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.

Examples:

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] == '[') 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; } // Driver code int main() { string s = "[]][]["; cout << swapCount(s) << "n"; s = "[[][]]"; cout << swapCount(s) << "n"; return 0; }

Output:

2 0

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