First Unique Character in a String - Practice Coding Problems

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: 2

💡 Note: The first unique character is 't' at index 2. 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

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Understanding the Visualization

Input

String with some repeated characters

Process

Find first character that appears exactly once

Output

Return index of that character, or -1 if none exists

Key Takeaway

🎯 Key Insight: Use frequency counting to efficiently identify unique characters in linear time

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

Approach	Time	Space	Notes
✓ Two-Pass Hash Map	O(n)	O(1)	First pass counts frequencies, second pass finds first character with count 1
Brute Force - Check Each Character	O(n²)	O(1)	For each character, count its occurrences in the entire string

Two-Pass Hash Map — Algorithm Steps

Step 1: Create frequency map by scanning entire string
Step 2: Scan string again from left to right
Step 3: Return index of first character with frequency 1
Step 4: Return -1 if no unique character exists

Visualization

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Step-by-Step Walkthrough

Count Frequencies

Build hash map: l:3, e:3, t:1, c:1, o:1, d:1

Find First Unique

Scan left to right, check frequency map

Result

First character with frequency 1 is 't' at index 2

Code -

solution.c — C

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

int solution(char* s) {
    // First pass: count frequencies (assuming only lowercase letters)
    int freq[26] = {0};
    int len = strlen(s);
    
    for (int i = 0; i < len; i++) {
        freq[s[i] - 'a']++;
    }
    
    // Second pass: find first unique character
    for (int i = 0; i < len; i++) {
        if (freq[s[i] - 'a'] == 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(n)

Two passes through string of length n, hash map operations are O(1)

✓ Linear Growth

Space Complexity

O(1)

Hash map stores at most 26 lowercase letters (constant space)

✓ Linear Space

125.0K Views

High Frequency

~15 min Avg. Time

4.9K Likes

Ln 1, Col 1

Smart Actions

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Input & Output

Constraints

Visualization

Related Problems

Common Approaches

Two-Pass Hash Map — Algorithm Steps

Visualization

Code -

Time & Space Complexity

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