2351. First Letter to Appear Twice

Description

Given a string s consisting of lowercase English letters, return the first letter to appear twice.

Note:

A letter a appears twice before another letter b if the second occurrence of a is before the second occurrence of b.
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.

Solutions

Solution 1: Array or Hash Table

Thinking

We want the first letter whose count reaches two. \(s\) is short and has a duplicate, so one scan suffices.

Increment a map or array; return as soon as some key becomes \(2\).

We traverse the string \(s\), using an array or hash table cnt to record the occurrence of each letter. When a letter appears twice, we return that letter.

The time complexity is \(O(n)\) and the space complexity is \(O(C)\). Here, \(n\) is the length of the string \(s\), and \(C\) is the size of the character set. In this problem, \(C = 26\).

Python3JavaC++GoTypeScriptRustPHPC

class Solution:
    def repeatedCharacter(self, s: str) -> str:
        cnt = Counter()
        for c in s:
            cnt[c] += 1
            if cnt[c] == 2:
                return c

class Solution {
    public char repeatedCharacter(String s) {
        int[] cnt = new int[26];
        for (int i = 0;; ++i) {
            char c = s.charAt(i);
            if (++cnt[c - 'a'] == 2) {
                return c;
            }
        }
    }
}

class Solution {
public:
    char repeatedCharacter(string s) {
        int cnt[26]{};
        for (int i = 0;; ++i) {
            if (++cnt[s[i] - 'a'] == 2) {
                return s[i];
            }
        }
    }
};

func repeatedCharacter(s string) byte {
    cnt := [26]int{}
    for i := 0; ; i++ {
        cnt[s[i]-'a']++
        if cnt[s[i]-'a'] == 2 {
            return s[i]
        }
    }
}

function repeatedCharacter(s: string): string {
    const vis = new Array(26).fill(false);
    for (const c of s) {
        const i = c.charCodeAt(0) - 'a'.charCodeAt(0);
        if (vis[i]) {
            return c;
        }
        vis[i] = true;
    }
    return ' ';
}

impl Solution {
    pub fn repeated_character(s: String) -> char {
        let mut vis = [false; 26];
        for &c in s.as_bytes() {
            if vis[(c - b'a') as usize] {
                return c as char;
            }
            vis[(c - b'a') as usize] = true;
        }
        ' '
    }
}

class Solution {
    /**
     * @param String $s
     * @return String
     */
    function repeatedCharacter($s) {
        for ($i = 0; ; $i++) {
            $hashtable[$s[$i]] += 1;
            if ($hashtable[$s[$i]] == 2) {
                return $s[$i];
            }
        }
    }
}

char repeatedCharacter(char* s) {
    int vis[26] = {0};
    for (int i = 0; s[i]; i++) {
        if (vis[s[i] - 'a']) {
            return s[i];
        }
        vis[s[i] - 'a']++;
    }
    return ' ';
}

Solution 2: Bit Manipulation

Thinking

Method 1 stores counts. We only need “seen or not”, so one bit per letter in an integer mask, in constant space.

We can also use an integer mask to record whether each letter has appeared, where the \(i\)-th bit of mask indicates whether the \(i\)-th letter has appeared. When a letter appears twice, we return that letter.

The time complexity is \(O(n)\) and the space complexity is \(O(1)\). Here, \(n\) is the length of the string \(s\).

Python3JavaC++GoTypeScriptRustC

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

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 {
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);
        }
        ' '
    }
}

char repeatedCharacter(char* s) {
    int mask = 0;
    for (int i = 0; s[i]; i++) {
        if (mask & (1 << s[i] - 'a')) {
            return s[i];
        }
        mask |= 1 << s[i] - 'a';
    }
    return ' ';
}