Minimum Swaps for Bracket Balancing

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  : []][][
Output : 2
First swap: Position 3 and 4
Second swap: Position 5 and 6

Input  : [[][]]
Output : 0
String is already balanced.

We can solve this problem using greedy strategies. If the first X characters form a balanced string, we can neglect these characters and continue on. If we encounter a ‘]’ before the required ‘[‘, then we must start swapping elements to balance the string.

Naive Approach
Initialize sum = 0 where sum stores result. Go through the string maintaining a count of the number of ‘[‘ brackets encountered. Reduce this count when we encounter a ‘]’ character. If the count hits negative, then we must start balancing the string.
Let index ‘i’ represents the position we are at. We now move forward to the next ‘[‘ at index j. Increase sum by j – i. Move the ‘[‘ at position j, to position i, and shift all other characters to the right. Set the count back to 0 and continue traversing the string. At the end ‘sum’ will have the required value.

Time Complexity = O(N^2)
Extra Space = O(1)

Optimized approach
We can initially go through the string and store the positions of ‘[‘ in a vector say ‘pos‘. Initialize ‘p’ to 0. We shall use p to traverse the vector ‘pos’. Similar to the naive approach, we maintain a count of encountered ‘[‘ brackets. When we encounter a ‘[‘ we increase the count, and increase ‘p’ by 1. When we encounter a ‘]’ we decrease the count. If the count ever goes negative, this means we must start swapping. The element pos[p] tells us the index of the next ‘[‘. We increase the sum by pos[p] – i, where i is the current index. We can swap the elements in the current index and pos[p] and reset count to 0.

Since we have converted a step that was O(N) in the naive approach, to an O(1) step, our new time complexity reduces.
Time Complexity = O(N)
Extra Space = O(N)

// Program to count swaps required to balance string
#include <iostream>
#include <vector>
#include <algorithm>
using namespace std;

// 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] == '[')

    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] == '[')
        else if (s[i] == ']')

        // 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]]);

            // Reset count to 1
            count = 1;
    return sum;

// Driver code
int main()
    string s = "[]][][";
    cout << swapCount(s) << "n";

    s = "[[][]]";
    cout << swapCount(s) << "n";
    return 0;



This article is contributed by Aditya Kamath. If you like GeeksforGeeks and would like to contribute, you can also write an article using or mail your article to See your article appearing on the GeeksforGeeks main page and help other Geeks.

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