A good way into this one is to ask: what do we need to remember so we never redo work blindly? For 89. Gray Code, the solution in this repository is mainly a bit manipulation 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: bit manipulation, backtracking.
The notes already sitting in the source point us in the right direction:
- backtracking
- TLE
- 4 / 12 test cases passed.
- testcase: 4
Guide
When?
This pattern shows up when the brute force version has too many repeated checks, too many possible branches, or too much bookkeeping to do by hand. The accepted code reduces that pressure by storing exactly the information that remains useful later.
The important function names to track are backtrack, grayCode, used, diffByOneBit.
Guide
Why?
The win comes from making each line carry responsibility: store the useful state, discard the rest, keep moving.
- 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:
- Read the setup variables first.
- Follow the main loop or recursive helper next.
- Watch where invalid states get skipped.
- Check which value survives to the return statement.
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) to O(n log n), depending on the dominant loop or data structure operation
- Space: O(n) in the usual case for auxiliary containers or recursion
Guide
C++ Solution
Your submission
The accepted solution
solution.cpp
01//backtracking
02//TLE
03//4 / 12 test cases passed.
04//testcase: 4
05class Solution {
06public:
07 int n;
08 vector<int> ans;
09
10 void backtrack(vector<int>& perm, vector<bool>& used){
11 if(all_of(used.begin(), used.end(), [](const bool& b){return b;})){
12 bool valid = true;
13 for(int i = 0; i+1 < n; ++i){
14 if(__builtin_popcount(perm[i]^perm[i+1]) != 1){
15 valid = false;
16 break;
17 }
18 }
19 if(valid){
20 ans = perm;
21 }
22 }else{
23 for(int i = 0; i < n; ++i){
24 if(used[i]) continue;
25 perm.push_back(i);
26 used[i] = true;
27 backtrack(perm, used);
28 if(!ans.empty()) return;
29 perm.pop_back();
30 used[i] = false;
31 }
32 }
33 }
34
35 vector<int> grayCode(int n) {
36 this->n = 1<<n;
37 vector<int> perm;
38 vector<bool> used(1<<n, false);
39 backtrack(perm, used);
40 return ans;
41 }
42};
43
44//speed up from above
45//6 / 12 test cases passed.
46//testcase: 10
47class Solution {
48public:
49 int n;
50 vector<int> ans;
51
52 void backtrack(vector<int>& perm, vector<bool>& used){
53 if(all_of(used.begin(), used.end(), [](const bool& b){return b;})){
54 bool valid = true;
55 // for(int i = 0; i+1 < n; ++i){
56 // if(__builtin_popcount(perm[i]^perm[i+1]) != 1){
57 // valid = false;
58 // break;
59 // }
60 // }
61 if(valid){
62 ans = perm;
63 }
64 }else{
65 for(int i = 0; i < n; ++i){
66 if(used[i]) continue;
67 if(!perm.empty() &&
68 __builtin_popcount(perm.back()^i) != 1) continue;
69 perm.push_back(i);
70 used[i] = true;
71 backtrack(perm, used);
72 if(!ans.empty()) return;
73 perm.pop_back();
74 used[i] = false;
75 }
76 }
77 }
78
79 vector<int> grayCode(int n) {
80 this->n = 1<<n;
81 vector<int> perm;
82 vector<bool> used(1<<n, false);
83 backtrack(perm, used);
84 return ans;
85 }
86};
87
88//backtracking, faster
89//https://leetcode.com/problems/gray-code/discuss/400651/Java-Solutions-with-Detailed-Comments-and-Explanations-(Backtracking-Prepending)
90//Runtime: 4 ms, faster than 25.43% of C++ online submissions for Gray Code.
91//Memory Usage: 7.8 MB, less than 6.92% of C++ online submissions for Gray Code.
92class Solution {
93public:
94 int p2n, n;
95 vector<int> ans;
96
97 bool diffByOneBit(int a, int b){
98 int x = a^b;
99
100 //x==0 <-> a==b
101 //x&(x-1): remove x's last bit
102 //((x&(x-1)) == 0): x has only one digit
103
104 return (x!=0) && ((x&(x-1)) == 0);
105 }
106
107 void backtrack(vector<int>& res, vector<bool>& used){
108 if(res.size() == p2n){
109 if(diffByOneBit(res[0], res.back())){
110 ans = res;
111 }
112 }else{
113 const int last = res.back();
114 for(int i = 0; i < n; ++i){
115 int cur = last ^ (1<<i);
116 if(used[cur]) continue;
117 res.push_back(cur);
118 used[cur] = true;
119 backtrack(res, used);
120 if(!ans.empty()) return;
121 res.pop_back();
122 used[cur] = false;
123 }
124 }
125 }
126
127 vector<int> grayCode(int n) {
128 if(n == 0) return {0};
129 this->n = n;
130 this->p2n = (1<<n);
131
132 vector<int> res;
133 vector<bool> used(this->p2n, false);
134
135 res = {0};
136 used[0] = true;
137
138 backtrack(res, used);
139
140 return ans;
141 }
142};
143
144//iterative
145//https://leetcode.com/problems/gray-code/discuss/400651/Java-Solutions-with-Detailed-Comments-and-Explanations-(Backtracking-Prepending)
146//Runtime: 0 ms, faster than 100.00% of C++ online submissions for Gray Code.
147//Memory Usage: 6.7 MB, less than 48.06% of C++ online submissions for Gray Code.
148class Solution {
149public:
150 vector<int> grayCode(int n) {
151 vector<int> res = {0};
152
153 for(int i = 0; i < n; ++i){
154 /*
155 in ith iteration,
156 we change res of size 1<<i to
157 res of size 1<<(i+1)
158
159 in ith iteration,
160 the range of res is [0,1<<(i+1)),
161 we already have range [0,1<<i),
162 now we need to add range[1<<i,1<<(i+1)),
163 and we can create it by
164 adding 1<<i to the old range
165 */
166 int prependVal = 1 << i;
167 int oldSize = res.size();
168 //double the size of res
169 for(int j = oldSize-1; j >= 0; --j){
170 res.push_back(prependVal + res[j]);
171 }
172 /*
173 the numbers in the later half all diff by one bit,
174 because they are created from the old res
175
176 res[oldSize-1] and res[oldSize] diff by one bit
177 (the most significant bit)
178
179 res[newSize-1] and res[0] diff by one bit
180 (the most significant bit)
181 */
182 }
183
184 return res;
185 }
186};Cost
Complexity
Time
O(n) to O(n log n), depending on the dominant loop or data structure operation
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.