1516. The K Th Lexicographical String Of All Happy Strings Of Length N¶

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1516. The k-th Lexicographical String of All Happy Strings of Length n

Medium

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

Solution¶

class Solution {
    private ArrayList<String> res;
    public String getHappyString(int n, int k) {
        res = new ArrayList<>();
        solve(n, "");
        Collections.sort(res);
        if (res.size() < k) return "";
        return res.get(k - 1);
    }
    private void solve(int n, String current) {
        if (current.length() == n) {
            res.add(current);
            return;
        }
        if (current.length() == 0) {
            solve(n, current + "a");
            solve(n, current + "b");
            solve(n, current + "c");
        }
        else {
            if (current.charAt(current.length() - 1) == 'a') {
                solve(n, current + "b");
                solve(n, current + "c");
            }
            else if (current.charAt(current.length() - 1) == 'b') {
                solve(n, current + "a");
                solve(n, current + "c");
            }
            else if (current.charAt(current.length() - 1) == 'c') {
                solve(n, current + "a");
                solve(n, current + "b");
            }
        }
    }
}

Complexity Analysis¶

• Time Complexity: O(?)
• Space Complexity: O(?)

Approach¶

Detailed explanation of the approach will be added here