First Unique Character in String — Two-Pass Frequency Map [Amazon Easy]

Problem Statement

Input: s = 'leetcode' → 0 (l appears once). Input: s = 'aabb' → -1 (no unique char).

Key Insight: First pass: count frequency of every character. Second pass: return the first index where frequency equals 1.

C Solution
C++ Solution
Java Solution
JavaScript Solution
Python Solution
Complexity Analysis

C Solution

#include <string.h>
int firstUniqChar(char* s) {
    int freq[26] = {0};
    int n = strlen(s);
    for (int i = 0; i < n; i++) freq[s[i]-'a']++;
    for (int i = 0; i < n; i++) if (freq[s[i]-'a'] == 1) return i;
    return -1;
}

C++ Solution

#include <string>
using namespace std;
class Solution {
public:
    int firstUniqChar(string s) {
        int freq[26] = {};
        for (char c : s) freq[c-'a']++;
        for (int i = 0; i < (int)s.size(); i++)
            if (freq[s[i]-'a'] == 1) return i;
        return -1;
    }
};

Java Solution

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

JavaScript Solution

function firstUniqChar(s) {
    const freq = new Array(26).fill(0);
    for (const c of s) freq[c.charCodeAt(0)-97]++;
    for (let i = 0; i < s.length; i++)
        if (freq[s.charCodeAt(i)-97] === 1) return i;
    return -1;
}

Python Solution

def firstUniqChar(s):
    from collections import Counter
    freq = Counter(s)
    for i, c in enumerate(s):
        if freq[c] == 1:
            return i
    return -1

Complexity Analysis

| O(n) time | O(1) space (26-element array) |

Streaming variant: Use LinkedHashMap to maintain insertion order while tracking counts — first key with count 1 is the answer.