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Formatted question description: https://leetcode.ca/all/1415.html
1415. 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 ofifrom1tos.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"
Example 4:
Input: n = 2, k = 7 Output: ""
Example 5:
Input: n = 10, k = 100 Output: "abacbabacb"
Constraints:
1 <= n <= 101 <= k <= 100
Related Topics:
Backtracking
Solution 1. Naive Solution
// OJ: https://leetcode.com/problems/the-k-th-lexicographical-string-of-all-happy-strings-of-length-n/
// Time: O(3^N)
// Space: O(1)
class Solution {
string addOne(string &s) {
int i = s.size() - 1;
for (; i >= 0; --i) {
if (s[i] == 'c') {
s[i] = 'a';
continue;
}
s[i]++;
break;
}
return i == -1 ? (s = "") : s;
}
bool valid(string &s) {
int i = s.size() - 1;
for (; i > 0; --i) {
if (s[i] == s[i - 1]) return false;
}
return true;
}
string next(string &s) {
do {
addOne(s);
} while (s != "" && !valid(s));
return s;
}
public:
string getHappyString(int n, int k) {
string ans;
for (int i = 0; i < n; ++i) ans.push_back(i % 2 == 0 ? 'a' : 'b');
while (--k) {
ans = next(ans);
if (ans == "") return "";
}
return ans;
}
};
Solution 2. DFS
// OJ: https://leetcode.com/problems/the-k-th-lexicographical-string-of-all-happy-strings-of-length-n/
// Time: O(3^N). It should be smaller than O(3^N) since there are cases skipped earlier, but should be greater than O(NK).
// Space: O(NK)
// Ref: https://leetcode.com/problems/the-k-th-lexicographical-string-of-all-happy-strings-of-length-n/discuss/585557/C%2B%2B-Straightforward-DFS.-Skip-appending-same-char.
class Solution {
vector<string> ans;
void dfs(string &cur, int n, int k) {
if (ans.size() == k) return;
if (cur.size() == n) {
ans.push_back(cur);
return;
}
for (int i = 0; i < 3; ++i) {
if (cur.size() && cur.back() == 'a' + i) continue;
cur.push_back('a' + i);
dfs(cur, n, k);
cur.pop_back();
}
}
public:
string getHappyString(int n, int k) {
string cur;
dfs(cur, n, k);
return ans.size() == k ? ans.back() : "";
}
};
-
class Solution { public String getHappyString(int n, int k) { if (n == 1) { if (k <= 3) return String.valueOf((char) ('a' + k - 1)); else return ""; } int max = 3 * (int) Math.pow(2, n - 1); if (k > max) return ""; int groupSize = max / 3; int highest = (k - 1) / groupSize; int index = (k - 1) % groupSize; int[] array = toArray(highest, index, n); StringBuffer sb = new StringBuffer(); char letter0 = (char) ('a' + array[0]); sb.append(letter0); char prevLetter = letter0; for (int i = 1; i < n; i++) { char letter = getNextLetter(array[i], prevLetter); sb.append(letter); prevLetter = letter; } return sb.toString(); } public int[] toArray(int highest, int index, int length) { int[] array = new int[length]; array[0] = highest; int position = length - 1; while (index > 0) { array[position] = index % 2; index /= 2; position--; } return array; } public char getNextLetter(int num, char prevLetter) { char letter = (char) ('a' + num); if (letter >= prevLetter) letter++; return letter; } } ############ 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); } } } -
// OJ: https://leetcode.com/problems/the-k-th-lexicographical-string-of-all-happy-strings-of-length-n/ // Time: O(3^N) // Space: O(1) class Solution { string addOne(string &s) { int i = s.size() - 1; for (; i >= 0; --i) { if (s[i] == 'c') { s[i] = 'a'; continue; } s[i]++; break; } return i == -1 ? (s = "") : s; } bool valid(string &s) { int i = s.size() - 1; for (; i > 0; --i) { if (s[i] == s[i - 1]) return false; } return true; } string next(string &s) { do { addOne(s); } while (s != "" && !valid(s)); return s; } public: string getHappyString(int n, int k) { string ans; for (int i = 0; i < n; ++i) ans.push_back(i % 2 == 0 ? 'a' : 'b'); while (--k) { ans = next(ans); if (ans == "") return ""; } return ans; } }; -
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] ############ class Solution: def getHappyString(self, n: int, k: int) -> str: happies = [] self.genHappies(n, "", happies) if k > len(happies): return "" return happies[k - 1] def genHappies(self, n, path, happies): if len(path) == n: happies.append(path) return for x in ['a', 'b', 'c']: if not path or path[-1] != x: self.genHappies(n, path + x, happies)