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541. 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

Solutions

Solution 1: Two Pointers

We can traverse the string \(\textit{s}\), iterating over every \(\textit{2k}\) characters, and then use the two-pointer technique to reverse the first \(\textit{k}\) characters among these \(\textit{2k}\) characters.

The time complexity is \(O(n)\), and the space complexity is \(O(n)\). Here, \(n\) is the length of the string \(\textit{s}\).

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

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class Solution {
    public String reverseStr(String s, int k) {
        char[] cs = s.toCharArray();
        int n = cs.length;
        for (int i = 0; i < n; i += k * 2) {
            for (int l = i, r = Math.min(i + k - 1, n - 1); l < r; ++l, --r) {
                char t = cs[l];
                cs[l] = cs[r];
                cs[r] = t;
            }
        }
        return new String(cs);
    }
}

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class Solution {
public:
    string reverseStr(string s, int k) {
        int n = s.size();
        for (int i = 0; i < n; i += 2 * k) {
            reverse(s.begin() + i, s.begin() + min(i + k, n));
        }
        return s;
    }
};

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func reverseStr(s string, k int) string {
    cs := []byte(s)
    n := len(cs)
    for i := 0; i < n; i += 2 * k {
        for l, r := i, min(i+k-1, n-1); l < r; l, r = l+1, r-1 {
            cs[l], cs[r] = cs[r], cs[l]
        }
    }
    return string(cs)
}

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function reverseStr(s: string, k: number): string {
    const n = s.length;
    const cs = s.split('');
    for (let i = 0; i < n; i += 2 * k) {
        for (let l = i, r = Math.min(i + k - 1, n - 1); l < r; l++, r--) {
            [cs[l], cs[r]] = [cs[r], cs[l]];
        }
    }
    return cs.join('');
}

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