First Unique Character in a String
Given a string, find the index of the first character that appears only once. If no such character exists, return -1.
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Problem
Given a string s, find the first non-repeating character in it and return its index. If it does not exist, return -1.
Input
A string `s`.
Output
The index of the first non-repeating character, or -1 if every character appears more than once.
Examples
Input: s = "leetcode"
Output: 0
'l' appears only once and is at index 0.
Input: s = "loveleetcode"
Output: 2
'l' and 'o' repeat. 'v' at index 2 is the first unique character.
Input: s = "aabb"
Output: -1
Every character repeats. No unique character exists.
The brute-force approach
For each character, scan the entire string to count how many times it appears. Return the index of the first character whose count is exactly 1.
for i in range(len(s)):
count = 0
for j in range(len(s)): # scan entire string for every char
if s[j] == s[i]:
count += 1
if count == 1:
return i
return -1For each of the n characters you scan all n characters again. O(n²). You can get the counts for every character in one pass through the string.
Spotting the pattern
This is a Frequency Map problem. The key question to ask yourself:
Can I count every character's frequency in one pass and check in O(1) on the second pass?
Answering that is where it clicks, and it's exactly what the guided walkthrough below builds with you: the pattern reasoning, a progressive hint ladder that never spoils the answer, a row-by-row dry run, the optimized solution, and an in-browser editor to run your code against real test cases.
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