This problem looks busy at first, but the accepted solution is built around one steady invariant. For 856. Score of Parentheses, the solution in this repository is mainly a stack solution.
Guide
What?
We want to turn the problem statement into a smaller set of decisions the computer can repeat safely. 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: stack.
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 scoreOfParentheses, F.
Guide
Why?
The win comes from making each line carry responsibility: store the useful state, discard the rest, keep moving.
- The stack stores unfinished context, which is usually the cleanest way to handle nested or monotonic structure.
- 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:
- Start from the smallest reliable state.
- Expand one legal move at a time.
- Cache, count, or merge information as soon as it becomes settled.
- Let the final stored value answer the original question.
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^2), space: O(N)
- Space: O(n) in the usual case for auxiliary containers or recursion
Guide
C++ Solution
Your submission
The accepted solution
01//Runtime: 0 ms, faster than 100.00% of C++ online submissions for Score of Parentheses.
02//Memory Usage: 6 MB, less than 100.00% of C++ online submissions for Score of Parentheses.
03class Solution {
04public:
05 int scoreOfParentheses(string S) {
06 stack<int> stk;
07 for(char c : S){
08 // cout << c << " " << stk.size() << endl;
09 switch(c){
10 case '(':
11 //use -1 to represent '('
12 stk.push(-1);
13 break;
14 case ')':
15 //stk.size() must > 0
16 if(stk.top() == -1){
17 stk.pop();
18 stk.push(1);
19 }else{
20 int tmp = 0;
21 while(stk.top() != -1){
22 tmp += stk.top(); stk.pop();
23 }
24 stk.pop(); //'('
25 stk.push(tmp * 2);
26 }
27 break;
28 }
29 }
30
31 int ans = 0;
32 while(!stk.empty()){
33 ans += stk.top(); stk.pop();
34 }
35 return ans;
36 }
37};
38
39//Approach 1: Divide and Conquer
40//Runtime: 0 ms, faster than 100.00% of C++ online submissions for Score of Parentheses.
41//Memory Usage: 6.2 MB, less than 100.00% of C++ online submissions for Score of Parentheses.
42//time: O(N^2), space: O(N)
43class Solution {
44public:
45 int F(string S, int start, int end){
46 //end is exclusive
47 int ans = 0, balance = 0;
48
49 for(int i = start; i < end; i++){
50 balance += (S[i] == '(') ? 1 : -1;
51 if(balance == 0){
52 if(i - start == 1){
53 //')' 's mathcing '(' is its previous char
54 ans++;
55 }else{
56 //')' matches to the '(' far before
57 ans += 2 * F(S, start+1, i);
58 }
59 //the start position of new '('
60 start = i+1;
61 }
62 }
63
64 return ans;
65 };
66
67 int scoreOfParentheses(string S) {
68 return F(S, 0, S.size());
69 }
70};
71
72//Approach 2: Stack
73//Runtime: 0 ms, faster than 100.00% of C++ online submissions for Score of Parentheses.
74//Memory Usage: 6.1 MB, less than 100.00% of C++ online submissions for Score of Parentheses.
75//time: O(N), space: O(N)
76class Solution {
77public:
78 int scoreOfParentheses(string S) {
79 stack<int> stk;
80 stk.push(0);
81
82 for(char c : S){
83 if(c == '('){
84 stk.push(0);
85 }else{
86 int v = stk.top(); stk.pop();
87 int w = stk.top(); stk.pop();
88 //max(2*v, 1): v could be
89 /*
90 if v is 0, a.k.a. last char is '(',
91 then score becomes 1
92 else if v is not 0
93 then use score specified by v
94 */
95 stk.push(w + max(2*v, 1));
96 // cout << v << " " << w << " " << w + max(2*v, 1) << endl;
97 }
98 }
99
100 return stk.top();
101 }
102};
103
104//Approach 3: Count Cores
105//Runtime: 0 ms, faster than 100.00% of C++ online submissions for Score of Parentheses.
106//Memory Usage: 6.2 MB, less than 100.00% of C++ online submissions for Score of Parentheses.
107//time: O(N), space: O(1)
108class Solution {
109public:
110 int scoreOfParentheses(string S) {
111 int ans = 0, balance = 0;
112
113 for(int i = 0; i < S.size(); i++){
114 if(S[i] == '('){
115 balance++;
116 }else{
117 balance--;
118 if(S[i-1] == '('){
119 ans += (1 << balance);
120 }
121 }
122 // cout << balance << " " << ans << endl;
123 }
124
125 return ans;
126 }
127};Cost