387. First Unique Character in a String 

Description

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 <= 10⁵
s consists of only lowercase English letters.

Solutions

Solution 1: Counting

We use a hash table or an array of length $26$ $\text{cnt}$ to store the frequency of each character. Then, we traverse each character $\text{s[i]}$ from the beginning. If $\text{cnt[s[i]]}$ is $1$, we return $i$.

If no such character is found after the traversal, we return $-1$.

The time complexity is $O(n)$, where $n$ is the length of the string. The space complexity is $O(|\Sigma|)$, where $\Sigma$ is the character set. In this problem, the character set consists of lowercase letters, so $|\Sigma|=26$.

Python3

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

Java

class Solution {
    public int firstUniqChar(String s) {
        int[] cnt = new int[26];
        int n = s.length();
        for (int i = 0; i < n; ++i) {
            ++cnt[s.charAt(i) - 'a'];
        }
        for (int i = 0; i < n; ++i) {
            if (cnt[s.charAt(i) - 'a'] == 1) {
                return i;
            }
        }
        return -1;
    }
}

C++

class Solution {
public:
    int firstUniqChar(string s) {
        int cnt[26]{};
        for (char& c : s) {
            ++cnt[c - 'a'];
        }
        int n = s.size();
        for (int i = 0; i < n; ++i) {
            if (cnt[s[i] - 'a'] == 1) {
                return i;
            }
        }
        return -1;
    }
};

Go

func firstUniqChar(s string) int {
	cnt := [26]int{}
	for _, c := range s {
		cnt[c-'a']++
	}
	for i, c := range s {
		if cnt[c-'a'] == 1 {
			return i
		}
	}
	return -1
}

TypeScript

function firstUniqChar(s: string): number {
    const cnt = new Map<string, number>();
    for (const c of s) {
        cnt.set(c, (cnt.get(c) || 0) + 1);
    }
    for (let i = 0; i < s.length; ++i) {
        if (cnt.get(s[i]) === 1) {
            return i;
        }
    }
    return -1;
}

JavaScript

/**
 * @param {string} s
 * @return {number}
 */
var firstUniqChar = function (s) {
    const cnt = new Map();
    for (const c of s) {
        cnt.set(c, (cnt.get(c) || 0) + 1);
    }
    for (let i = 0; i < s.length; ++i) {
        if (cnt.get(s[i]) === 1) {
            return i;
        }
    }
    return -1;
};

PHP

class Solution {
    /**
     * @param String $s
     * @return Integer
     */
    function firstUniqChar($s) {
        for ($i = 0; $i < strlen($s); $i++) {
            $hashtable[$s[$i]]++;
        }
        for ($i = 0; $i < strlen($s); $i++) {
            if ($hashtable[$s[$i]] == 1) {
                return $i;
            }
        }
        return -1;
    }
}