A happy string is a string that:
consists only of letters of the set ['a', 'b', 'c'].
s[i] != s[i + 1] for all values of i from 1 to s.length - 1 (string is 1-indexed).
For example, strings "abc", "ac", "b" and "abcbabcbcb" are all happy strings and strings "aa", "baa" and "ababbc" are not happy strings.
Given two integers n and k, consider a list of all happy strings of length n sorted in lexicographical order.
Return the kth string of this list or return an empty string if there are less than k happy strings of length n.
Example 1:
Input: n = 1, k = 3
Output: "c"
Explanation: The list ["a", "b", "c"] contains all happy strings of length 1. The third string is "c".
Example 2:
Input: n = 1, k = 4
Output: ""
Explanation: There are only 3 happy strings of length 1.
Example 3:
Input: n = 3, k = 9
Output: "cab"
Explanation: There are 12 different happy string of length 3 ["aba", "abc", "aca", "acb", "bab", "bac", "bca", "bcb", "cab", "cac", "cba", "cbc"]. You will find the 9th string = "cab"
Example 4:
Input: n = 2, k = 7
Output: ""
Example 5:
Input: n = 10, k = 100
Output: "abacbabacb"
Constraints:
1 <= n <= 10
1 <= k <= 100
class Solution:
    def getHappyString(self, n: int, k: int) -> str:
        N = 3 << (n - 1)
        res = ''
        if k > N:
            return res
        k -= 1
        letters = ('a', 'b', 'c')
        res += letters[k * 3 // N]
        N //= 3
        k %= N
        for i in range(1, n):
            prev = res[i - 1]
            index = k * 2 // N
            offset = int(prev <= letters[index])
            res += letters[index + offset]
            N >>= 1
            k %= N
        return res

class Solution:
    def getHappyString(self, n: int, k: int) -> str:
        N = 3 << (n - 1)
        res = ''
        if k > N:
            return res
        k -= 1
        letters = ('a', 'b', 'c')
        res += letters[k * 3 // N]
        N //= 3
        k %= N
        for i in range(1, n):
            prev = res[i - 1]
            index = k * 2 // N
            offset = int(prev <= letters[index])
            res += letters[index + offset]
            N >>= 1
            k %= N
        return res
