https://practice.geeksforgeeks.org/problems/minimum-swaps-for-bracket-balancing/0

You are given a string of 2N characters consisting of N ‘[‘ brackets and N ‘]’ brackets. A string is considered balanced if it can be represented in the for S2[S1] where S1 and S2 are balanced strings. We can make an unbalanced string balanced by swapping adjacent characters. Calculate the minimum number of swaps necessary to make a string balanced.

Input:

The first line of input contains an integer T denoting the number of test cases. Then T test cases follow. The first line of each test case contains an integer N denoting the length of the string.

The second line of each test case contains the string consisting of ‘[‘ and ‘]’.

Output:

Print the minimum number of swaps to make the string balanced for each test case in a new line.

Example:
Input : []][][
Output : 2
First swap: Position 3 and 4
[][]][
Second swap: Position 5 and 6
[][][]
Input : [[][]]
Output : 0
String is already balanced.

方法一：每碰到一个 ‘ ] ‘ ，抵消左侧的一个 ‘ [ ‘ ,如果左侧没有 ，则抵消右侧的，此时需要要 swap ,sum += pos(‘ [ ‘) -pos (‘ ] ‘)。但是此时 应回到交换后的 ‘ [ ‘ 重新开始，因为可能已影响到交换前 ‘ ] ‘ 之后符号的决策。

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#include <bits/stdc++.h>
using namespace std;
int n,t;
string s;
// Function to calculate swaps required 
long swapCount(string s) 
{ 
    // Keep track of '[' 
    vector<int> pos; 
    for (int i = 0; i < s.length(); ++i) 
        if (s[i] == '[') 
            pos.push_back(i); 
  
    int count = 0; // To count number of encountered '[' 
    int p = 0;     // To track position of next '[' in pos 
    long sum = 0; // To store result 
  
    for (int i = 0; i < s.length(); ++i) 
    { 
        // Increment count and move p to next position 
        if (s[i] == '[') 
        { 
            ++count; 
            ++p; 
        } 
        else if (s[i] == ']') 
            --count; 
  
        // We have encountered an unbalanced part of string 
        if (count < 0) 
        { 
            // Increment sum by number of swaps required 
            // i.e. position of next '[' - current position 
            sum += pos[p] - i; 
            swap(s[i], s[pos[p]]); 
            ++p; 
  
            // Reset count to 1 
            count = 1; 
        } 
    } 
    return sum; 
} 
int main(){
	freopen("in","r",stdin);
	cin>>t;
	while(t--){
		cin>>n;
		cin>>s;
		cout<<swapCount(s)<<endl; 
	}
	return 0;
}

方法二：记录 ‘ ] ‘ 不匹配的个数 imbalance，遇见 ‘ [ ‘ 时，优先配对最远的 ‘ ] ‘,且 imbalance –，依序往下。

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#include <bits/stdc++.h>
using namespace std;
const int MAX=100000+10;
int n,t;
string s;
long swapCount(string s)  
{      
    // stores total number of Left and Right  
    // brackets encountered 
    int countLeft = 0, countRight = 0;  
    // swap stores the number of swaps required 
    //imbalance maintains the number of imbalance pair 
    int swap = 0 , imbalance = 0;  
      
    for(int i =0; i< s.length(); i++)  
    { 
        if(s[i] == '[')  
        { 
            // increment count of Left bracket 
            countLeft++;  
            if(imbalance > 0)  
            { 
                // swaps count is last swap count + total  
                // number imbalanced brackets 
                swap += imbalance;  
                // imbalance decremented by 1 as it solved 
                // only one imbalance of Left and Right 
                imbalance--;      
            } 
        } else if(s[i] == ']' )  
        { 
            // increment count of Right bracket 
            countRight++;  
            // imbalance is reset to current difference  
            // between Left and Right brackets 
            imbalance = (countRight-countLeft);  
        } 
    } 
    return swap; 
} 
int main(){
	freopen("in","r",stdin);
	cin>>t;
	while(t--){
		cin>>n;
		cin>>s;
		cout<<swapCount(s)<<endl; 
	}
	return 0;
}