# LeetCode 387. First Unique Character in a String

## Description

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

Examples:

```s = "leetcode"
return 0.

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

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

## Java Solution

``````class Solution {
public int firstUniqChar(String s) {
if (s == null || s.length() == 0) {
return -1;
}

int[] charCounts = new int[26];

for (int i = 0; i < s.length(); i++) {
charCounts[s.charAt(i) - 'a']++;
}

for (int i = 0; i < s.length(); i++) {
if (charCounts[s.charAt(i) - 'a'] == 1) {
return i;
}
}

return -1;
}
}``````

## Python Solution

``````class Solution:
def firstUniqChar(self, s: str) -> int:
count = {}
for ch in s:
count[ch] = count.get(ch, 0) + 1

for i, ch in enumerate(s):
if count[ch] == 1:
return i

return -1``````
• Time complexity: O(N) since we go through the string of length `N` two times.
• Space complexity: O(N) since we have to keep a hash map with `N` elements.

## 2 Thoughts to “LeetCode 387. First Unique Character in a String”

1. Abbos says:

here I don’t understand why minus “a”

2. Abhishek says:

Simple solution

public class FirstNonRepetativeChar {

public static void main(String[] args) {
String input = “geeksforgeeks”;
char[] charr = input.toCharArray();
for (int i = 0; i < input.length(); i++) {
if (input.indexOf(charr[i]) == input.lastIndexOf(charr[i])) {
System.out.println(charr[i]);
break;
}
}

}

}