LeetCode - First Unique Character in a String

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.

Solution

Create an Array of length 26 and maintain frequency of the each character. Here ‘a’ having index as 0 and ‘z’ having index as 25. Iterate over the String and increament the value in the array.

After we build frequency Array, Just iterate a String one more time and check for each character, if frequency is 1 then that index is the answer.

Code


 public class FirstUniqueCharacter {

	public int firstUniqChar(String s) {

		int a[] = new int[26];

		for (char c : s.toCharArray()) {
			a[c - 'a']++;
		}

		for (int i = 0; i < s.length(); i++) {

			if (a[s.charAt(i) - 'a'] == 1)
				return i;
		}

		return -1;
	}

	public static void main(String[] args) {
		FirstUniqueCharacter frstUncCharObj=new FirstUniqueCharacter();
		System.out.println(frstUncCharObj.firstUniqChar("leetcode"));

	}

}