First Unique Character in a String

First Unique Character in a String - Problem

Given a string s, find the first non-repeating character in it and return its index. If it does not exist, return -1.

A non-repeating character appears exactly once in the string.

Input & Output

Example 1 — Basic Case

$ Input: s = "leetcode"

› Output: 4

💡 Note: The first unique character is 't' at index 4. Characters 'l' and 'e' appear multiple times.

Example 2 — No Unique Character

$ Input: s = "loveleetcode"

› Output: 2

💡 Note: The first unique character is 'v' at index 2.

Example 3 — All Characters Repeated

$ Input: s = "aabb"

› Output: -1

💡 Note: All characters appear more than once, so return -1.

Constraints

1 ≤ s.length ≤ 10⁵
s consists of only lowercase English letters.

Visualization

Tap to expand

Asked in

a Amazon 45 G Google 35 M Microsoft 28 🍎 Apple 20

Brief HTML summary: The key insight is to use frequency counting with a hash map. Best approach is the two-pass hash map solution. Time: O(n), Space: O(1)

Common Approaches

✓ Brute Force - Check Each Character

⏱️ Time: O(n²) Space: O(1)

For each position, scan the entire string to count how many times that character appears. Return the first position where count equals 1.

Two-Pass Hash Map

⏱️ Time: O(n) Space: O(1)

Use a hash map to count the frequency of each character in the first pass. Then iterate through the string again to find the first character with frequency 1.

Brute Force - Check Each Character — Algorithm Steps

Step 1: For each character at position i
Step 2: Count occurrences of s[i] in entire string
Step 3: If count is 1, return i
Step 4: If no unique character found, return -1

Visualization

Tap to expand

Step-by-Step Walkthrough

Check l

Count 'l' in entire string: appears 3 times

Check e

Count 'e' in entire string: appears 3 times

Check t

Count 't' in entire string: appears 1 time - FOUND!

Code -

solution.c — C

#include <stdio.h>
#include <string.h>

int solution(char* s) {
    int len = strlen(s);
    for (int i = 0; i < len; i++) {
        int count = 0;
        for (int j = 0; j < len; j++) {
            if (s[j] == s[i]) {
                count++;
            }
        }
        if (count == 1) {
            return i;
        }
    }
    return -1;
}

int main() {
    char s[100001];
    fgets(s, sizeof(s), stdin);
    s[strcspn(s, "\n")] = 0;
    int result = solution(s);
    printf("%d\n", result);
    return 0;
}

Time & Space Complexity

Time Complexity

⏱️

O(n²)

For each of n characters, we scan the entire string of n characters

✓ Linear Growth

Space Complexity

O(1)

Only using constant extra variables

✓ Linear Space

125.0K Views

High Frequency

~15 min Avg. Time

4.9K Likes

Ln 1, Col 1

Smart Actions

💡 Explanation

AI Ready

💡 Suggestion Tab to accept Esc to dismiss

// Output will appear here after running code

Code Editor Closed

Click the red button to reopen

Input & Output

Constraints

Visualization

Related Problems

Common Approaches

Brute Force - Check Each Character — Algorithm Steps

Visualization

Code -

Time & Space Complexity

Select Compiler