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Formatted question description: https://leetcode.ca/all/2351.html
2351. First Letter to Appear Twice
 Difficulty: Easy.
 Related Topics: Hash Table, String, Counting.
 Similar Questions: Two Sum, First Unique Character in a String.
Problem
Given a string s
consisting of lowercase English letters, return the first letter to appear **twice**.
Note:

A letter
a
appears twice before another letterb
if the second occurrence ofa
is before the second occurrence ofb
. 
s
will contain at least one letter that appears twice.
Example 1:
Input: s = "abccbaacz"
Output: "c"
Explanation:
The letter 'a' appears on the indexes 0, 5 and 6.
The letter 'b' appears on the indexes 1 and 4.
The letter 'c' appears on the indexes 2, 3 and 7.
The letter 'z' appears on the index 8.
The letter 'c' is the first letter to appear twice, because out of all the letters the index of its second occurrence is the smallest.
Example 2:
Input: s = "abcdd"
Output: "d"
Explanation:
The only letter that appears twice is 'd' so we return 'd'.
Constraints:

2 <= s.length <= 100

s
consists of lowercase English letters. 
s
has at least one repeated letter.
Solution (Java, C++, Python)

class Solution { public char repeatedCharacter(String s) { int n = s.length(); int[] hm = new int[26]; for (int i = 0; i < n; i++) { char c = s.charAt(i); hm[c  'a']++; if (hm[c  'a'] > 1) { return c; } } return '0'; } } ############ class Solution { public char repeatedCharacter(String s) { int mask = 0; for (int i = 0;; ++i) { char c = s.charAt(i); if ((mask >> (c  'a') & 1) == 1) { return c; } mask = 1 << (c  'a'); } } }

class Solution: def repeatedCharacter(self, s: str) > str: mask = 0 for c in s: i = ord(c)  ord('a') if mask >> i & 1: return c mask = 1 << i ############ # 2351. First Letter to Appear Twice # https://leetcode.com/problems/firstlettertoappeartwice/ class Solution: def repeatedCharacter(self, s: str) > str: seen = set() for x in s: if x in seen: return x seen.add(x)

class Solution { public: char repeatedCharacter(string s) { int mask = 0; for (int i = 0;; ++i) { if (mask >> (s[i]  'a') & 1) { return s[i]; } mask = 1 << (s[i]  'a'); } } };

func repeatedCharacter(s string) byte { mask := 0 for i := 0; ; i++ { if mask>>(s[i]'a')&1 == 1 { return s[i] } mask = 1 << (s[i]  'a') } }

function repeatedCharacter(s: string): string { let mask = 0; for (const c of s) { const i = c.charCodeAt(0)  'a'.charCodeAt(0); if (mask & (1 << i)) { return c; } mask = 1 << i; } return ' '; }

impl Solution { pub fn repeated_character(s: String) > char { let mut mask = 0; for &c in s.as_bytes() { if mask & 1 << (c  b'a') as i32 != 0 { return c as char; } mask = 1 << (c  b'a') as i32; } ' ' } }
Explain:
nope.
Complexity:
 Time complexity : O(n).
 Space complexity : O(n).