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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) {
            ans.add(t);
            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]


  • function getHappyString(n: number, k: number): string {
    const ans: string[] = [];

    const dfs = (s = '') => {
        if (s.length === n) {
            ans.push(s);
            return;
        }

        for (const ch of 'abc') {
            if (s.at(-1) === ch) continue;
            dfs(s + ch);
        }
    };

    dfs();

    return ans[k - 1] ?? '';
}


  • function getHappyString(n, k) {
    const ans = [];

    const dfs = (s = '') => {
        if (s.length === n) {
            ans.push(s);
            return;
        }

        for (const ch of 'abc') {
            if (s.at(-1) === ch) continue;
            dfs(s + ch);
        }
    };

    dfs();

    return ans[k - 1] ?? '';
}


  • func getHappyString(n int, k int) string {
    ans := []string{}
    var s []byte

    var dfs func()
    dfs = func() {
        if len(s) == n {
            ans = append(ans, string(s))
            return
        }
        if len(ans) >= k {
            return
        }
        for c := byte('a'); c <= 'c'; c++ {
            if len(s) == 0 || s[len(s)-1] != c {
                s = append(s, c)
                dfs()
                s = s[:len(s)-1]
            }
        }
    }

    dfs()
    if len(ans) < k {
        return ""
    }
    return ans[k-1]
}


  • public class Solution {
    public string GetHappyString(int n, int k) {
        List<string> ans = new List<string>();
        StringBuilder s = new StringBuilder();

        void Dfs() {
            if (s.Length == n) {
                ans.Add(s.ToString());
                return;
            }
            if (ans.Count >= k) {
                return;
            }
            foreach (char c in "abc") {
                if (s.Length == 0 || s[s.Length - 1] != c) {
                    s.Append(c);
                    Dfs();
                    s.Length--;
                }
            }
        }

        Dfs();
        return ans.Count < k ? "" : ans[k - 1];
    }
}


  • impl Solution {
    pub fn get_happy_string(n: i32, k: i32) -> String {
        let mut ans = Vec::new();
        let mut s = String::new();
        let mut k = k;

        fn dfs(n: i32, s: &mut String, ans: &mut Vec<String>, k: &mut i32) {
            if s.len() == n as usize {
                ans.push(s.clone());
                return;
            }
            if ans.len() >= *k as usize {
                return;
            }
            for c in "abc".chars() {
                if s.is_empty() || s.chars().last() != Some(c) {
                    s.push(c);
                    dfs(n, s, ans, k);
                    s.pop();
                }
            }
        }

        dfs(n, &mut s, &mut ans, &mut k);
        if ans.len() < k as usize {
            "".to_string()
        } else {
            ans[(k - 1) as usize].clone()
        }
    }
}

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