Formatted question description: https://leetcode.ca/all/2000.html

Description

LeetCode Problem 2000.

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.

Solution

  1. find ch index
  2. all chars before index -> reverse
class Solution {
    public String reversePrefix(String word, char ch) {
        
        int pos = word.indexOf(ch);
        
        if (pos < 0) {
            return word;
        }
        
        return new StringBuilder(word.substring(0, pos + 1)).reverse().toString() + word.substring(pos + 1);
    }
}

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