Solution 1 - Frequency Counting

Count each character, then return the first index whose character occurs once.

TimeComplexity: O(n)

SpaceComplexity: O(k)

""" 0387.1 - First Unique Character in a String - Solution 1 - Frequency Counting """

from collections import Counter

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

if __name__ == "__main__":
    assert Solution().firstUniqChar("leetcode") == 0

function firstUniqChar(s:string):number{
 const m:Record<string,number>={};
 for(const ch of s) m[ch]=(m[ch]??0)+1;
 for(let i=0;i<s.length;i++) if(m[s[i]]===1) return i;
 return -1;
}
console.log(firstUniqChar("leetcode"));

package main

func firstUniqChar(s string) int {
	cnt := map[byte]int{}
	for i := 0; i < len(s); i++ {
		cnt[s[i]]++
	}
	for i := 0; i < len(s); i++ {
		if cnt[s[i]] == 1 {
			return i
		}
	}
	return -1
}

func main() {}

fn first_uniq_char(s: &str) -> i32 {
	let mut cnt = [0i32; 26];
	for b in s.bytes() { cnt[(b - b'a') as usize] += 1; }
	for (i, b) in s.bytes().enumerate() {
		if cnt[(b - b'a') as usize] == 1 { return i as i32; }
	}
	-1
}
fn main() {}

0387 - First Unique Character in a String

Solution 1 - Frequency Counting