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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 ofi
from1
tos.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(); } } };
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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]
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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] ?? ''; }
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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] }
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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]; } }
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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() } } }