README_EN.md

comments	difficulty	edit_url	rating	source	tags
true	Easy	https://github.com/doocs/leetcode/edit/main/solution/2300-2399/2351.First%20Letter%20to%20Appear%20Twice/README_EN.md	1155	Weekly Contest 303 Q1	Bit Manipulation Hash Table String Counting

2351. First Letter to Appear Twice

中文文档

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

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, $C = 26$.

Python3

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

Java

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;
            }
        }
    }
}

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];
            }
        }
    }
};

Go

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]
		}
	}
}

TypeScript

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 ' ';
}

Rust

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;
        }
        ' '
    }
}

PHP

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];
            }
        }
    }
}

C

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

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 $s$.

Python3

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

Java

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');
        }
    }
}

C++

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');
        }
    }
};

Go

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')
	}
}

TypeScript

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 ' ';
}

Rust

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);
        }
        ' '
    }
}

C

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 ' ';
}