1415. The k-th Lexicographical String of All Happy Strings of Length n

Description

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"


Constraints:

• 1 <= n <= 10
• 1 <= k <= 100

Solutions

• class Solution {
private List<String> ans = new ArrayList<>();

public String getHappyString(int n, int k) {
dfs("", n);
return ans.size() < k ? "" : ans.get(k - 1);
}

private void dfs(String t, int n) {
if (t.length() == n) {
return;
}
for (char c : "abc".toCharArray()) {
if (t.length() > 0 && t.charAt(t.length() - 1) == c) {
continue;
}
dfs(t + c, n);
}
}
}

• class Solution {
public:
vector<string> ans;
string getHappyString(int n, int k) {
dfs("", n);
return ans.size() < k ? "" : ans[k - 1];
}

void dfs(string t, int n) {
if (t.size() == n) {
ans.push_back(t);
return;
}
for (int c = 'a'; c <= 'c'; ++c) {
if (t.size() && t.back() == c) continue;
t.push_back(c);
dfs(t, n);
t.pop_back();
}
}
};

• class Solution:
def getHappyString(self, n: int, k: int) -> str:
def dfs(t):
if len(t) == n:
ans.append(t)
return
for c in 'abc':
if t and t[-1] == c:
continue
dfs(t + c)

ans = []
dfs('')
return '' if len(ans) < k else ans[k - 1]