LeetCode 2000. Reverse Prefix of Word Solution in Java, C++, Python & More | Explanation + Code

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.

PythonJavaC++GoTypeScriptRustPHP

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 :](code-box)

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); } }(code-box)

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; } };(code-box)

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) }(code-box)

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); }(code-box)

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, } } }(code-box)

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; } }(code-box)