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 The string is already balanced.
We can solve this problem by 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.
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. In the end, ‘sum’ will have the required value.
Time Complexity = O(N^2)
Extra Space = O(1)
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 the count to 0 and increment p so that it pos[p] indicates to the next ‘[‘.
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)
Time Complexity = O(N)
Extra Space = O(1)
We can do without having to store the positions of ‘[‘.
Below is the implementation :
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