Coding problem · Easy · Basics

First Unique Character

Easy Algorithms & data structures

Given a string s of lower-case letters, return the index of the first character that appears exactly once. If there's none, return -1.

Example 1

Input: s = "leetcode"
Output: 0

Example 2

Input: s = "loveleetcode"
Output: 2

Example 3

Input: s = "aabb"
Output: -1

Hints

Hint 1

First count how many times each character appears.

Hint 2

Then scan again and return the first index whose character has count 1.

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Solution and explanation

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Time O(n) · Space O(1)

Two passes: build a frequency map, then find the first character with frequency 1. Both passes are O(n).

Reference solution

function firstUniqueChar(s) {
  const count = new Map();
  for (const ch of s) count.set(ch, (count.get(ch) ?? 0) + 1);
  for (let i = 0; i < s.length; i++) if (count.get(s[i]) === 1) return i;
  return -1;
}

function firstUniqueChar(s: string): number {
  const count = new Map<string, number>();
  for (const ch of s) count.set(ch, (count.get(ch) ?? 0) + 1);
  for (let i = 0; i < s.length; i++) if (count.get(s[i]) === 1) return i;
  return -1;
}

from collections import Counter

def first_unique_char(s: str) -> int:
    count = Counter(s)
    for i, ch in enumerate(s):
        if count[ch] == 1:
            return i
    return -1

<?php

function firstUniqueChar(string $s): int
{
    $count = count_chars($s, 1);
    for ($i = 0; $i < strlen($s); $i++) {
        if ($count[ord($s[$i])] === 1) {
            return $i;
        }
    }
    return -1;
}

class Solution {
    public int firstUniqueChar(String s) {
        int[] count = new int[26];
        for (char ch : s.toCharArray()) count[ch - 'a']++;
        for (int i = 0; i < s.length(); i++) if (count[s.charAt(i) - 'a'] == 1) return i;
        return -1;
    }
}

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