I like to read this solution as a small machine: keep the useful information, throw away the noise. For 1415. The k-th Lexicographical String of All Happy Strings of Length n, the solution in this repository is mainly a backtracking solution.
Guide
What?
The first job is to translate the English prompt into state, transition, and stopping conditions. Instead of trying to be clever immediately, read the code as a sequence of questions:
- What state are we keeping?
- How do we move from one state to the next?
- When do we know the answer is already determined?
For this file, the main tools are: backtracking.
Guide
When?
This is the kind of solution you want when the problem has structure hiding inside a messy-looking input. The accepted code reduces that pressure by storing exactly the information that remains useful later.
The important function names to track are backtrack, getHappyString.
Guide
Why?
The code is doing bookkeeping so your brain does not have to keep the entire search space open at once.
- The two-dimensional vector is the memory of the solution: grid state, DP state, or adjacency shape.
- The final return is not magic; it is the invariant after the loops or recursion have finished doing their accounting.
Guide
How?
Walk through the solution in this order:
- Initialize the memory or helper structure.
- Process candidates in the order the invariant expects.
- Update the answer only when the current state is valid.
- Return the value that represents the fully processed input.
The most important competitive-programming habit here is to trust the invariant. Once the invariant is right, the loops become much less scary.
Guide
Complexity
- Time: O(N), space: O(1)
- Space: O(n) in the usual case for auxiliary containers or recursion
Guide
C++ Solution
Your submission
The accepted solution
solution.cpp
01//Runtime: 92 ms, faster than 43.96% of C++ online submissions for The k-th Lexicographical String of All Happy Strings of Length n.
02//Memory Usage: 24.6 MB, less than 100.00% of C++ online submissions for The k-th Lexicographical String of All Happy Strings of Length n.
03class Solution {
04public:
05 int n;
06
07 void backtrack(vector<vector<char>>& pers, vector<char>& per, vector<char>& chars){
08 if(per.size() == n){
09 pers.push_back(per);
10 }else{
11 for(char c : chars){
12 if(per.size() > 0 && per[per.size()-1] == c){
13 continue;
14 }
15 per.push_back(c);
16 backtrack(pers, per, chars);
17 per.pop_back();
18 }
19 }
20 };
21
22 string getHappyString(int n, int k) {
23 int count = 3 * (1 << (n-1));
24 if(k > count) return "";
25
26 this->n = n;
27
28 vector<vector<char>> pers;
29 vector<char> per;
30 vector<char> chars = {'a', 'b', 'c'};
31
32 backtrack(pers, per, chars);
33
34 // for(vector<char> p : pers){
35 // for(char c : p){
36 // cout << c;
37 // }
38 // cout << endl;
39 // }
40 // cout << endl;
41
42 string ans = "";
43 for(char c : pers[k-1]){
44 ans += c;
45 }
46
47 return ans;
48 }
49};
50
51//Math
52//https://leetcode.com/problems/the-k-th-lexicographical-string-of-all-happy-strings-of-length-n/discuss/585590/C%2B%2B-DFS-and-Math
53//Runtime: 0 ms, faster than 100.00% of C++ online submissions for The k-th Lexicographical String of All Happy Strings of Length n.
54//Memory Usage: 5.8 MB, less than 100.00% of C++ online submissions for The k-th Lexicographical String of All Happy Strings of Length n.
55//time: O(N), space: O(1)
56class Solution {
57public:
58 string getHappyString(int n, int k) {
59 //there are total 3*pow(2, n-1) happy strings
60 int prem = 1 << (n-1);
61 if(k > 3 * prem) return "";
62
63 //k falls in (1) [1, prem] -> 'a' (2) [prem+1, prem*2] -> 'b' (3) [prem*2+1, prem*3] -> 'c'
64 //(k-1)/prem: which interval(0,1 or 2) current char is in
65 string ans(1, 'a' + (k-1)/prem);
66
67 // cout << k << " " << prem << " " << (char)('a'+(k-1)/prem) << endl;
68
69 // while(prem > 1){
70 while(--n > 0){
71 //we have consumed one position
72 /*
73 (k-1)/prem: which interval current char is in
74 prem: each interval's length
75 previous intervals have taken "(k - 1) / prem * prem" spaces,
76 so we need to remove them
77
78 for 1st iteration, possible intervals are (0,1,2) (we can choose from a, b or c)
79 for later iteration, possible intervals are (0,1) (we can only choose two character different from last character)
80 */
81 k -= (k - 1) / prem * prem;
82 prem >>= 1;
83 /*
84 n = 3
85 ["aba", "abc", "aca", "acb", "bab", "bac", "bca", "bcb", "cab", "cac", "cba", "cbc"]
86
87 for interval 0, choose from a or b
88 for interval 1, choose from b or c
89 */
90 // cout << k << " " << prem << " " << (k-1)/prem << endl;
91 ans += (k - 1) / prem == 0 ? 'a' + (ans.back() == 'a') : 'b' + (ans.back() != 'c');
92 }
93
94 return ans;
95 }
96};
97
98//DFS
99//https://leetcode.com/problems/the-k-th-lexicographical-string-of-all-happy-strings-of-length-n/discuss/585590/C%2B%2B-DFS-and-Math
100//Runtime: 0 ms, faster than 100.00% of C++ online submissions for The k-th Lexicographical String of All Happy Strings of Length n.
101//Memory Usage: 5.9 MB, less than 100.00% of C++ online submissions for The k-th Lexicographical String of All Happy Strings of Length n.
102//time: O(N*k), space: O(N)
103class Solution {
104public:
105 string getHappyString(int n, int& k, int l = 0, char last = '\0') {
106 //l: the length of building string
107 //last: last character in the building string
108 //note that k is passed by reference
109 if(l == n){
110 //we have built a happy string
111 k--;
112 }else{
113 for(char cur = 'a'; cur <= 'c'; cur++){
114 if(cur == last) continue;
115 string res = getHappyString(n, k, l+1, cur);
116 if(k == 0){
117 return string(1, cur) + res;
118 }
119 }
120 }
121 return "";
122 }
123};Cost
Complexity
Time
O(N), space: O(1)
Dominated by the main traversal, recursion, or data-structure operations in the code.
Space
O(n) in the usual case for auxiliary containers or recursion
Auxiliary state plus the answer structure where the problem requires one.