A good way into this one is to ask: what do we need to remember so we never redo work blindly? For 1410. HTML Entity Parser, the solution in this repository is mainly a trie 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: trie.
The notes already sitting in the source point us in the right direction:
- Once accepted answer
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 entityParser, add.
Guide
Why?
The win comes from making each line carry responsibility: store the useful state, discard the rest, keep moving.
- A map keeps the lookup side cheap; the code pays a little memory to avoid repeated searching.
- Substring checks are convenient but not free, so they are part of the real complexity story.
- 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//Once accepted answer
02/*
03151 / 154 test cases passed.
04Input: "&"
05Output: "&"
06Expected: "&"
07*/
08//Runtime: 1204 ms
09//Memory Usage: 16.4 MB
10class Solution {
11public:
12 string entityParser(string text) {
13 map<string, string> entityMap = {
14 {""" , "\""},
15 {"'", "'"},
16 {"&" , "&"},
17 {">" , ">"},
18 {"<" , "<"},
19 {"⁄" , "/"}
20 };
21
22 for(auto it = entityMap.begin(); it != entityMap.end(); it++){
23 size_t pos = 0;
24
25 while((pos = text.find(it->first, pos)) != string::npos){
26 text.replace(pos, (it->first).length(), it->second);
27 }
28 }
29
30 return text;
31 }
32};
33
34//Two pointer
35//https://leetcode.com/problems/html-entity-parser/discuss/575416/C%2B%2B-two-pointers-O(n)-or-O(1)
36//Runtime: 196 ms, faster than 93.55% of C++ online submissions for HTML Entity Parser.
37//Memory Usage: 16.3 MB, less than 100.00% of C++ online submissions for HTML Entity Parser.
38class Solution {
39public:
40 string entityParser(string text) {
41 map<string, char> entityMap = {
42 {""" , '\"'},
43 {"'", '\''},
44 {"&" , '&'},
45 {">" , '>'},
46 {"<" , '<'},
47 {"⁄" , '/'}
48 };
49
50 int slow = 0;
51 map<string, char>::iterator it;
52
53 //it ends when we examine the whole string using "fast"
54 for (int fast = 0, lastAnd = 0; fast < text.size(); ++fast, ++slow) {
55 text[slow] = text[fast];
56
57 if (text[slow] == '&')
58 lastAnd = slow; //will be used when we meet ';'
59
60 if (text[slow] == ';') {
61 // cout << text << " " << lastAnd << " " << slow << endl;
62 if((it = entityMap.find(text.substr(lastAnd, slow - lastAnd + 1))) != entityMap.end()){
63 //suppose we meet ">", now we will write '>' in the position of '&'
64 slow = lastAnd;
65 //modify the "text" in-place
66 text[slow] = it->second;
67
68 }
69 /*an '&' can be used only once,
70 here we move lastAnd forward to
71 avoid the last '&' be used next time
72 Example: "&amp;"
73 */
74 lastAnd = slow + 1;
75 }
76 }
77
78 //now "slow" is the index just after last write
79 text.resize(slow);
80 return text;
81 }
82};
83
84//Two pointer + Trie
85//https://leetcode.com/problems/html-entity-parser/discuss/575416/C%2B%2B-two-pointers-O(n)-or-O(1)
86//Runtime: 180 ms, faster than 95.80% of C++ online submissions for HTML Entity Parser.
87//Memory Usage: 18 MB, less than 100.00% of C++ online submissions for HTML Entity Parser.
88//time: O(N)
89class TrieNode{
90public:
91 vector<TrieNode*> children;
92 char mappedSymbol;
93
94 TrieNode(){
95 children = vector<TrieNode*>(26, nullptr);
96 mappedSymbol = '\0';
97 }
98};
99
100class Trie{
101public:
102 TrieNode* root;
103
104 Trie(){
105 root = new TrieNode();
106 };
107
108 void add(string word, char symbol){
109 TrieNode* cur = root;
110 for(char c : word){
111 if(cur->children[c-'a'] == nullptr){
112 cur->children[c-'a'] = new TrieNode();
113 }
114 cur = cur->children[c-'a'];
115 }
116 //the trie contains the info mapping by itself
117 cur->mappedSymbol = symbol;
118 };
119
120 char search(string word){
121 TrieNode* cur = root;
122 for(char c : word){
123 if(cur->children[c-'a'] == nullptr)
124 return '\0';
125 cur = cur->children[c-'a'];
126 }
127 return cur->mappedSymbol;
128 };
129};
130
131class Solution {
132public:
133 string entityParser(string text) {
134 map<string, char> entityMap = {
135 {""" , '\"'},
136 {"'", '\''},
137 {"&" , '&'},
138 {">" , '>'},
139 {"<" , '<'},
140 {"⁄" , '/'}
141 };
142
143 int slow = 0;
144 map<string, char>::iterator it;
145 Trie* trie = new Trie();
146
147 for(it = entityMap.begin(); it != entityMap.end(); it++){
148 //remove key's head's & and tail's ;
149 // cout << it->first.substr(1, it->first.size()-2) << endl;
150 trie->add(it->first.substr(1, it->first.size()-2), it->second);
151 }
152
153 for (int fast = 0, lastAnd = 0; fast < text.size(); ++fast, ++slow) {
154 text[slow] = text[fast];
155
156 if (text[slow] == '&')
157 lastAnd = slow;
158
159 if (text[slow] == ';') {
160 // cout << text << " " << lastAnd << " " << slow << endl;
161 char c;
162 if((c = trie->search(text.substr(lastAnd+1, slow-1-lastAnd))) != '\0'){
163 // cout << text.substr(lastAnd+1, slow-1-lastAnd) << ", " << c << endl;
164 slow = lastAnd;
165 text[slow] = c;
166
167 }
168 lastAnd = slow + 1;
169 }
170 }
171
172 text.resize(slow);
173 return text;
174 }
175};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.