1415. The k-th Lexicographical String of All Happy Strings of Length n -----M

 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

A:

就是实现题。  一个个数呗  （唯一可能tricky的地方就是 需要k--, 来变成 0 based

class Solution {
public:
    string getHappyString(int n, int k) {
        vector<int> c(n+1,1);
        for(int j = n-1; j>=0; j--){
            c[j] = (j==0?3:2) * c[j+1];
        }
        string res = "";
        k--;// zero index
        if(c[0] <= k){ // since we have k-- above
            return res;
        }
        for(int i =0;i<n;i++){
            // find how many next iter count
            int count = k/c[i+1]; // count can be 0, 1,2
            if(i == 0){
                res+= ('a'+count);
            }else{
                res += getNext(res.back(), count);
            }
            k -= count * c[i+1];
        }
        return res;
    }
private:
    char getNext(char pre, int count){ // count can only be 0 or 1
        if(pre == 'a'){
            return count==0? 'b' : 'c';
        }else if(pre == 'b'){
            return count==0? 'a' : 'c';
        }else{// if(pre == 'c'){
            return count==0? 'a' : 'b';
        }
    }
};