Reverse String II

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 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 leave the other as original.

Example 1:

Input: s = “abcdefg”, k = 2 Output: “bacdfeg”

Example 2:

Input: s = “abcd”, k = 2 Output: “bacd”

Constraints:

• 1 <= s.length <= 104
• s consists of only lowercase English letters.
• 1 <= k <= 104

Code

Time Complexity: $O(n)$, Space Complexity: $O(1)$

class Solution {
public:
    string reverseStr(string s, int k) {
        int i = 0, j = 0;
        int n = s.size();
        while(j < n) {
            j += 2 * k;
            if(j >= n) break;
            reverse(s.begin() + i, s.begin() + j - k);
            i = j;
        }
        
        if((n - i) < k) reverse(s.begin() + i, s.end());
        else {
            reverse(s.begin() + i, s.begin() + i + k);
        }
        return s;
    }   
};

Source

• Reverse String II - LeetCode