Welcome to Subscribe On Youtube

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

• Java
• C++
• Python
• Go
• TypeScript
• Php
• RenderScript

• 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);
      }
  }
  
  ############
  
  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);
      }
  }
  

• // OJ: https://leetcode.com/problems/reverse-prefix-of-word/
  // Time: O(N)
  // Space: O(1)
  class Solution {
  public:
      string reversePrefix(string s, char ch) {
          int i = s.find(ch);
          if (i == string::npos) return s;
          reverse(begin(s), begin(s) + i + 1);
          return s;
      }
  };
  

• 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 :]
  
  ############
  
  # 2000. Reverse Prefix of Word
  # https://leetcode.com/problems/reverse-prefix-of-word/
  
  class Solution:
      def reversePrefix(self, word: str, ch: str) -> str:
          i = -1
          
          for index,x in enumerate(word):
              if x == ch:
                  i = index
                  break
          
          if i == -1: return word
          
          return word[:i + 1][::-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,
          }
      }
  }
  

All Problems

All Solutions