First Unique Character in a String – Solution & Complexity

Q: What is the time and space complexity of First Unique Character in a String?

The optimal First Unique Character in a String solution runs in `O(n)` time and `O(1)` space. Count character frequencies, then return the first index with a count of one.

Q: What is the most efficient way to solve First Unique Character in a String?

Count character frequencies, then return the first index with a count of one. This gives `O(n)` time complexity, and is a classic strings and hashmap problem.

Q: How do you solve First Unique Character in a String in Python & JavaScript?

We provide a worked First Unique Character in a String solution in Python & JavaScript with a step-by-step explanation above. You can also open the interactive editor to write and run your own Python solution against the test cases in the browser.

Q: Is First Unique Character in a String an easy or hard interview question?

First Unique Character in a String is generally rated **Easy**. It tests strings and hashmap, which are common coding-interview patterns.

1. Understand the Goal

Return the index of the first character that appears exactly once.
If every character repeats, return -1.
Since the first unique character depends on original order, we must preserve or revisit indices.

2. Consider the Naive Approach

For each character, scan the whole string to count how many times it appears.
The first character with count one is the answer.
This works, but it can take O(n²) time for long strings.

def first_uniq_char(s):
    for i, char in enumerate(s):
        if s.count(char) == 1:
            return i
    return -1

3. Count Frequencies First

Build a frequency map in one pass so each character count is available quickly.
Then make a second pass through the string in original order.
The first index whose character has count one is the required answer.

def first_uniq_char(s):
    counts = {}
    for char in s:
        counts[char] = counts.get(char, 0) + 1
    for i, char in enumerate(s):
        if counts[char] == 1:
            return i
    return -1

4. Why Two Passes Are Useful

The first pass answers the question: how many times does each character appear?
The second pass answers the question: which unique character appears first?
Separating these keeps the algorithm simple and linear.

5. Final Complete Solution

Time complexity is O(n) because the string is traversed twice.
Space complexity is O(1) if limited to lowercase English letters, otherwise O(k) for distinct characters.
Returning -1 covers the case where no unique character exists.

def first_uniq_char(s: str) -> int:
    counts = {}

    for char in s:
        counts[char] = counts.get(char, 0) + 1

    for i, char in enumerate(s):
        if counts[char] == 1:
            return i

    return -1

First Unique Character in a String – Solution & Complexity

1. Understand the Goal

2. Consider the Naive Approach

3. Count Frequencies First

4. Why Two Passes Are Useful

5. Final Complete Solution

FAQ

What is the time and space complexity of First Unique Character in a String?

What is the most efficient way to solve First Unique Character in a String?

How do you solve First Unique Character in a String in Python & JavaScript?

Is First Unique Character in a String an easy or hard interview question?