return the index. 

**Time Complexity**: $$O(n^2)$$

**Space Complexity**: $$O(1)$$

```python
def firstUniqChar(s):
    """
    :type s: str
    :rtype: int
    """

    for i in range(len(s)):
        c = s[i]
        if s.count(c) == 1: # can write count as a helper function if needed
            return i
    return -1
```

## Solution 2: Use dictionary

**Idea**:

* Loop over each character from left to right.
* Store the character and its index in a dictionary if it's the first time it occur.
* If it occur more than once, store the character in another set.
* return the minimum index in the dictionary if exists.

**Time Complexity**: $$O(n)$$

**Space Complexity**: $$O(n)$$

```python
def firstUniqChar(s):
    """
    :type s: str
    :rtype: int
    """

    dict1 = {} # for characters appear once
    dict2 = set() # for characters appear more than once
    for i in range(len(s)):
        c = s[i]
        if c not in dict1 and c not in dict2:
            dict1[c] = i
        elif c in dict1:
            del(dict1[c])
            dict2.add(c)
    return min(dict1.values()) if len(dict1) > 0 else -1
```