LeetCode 387. First Unique Character in a String

Description

Given a string, find the first non-repeating character in it and return it’s index. If it doesn’t exist, return -1.

Examples:

s = "leetcode"
return 0.

s = "loveleetcode",
return 2.

Note: You may assume the string contain only lowercase letters.

Java Solution

class Solution {
    public int firstUniqChar(String s) {
        if (s == null || s.length() == 0) {
            return -1;
        }
        
        int[] charCounts = new int[26];
        
        for (int i = 0; i < s.length(); i++) {
            charCounts[s.charAt(i) - 'a']++;            
        }
        
        for (int i = 0; i < s.length(); i++) {
            if (charCounts[s.charAt(i) - 'a'] == 1) {
                return i;
            }
        }
        
        return -1;
    }
}

Python Solution

class Solution:
    def firstUniqChar(self, s: str) -> int:        
        count = {}
        for ch in s:
            count[ch] = count.get(ch, 0) + 1
        
        for i, ch in enumerate(s):
            if count[ch] == 1:
                return i
        
        return -1
  • Time complexity : O(N) since we go through the string of length N two times.
  • Space complexity : O(N) since we have to keep a hash map with N elements.

One Thought to “LeetCode 387. First Unique Character in a String”

  1. Simple solution

    public class FirstNonRepetativeChar {

    public static void main(String[] args) {
    String input = “geeksforgeeks”;
    char[] charr = input.toCharArray();
    for (int i = 0; i < input.length(); i++) {
    if (input.indexOf(charr[i]) == input.lastIndexOf(charr[i])) {
    System.out.println(charr[i]);
    break;
    }
    }

    }

    }

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