README_EN.md

comments	difficulty	edit_url	rating	source	tags
true	Easy	https://github.com/doocs/leetcode/edit/main/solution/3200-3299/3210.Find%20the%20Encrypted%20String/README_EN.md	1179	Weekly Contest 405 Q1	String

3210. Find the Encrypted String

中文文档

Description

You are given a string s and an integer k. Encrypt the string using the following algorithm:

• For each character c in s, replace c with the kth character after c in the string (in a cyclic manner).

Return the encrypted string.

 

Example 1:

Input: s = "dart", k = 3

Output: "tdar"

Explanation:

• For i = 0, the 3^rd character after 'd' is 't'.
• For i = 1, the 3^rd character after 'a' is 'd'.
• For i = 2, the 3^rd character after 'r' is 'a'.
• For i = 3, the 3^rd character after 't' is 'r'.

Example 2:

Input: s = "aaa", k = 1

Output: "aaa"

Explanation:

As all the characters are the same, the encrypted string will also be the same.

 

Constraints:

• 1 <= s.length <= 100
• 1 <= k <= 104
• s consists only of lowercase English letters.

Solutions

Solution 1: Simulation

We can use the simulation method. For the $i^{th}$ character of the string, we replace it with the character at position $(i + k) \bmod n$ of the string.

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

Python3

class Solution:
    def getEncryptedString(self, s: str, k: int) -> str:
        cs = list(s)
        n = len(s)
        for i in range(n):
            cs[i] = s[(i + k) % n]
        return "".join(cs)

Java

class Solution {
    public String getEncryptedString(String s, int k) {
        char[] cs = s.toCharArray();
        int n = cs.length;
        for (int i = 0; i < n; ++i) {
            cs[i] = s.charAt((i + k) % n);
        }
        return new String(cs);
    }
}

C++

class Solution {
public:
    string getEncryptedString(string s, int k) {
        int n = s.length();
        string cs(n, ' ');
        for (int i = 0; i < n; ++i) {
            cs[i] = s[(i + k) % n];
        }
        return cs;
    }
};

Go

func getEncryptedString(s string, k int) string {
	cs := []byte(s)
	for i := range s {
		cs[i] = s[(i+k)%len(s)]
	}
	return string(cs)
}

TypeScript

function getEncryptedString(s: string, k: number): string {
    const cs: string[] = [];
    const n = s.length;
    for (let i = 0; i < n; ++i) {
        cs[i] = s[(i + k) % n];
    }
    return cs.join('');
}