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.
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;
}
}
}
}