main.go

// Source: https://leetcode.com/problems/the-k-th-lexicographical-string-of-all-happy-strings-of-length-n
// Title: The k-th Lexicographical String of All Happy Strings of Length n
// Difficulty: Medium
// Author: Mu Yang <http://muyang.pro>

////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////
// 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 9^th string = "cab"
// ```
//
// **Constraints:**
//
// - `1 <= n <= 10`
// - `1 <= k <= 100`
//
////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////

package main

func getHappyString(n int, k int) string {
	res := make([]byte, n)

	var recursion func(idx int) bool
	recursion = func(idx int) bool {
		if idx == n {
			k--
			if k == 0 {
				return true
			}
			return false
		}

		for ch := byte('a'); ch <= 'c'; ch++ {
			if idx > 0 && ch == res[idx-1] {
				continue
			}
			res[idx] = ch
			if recursion(idx + 1) {
				return true
			}
		}
		return false
	}

	if recursion(0) {
		return string(res)
	}
	return ""
}