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3794. Reverse String Prefix

Description

You are given a string s and an integer k.

Reverse the first k characters of s and return the resulting string.

 

Example 1:

Input: s = "abcd", k = 2

Output: "bacd"

Explanation:​​​​​​​

The first k = 2 characters "ab" are reversed to "ba". The final resulting string is "bacd".

Example 2:

Input: s = "xyz", k = 3

Output: "zyx"

Explanation:

The first k = 3 characters "xyz" are reversed to "zyx". The final resulting string is "zyx".

Example 3:

Input: s = "hey", k = 1

Output: "hey"

Explanation:

The first k = 1 character "h" remains unchanged on reversal. The final resulting string is "hey".

 

Constraints:

  • 1 <= s.length <= 100
  • s consists of lowercase English letters.
  • 1 <= k <= s.length

Solutions

Solution 1: Simulation

We reverse the first $k$ characters of the string according to the problem description, and then concatenate them with the remaining characters.

The time complexity is $O(n)$ and the space complexity is $O(n)$, where $n$ is the length of the string.

  • class Solution {
    public String reversePrefix(String s, int k) {
        StringBuilder sb = new StringBuilder(s.substring(0, k));
        return sb.reverse().toString() + s.substring(k);
    }
}


  • class Solution {
public:
    string reversePrefix(string s, int k) {
        string t = s.substr(0, k);
        reverse(t.begin(), t.end());
        return t + s.substr(k);
    }
};


  • class Solution:
    def reversePrefix(self, s: str, k: int) -> str:
        return s[:k][::-1] + s[k:]


  • func reversePrefix(s string, k int) string {
	t := []byte(s[:k])
	slices.Reverse(t)
	return string(t) + s[k:]
}


  • function reversePrefix(s: string, k: number): string {
    return s.slice(0, k).split('').reverse().join('') + s.slice(k);
}

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