First Unique Character in a String - Leet Code Solution

Problem Statement

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

Example

s = "leetcode"
return 0.

s = "loveleetcode"
return 2.

Note: You may assume the string contains only lowercase English letters.

Solution Brute Force

Lets take a look at the simple solution.

Have two loops. first will iterate till length of string
in second loop, iterate from beginning to end.
And check if the character is found anywhere
We can keep track of duplicate found by a boolean flag
And if during our inner loop, if we found that character is not found. This is our answer

Code

public int firstUniqChar_bruteforce(String s) {
   for (int i=0; i<s.length(); i++) {
      boolean unique = true;
      for (int j=0; j<s.length(); j++) {
         if (i != j && s.charAt(i) == s.charAt(j)) {
            unique = false;
            break;
         }
      }
      
      if (unique) {
         return i;
      }
   }
   
   return -1;
}

Complexity

Its O(n^2)

Another Solution using a HashMap

Iterate over string, and keep track of count of each character
Maintain a HashMap<Character, Integer>
Now, iterate over string again
For each character, lookup in our HashMap
If the count of that character is only 1, this is our answer
Since, 1 means this character is in the string only 1 times.

Code

public int firstUniqChar(String s) {
   Map<Character, Integer> map = new HashMap<Character, Integer>();
   for (int i=0; i<s.length(); i++) {
      int count = map.getOrDefault(s.charAt(i), 0);
      count ++;
      map.put(s.charAt(i), count);
   }
   
   for (int i=0; i<s.length(); i++) {
      if (map.get(s.charAt(i)) == 1) {
         return i;
      }
   }
   return -1;
}