541 Reverse String II

1. Description

Given a string s and an integer k, reverse the first k characters for every 2k characters counting from the start of the string.

 

If there are fewer than k character left, reverse all of them. If there are less than 2k but greater than or eqaul to k characters, then reverse the first k characters and leave the other as original.

 

constraints :

• 1 <= s.length <= \(10^{4}\)
• s consists of only lowercase English letters.
• 1 <= k <= \(10^{4}\)

 

2. Algorithms

Let's consider the problem when we got input s as "abcdefghijk" and k as 3. Then we have to reverse the first 3 character s for every 6 characters from the start of the string.

 

If there are fewer than 3 characters left, we have to reverse all of them. If there are less than 6 but greater than or eqaul to 3 characters, then we have to reverse the first 3 characters and leave other as original.

 

"abcdef" -> "cbadef"

"ghijk" -> "ihgjk"

 

1. Start
2. Convert input string into list.
3. Repeat the steps through integers 0 - len(s) incrementing by 2k.
   1. Reverse the values in index i : i + k
4. Return the joined list.
5. End

 

3. Codes

class Solution:
    def reverseStr(self, s: str, k: int) -> str:
        s = list(s)
        for i in range(0, len(s), 2*k) : 
            s[i:i+k] = s[i:i+k][::-1]
        return "".join(s)

 

4. Conclusion