This problem looks busy at first, but the accepted solution is built around one steady invariant. For 1541. Minimum Insertions to Balance a Parentheses String, the solution in this repository is mainly a stack solution.
Guide
What?
Before optimizing anything, pin down what information is still useful after each move. 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.
The notes already sitting in the source point us in the right direction:
- Stack
Guide
When?
Use this approach when the hard part is not syntax, but deciding what must stay true after every update. The accepted code reduces that pressure by storing exactly the information that remains useful later.
The important function names to track are minInsertions.
Guide
Why?
The code is doing bookkeeping so your brain does not have to keep the entire search space open at once.
- 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) 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
01//Stack
02/*
03Hint 1: Use a stack to keep opening brackets. If you face single closing ')' add 1 to the answer and consider it as '))'.
04Hint 2: If you have '))' with empty stack, add 1 to the answer, If after finishing you have x opening remaining in the stack, add 2x to the answer.
05*/
06//Runtime: 108 ms, faster than 100.00% of C++ online submissions for Minimum Insertions to Balance a Parentheses String.
07//Memory Usage: 12.9 MB, less than 50.00% of C++ online submissions for Minimum Insertions to Balance a Parentheses String.
08class Solution {
09public:
10 int minInsertions(string s) {
11 stack<char> stk;
12
13 int close = 0;
14 int ans = 0;
15
16 for(char c : s){
17 if(c == ')'){
18 ++close;
19 if(close >= 2){
20 if(!stk.empty()){
21 stk.pop();
22 }else{
23 ++ans;
24 }
25 close -= 2;
26 }
27 }else{
28 //c == '('
29 if(close > 0){
30 if(!stk.empty() && stk.top() == '('){
31 ans += (2-close);
32 stk.pop();
33 }else{
34 //add '(' and 2-close ')'
35 ans += 1 + (2-close);
36 }
37 close = 0;
38 }
39 stk.push('(');
40 }
41 // cout << "open: " << stk.size() << ", close: " << close << ", ans: " << ans << endl;
42 }
43
44 while(close > 0 && !stk.empty()){
45 if(close == 1){
46 close = 0;
47 //add a ')'
48 ++ans;
49 }else{
50 close -= 2;
51 }
52 stk.pop();
53 // cout << "open: " << stk.size() << ", close: " << close << ", ans: " << ans << endl;
54 }
55
56 if(stk.empty()){
57 //make it even
58 if(close&1){
59 ++close;
60 ++ans;
61 }
62 ans += close>>1;
63 }else{
64 /*
65 if there are k unmatched '(' in stack,
66 we will need k*2 ')'
67 */
68 ans += stk.size() * 2;
69 }
70
71 return ans;
72 }
73};
74
75//One pass
76//https://leetcode.com/problems/minimum-insertions-to-balance-a-parentheses-string/discuss/780199/JavaC%2B%2BPython-Straight-Forward-One-Pass
77//Runtime: 84 ms, faster than 100.00% of C++ online submissions for Minimum Insertions to Balance a Parentheses String.
78//Memory Usage: 12.6 MB, less than 50.00% of C++ online submissions for Minimum Insertions to Balance a Parentheses String.
79//time: O(N), space: O(1)
80class Solution {
81public:
82 int minInsertions(string s) {
83 int added = 0, right_to_add = 0;
84
85 for(char& c : s){
86 if(c == '('){
87 if(right_to_add&1){
88 //actually add one ')'
89 ++added;
90 //so the ')' needed to be added is decreased by one
91 --right_to_add;
92 }
93 //meet one '(', so we need two more ')'
94 right_to_add += 2;
95 }else{
96 //meet one ')', so need one less ')'
97 --right_to_add;
98 if(right_to_add < 0){
99 //add one '('
100 ++added;
101 /*
102 because we have added one '(',
103 we need two more ')'
104 */
105 right_to_add += 2;
106 }
107 }
108 }
109
110 return added + right_to_add;
111 }
112};Cost