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2000. Reverse Prefix of Word

Description

Given a 0-indexed string word and a character ch, reverse the segment of word that starts at index 0 and ends at the index of the first occurrence of ch (inclusive). If the character ch does not exist in word, do nothing.

  • For example, if word = "abcdefd" and ch = "d", then you should reverse the segment that starts at 0 and ends at 3 (inclusive). The resulting string will be "dcbaefd".

Return the resulting string.

 

Example 1:

Input: word = "abcdefd", ch = "d"
Output: "dcbaefd"
Explanation: The first occurrence of "d" is at index 3. 
Reverse the part of word from 0 to 3 (inclusive), the resulting string is "dcbaefd".

Example 2:

Input: word = "xyxzxe", ch = "z"
Output: "zxyxxe"
Explanation: The first and only occurrence of "z" is at index 3.
Reverse the part of word from 0 to 3 (inclusive), the resulting string is "zxyxxe".

Example 3:

Input: word = "abcd", ch = "z"
Output: "abcd"
Explanation: "z" does not exist in word.
You should not do any reverse operation, the resulting string is "abcd".

 

Constraints:

  • 1 <= word.length <= 250
  • word consists of lowercase English letters.
  • ch is a lowercase English letter.

Solutions

Solution 1: Simulation

First, we find the index $i$ where the character $ch$ first appears. Then, we reverse the characters from index $0$ to index $i$ (including index $i$). Finally, we concatenate the reversed string with the string starting from index $i + 1$.

The time complexity is $O(n)$, and the space complexity is $O(n)$. Here, $n$ is the length of the string $word$.

  • class Solution {
    public String reversePrefix(String word, char ch) {
        int j = word.indexOf(ch);
        if (j == -1) {
            return word;
        }
        char[] cs = word.toCharArray();
        for (int i = 0; i < j; ++i, --j) {
            char t = cs[i];
            cs[i] = cs[j];
            cs[j] = t;
        }
        return String.valueOf(cs);
    }
}

  • class Solution {
public:
    string reversePrefix(string word, char ch) {
        int i = word.find(ch);
        if (i != string::npos) {
            reverse(word.begin(), word.begin() + i + 1);
        }
        return word;
    }
};

  • class Solution:
    def reversePrefix(self, word: str, ch: str) -> str:
        i = word.find(ch)
        return word if i == -1 else word[i::-1] + word[i + 1 :]


  • func reversePrefix(word string, ch byte) string {
	j := strings.IndexByte(word, ch)
	if j < 0 {
		return word
	}
	s := []byte(word)
	for i := 0; i < j; i++ {
		s[i], s[j] = s[j], s[i]
		j--
	}
	return string(s)
}

  • function reversePrefix(word: string, ch: string): string {
    const i = word.indexOf(ch) + 1;
    if (!i) {
        return word;
    }
    return [...word.slice(0, i)].reverse().join('') + word.slice(i);
}


  • class Solution {
    /**
     * @param String $word
     * @param String $ch
     * @return String
     */
    function reversePrefix($word, $ch) {
        $len = strlen($word);
        $rs = '';
        for ($i = 0; $i < $len; $i++) {
            $rs = $rs . $word[$i];
            if ($word[$i] == $ch) {
                break;
            }
        }
        if (strlen($rs) == $len && $rs[$len - 1] != $ch) {
            return $word;
        }
        $rs = strrev($rs);
        $rs = $rs . substr($word, strlen($rs));
        return $rs;
    }
}


  • impl Solution {
    pub fn reverse_prefix(word: String, ch: char) -> String {
        match word.find(ch) {
            Some(i) => word[..=i].chars().rev().collect::<String>() + &word[i + 1..],
            None => word,
        }
    }
}

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