Leetcode: Reverse String II

Reverse string

Given a string and an integer k, you need to reverse the first k characters for every 2k characters counting from the start of the string. If there are less than k characters left, reverse all of them. If there are less than 2k but greater than or equal to k characters, then reverse the first k characters and left the other as original.

Input: s = "abcdefg", k = 2
Output: "bacdfeg"


  1. The string consists of lower English letters only.
  2. Length of the given string and k will in the range [1, 10000]

Github: code.dennyzhang.com

Credits To: leetcode.com

Hint: Time O(n), Space O(1). Moore voting

Leave me comments, if you have better ways to solve.

## Blog link: https://code.dennyzhang.com/reverse-string-ii
## Basic Ideas:
## Complexity: Time O(n), Space O(1)
## Assumptions:
class Solution(object):
    def reverseStr(self, s, k):
        :type s: str
        :type k: int
        :rtype: str
        length = len(s)
        l = list(s)
        should_reverse = True
        i = 0
        while i < length:
            if i+k-1 < length:
                j = i+k-1
                j = length - 1

            if should_reverse:
                # print("i: %d, j: %d" % (i, j))
                start = i
                end = j
                while start < end:
                    l[start],l[end] = l[end], l[start]
                    start += 1
                    end -= 1
            should_reverse = not should_reverse
            i = j + 1
        return ''.join(l)

Share It, If You Like It.

Leave a Reply

Your email address will not be published.