1374. Generate a String With Characters That Have Odd Counts

Description

Given an integer n, return a string with n characters such that each character in such string occurs an odd number of times.

The returned string must contain only lowercase English letters. If there are multiples valid strings, return any of them.  

Example 1:

Input: n = 4
Output: "pppz"
Explanation: "pppz" is a valid string since the character 'p' occurs three times and the character 'z' occurs once. Note that there are many other valid strings such as "ohhh" and "love".

Example 2:

Input: n = 2
Output: "xy"
Explanation: "xy" is a valid string since the characters 'x' and 'y' occur once. Note that there are many other valid strings such as "ag" and "ur".

Example 3:

Input: n = 7
Output: "holasss"

Constraints:

• 1 <= n <= 500

Solutions

Solution 1: Construction

If \(n\) is odd, then we can directly construct a string with \(n\) 'a' characters.

If \(n\) is even, then we can construct a string with \(n-1\) 'a' characters and \(1\) 'b' character.

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

1
2
3

class Solution:
    def generateTheString(self, n: int) -> str:
        return 'a' * n if n & 1 else 'a' * (n - 1) + 'b'

1
2
3
4
5

class Solution {
    public String generateTheString(int n) {
        return (n % 2 == 1) ? "a".repeat(n) : "a".repeat(n - 1) + "b";
    }
}

 1
 2
 3
 4
 5
 6
 7
 8
 9
10

class Solution {
public:
    string generateTheString(int n) {
        string ans(n, 'a');
        if (n % 2 == 0) {
            ans[0] = 'b';
        }
        return ans;
    }
};

1
2
3
4
5
6
7
8
9

func generateTheString(n int) string {
    ans := strings.Repeat("a", n-1)
    if n%2 == 0 {
        ans += "b"
    } else {
        ans += "a"
    }
    return ans
}

1
2
3
4
5
6
7

function generateTheString(n: number): string {
    const ans = Array(n).fill('a');
    if (n % 2 === 0) {
        ans[0] = 'b';
    }
    return ans.join('');
}

Comments