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 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"

Example 4:

Input: n = 2, k = 7
Output: ""

Example 5:

Input: n = 10, k = 100
Output: "abacbabacb"

 

Constraints:

  • 1 <= n <= 10
  • 1 <= 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() : "";
    }
};

Java

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;
    }
}

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