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"
andch = "d"
, then you should reverse the segment that starts at0
and ends at3
(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 $word$.
1 2 3 4 |
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1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 |
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1 2 3 4 5 6 7 8 9 10 |
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1 2 3 4 5 6 7 8 9 10 11 12 |
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1 2 3 4 5 6 7 |
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1 2 3 4 5 6 7 8 |
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1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 |
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