# 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,
}
}
}