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[BOJ] 괄호 제거
2025-11-06

괄호 제거

TimeLimitMemoryLimitConditionTAG
1s128MB(1 <= 괄호 개수 <= 10)Implementation

오랜만에 문제를 풀기 때문에 가벼운 문제 입니다. ㅠ

정답 코드


#include <iostream>
#include <queue>
#include <vector>
#include <algorithm>
#include <set>
#include <map>
#include <sstream>
#include <iomanip>
#include <string>
#include <numeric>
#include <unordered_set>
#include <unordered_map>
#include <climits>
#include <math.h>
using namespace std;
using ll = long long;
using ld = long double;
ll MOD = 1'000'000'000;

set<string> results;

string input;



set<int> visited;



void solution(int left){
    if(left >= input.size()) return;
    
    if(input[left] == '('){
        int right_idx = input.size()-1;
        visited.insert(left);
        int cnt = 1;
        for(int i = left+1; i < input.size(); i++){
            if(input[i] == ')') cnt--;
            else if(input[i] == '(') cnt++;
        
            if(cnt == 0){
                right_idx = i;
                break;
            }
        }
        visited.insert(right_idx);
        solution(left+1);
        
        
        string result = "";
        for(int i = 0; i < input.size(); i++){
            if(visited.find(i) == visited.end())
                result += input[i];
        }
        results.insert(result);
        
        visited.erase(left);
        visited.erase(right_idx);
        
        
        solution(left+1);
    }
    else{
        solution(left+1);
    }
}

int main() {
    ios_base::sync_with_stdio(0);
    cin.tie(0);
    cin >> input;
    
    
    solution(0);
    
    
    
    
    
    for(string result : results) cout << result << "\n";
   
        
        
    
    
    return 0;
}
[BOJ] 괄호 제거
https://compy07.github.io/Blog/posts/boj/2800/
Author
뒹굴뒹굴 이정훈 공부방
Published at
2025-11-06