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.
here I don’t understand why minus “a”
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;
}
}
}
}