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First Unique Character in a String

Find the index of the first non-repeating character with two passes: count every letter in a hash map, then return the first index with count 1.

EasySolve on LeetCode

Problem Understanding

First Unique Character in a String: given a string s of lowercase letters, return the index of the first character that does not repeat anywhere in s, or -1 if there is none.

Example: "loveleetcode" returns 2: l and o repeat, and v at index 2 appears once.

The Intuition: Count First, Decide Second

Checking each letter against the rest of the string is O(n²). Instead, use a hash map of counts:

  1. Pass 1: count every letter.
  2. Pass 2: walk s again in order and return the first index whose count is 1.

The second pass matters: the counts say which letters are unique, but only the original order says which comes first. The same count-then-use shape solves Valid Anagram and Ransom Note.

Interactive Walkthrough

Pass 1 fills each letter as it is counted, and the counts appear beside the figure. Pass 2 walks the string again, greying out repeated letters until it finds one with count 1. Try Late repeat, where letters that look unique early on turn out to repeat.

Dry Run Table

Input: s = "loveleetcode"

Counts after pass 1: l:2, o:2, v:1, e:4, t:1, c:1, d:1.

Index Letter Count Action
0 l 2 skip
1 o 2 skip
2 v 1 return 2

Answer: 2.

The Solution Template

Count, then scan. Python uses Counter; the others a 26-slot array.

First Unique Character in a String Code
1var firstUniqChar = function(s) {
2    const counts = new Array(26).fill(0);
3    for (const c of s) counts[c.charCodeAt(0) - 97]++;
4    for (let i = 0; i < s.length; i++) {
5        if (counts[s.charCodeAt(i) - 97] === 1) return i;
6    }
7    return -1;
8};

Edge Cases & Common Mistakes

  • Deciding during the first pass: a letter that looks unique so far may repeat later ("abcab"); wait for the full count.
  • Scanning the count table instead of the string: finds a unique letter, but not necessarily the first one in s.
  • No unique letter: return -1, as in "aabb".
  • Single character: it is unique; return 0.
  • Returning the letter: the problem asks for its index.

Summary

Count every letter in one pass, then walk the string again and return the first index whose count is 1, or -1. O(n) time and O(1) space for a fixed alphabet.

More practice: DSA interview questions with answers · patterns cheat sheet

Previous

Ransom Note

Next

Roman to Integer

The approach, step by step

  1. Count the letters

    Walk s once and count each letter in a hash map or 26-slot array.

  2. Scan in order

    Walk s again and return the first index whose letter has count 1.

  3. Return -1

    If no letter has count 1, there is no unique character.

Frequently asked questions

Why does First Unique Character need two passes?

A letter can only be called unique after the whole string has been seen, because a repeat might come at the very end. The first pass counts every letter; the second pass, in the original order, finds the first one with count 1.

Can it be done in one pass?

Yes: record each letter's first index and mark letters seen twice. Then the answer is the smallest first index among letters seen once, found by scanning the 26 letters instead of the string. That is useful when the string is a stream.

What is the space complexity?

O(1). With only lowercase letters, the count table has at most 26 entries no matter how long the string is.

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