387. 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:

• $1<=s.length<=10^5$
• s consists of only lowercase English letters.

From: LeetCode
Link: 387. First Unique Character in a String

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Solution:

Ideas:

1. Frequency Array: We use an array freq[26] to store the count of each character. Since there are only 26 lowercase English letters, we can map each character to an index (s[i] - ‘a’).

2. First Pass: We iterate through the string once and count how many times each character appears.

3. Second Pass: We iterate through the string again and check the frequency of each character. The first character with a frequency of 1 is the first unique character, and we return its index.

4. Edge Case: If no unique character is found, the function returns -1.

Code:

int firstUniqChar(char* s) {
    int freq[26] = {0};  // Array to store frequency of each character

    // First pass: count the frequency of each character
    for (int i = 0; s[i] != '\0'; i++) {
        freq[s[i] - 'a']++;
    }

    // Second pass: find the first character with frequency 1
    for (int i = 0; s[i] != '\0'; i++) {
        if (freq[s[i] - 'a'] == 1) {
            return i;  // Return the index of the first unique character
        }
    }

    return -1;  // If no unique character is found, return -1
}