# 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.

**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; ` `} ` |

*chevron_right*

*filter_none*

Output:

2 0

**Another Method:**

Time Complexity = O(N)

Extra Space = O(1)

We can do without having to store the positions of ‘[‘.

Below is the implementation :

## Java

`// Java Program to count swaps required to balance string ` `public` `class` `BalanceParan ` `{ ` ` ` ` ` `static` `long` `swapCount(String s) ` ` ` `{ ` ` ` `char` `[] chars = s.toCharArray(); ` ` ` ` ` `// 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< chars.length; i++) ` ` ` `{ ` ` ` `if` `(chars[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` `(chars[i] == ` `']'` `) ` ` ` `{ ` ` ` `// increment count of Right bracket ` ` ` `countRight++; ` ` ` `// imbalance is reset to current difference ` ` ` `// between Left and Right brackets ` ` ` `imbalance = (countRight-countLeft); ` ` ` `} ` ` ` `} ` ` ` `return` `swap; ` ` ` `} ` ` ` `// Driver code ` ` ` `public` `static` `void` `main(String args[]) ` ` ` `{ ` ` ` `String s = ` `"[]][]["` `; ` ` ` `System.out.println(swapCount(s) ); ` ` ` ` ` `s = ` `"[[][]]"` `; ` ` ` `System.out.println(swapCount(s) ); ` ` ` ` ` `} ` `} ` `// This code is contributed by Janmejaya Das. ` |

*chevron_right*

*filter_none*

## C#

`// C# Program to count swaps required ` `// to balance string ` `using` `System; ` ` ` `class` `GFG ` `{ ` ` ` `public` `static` `long` `swapCount(` `string` `s) ` `{ ` ` ` `char` `[] chars = s.ToCharArray(); ` ` ` ` ` `// 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 < chars.Length; i++) ` ` ` `{ ` ` ` `if` `(chars[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` `(chars[i] == ` `']'` `) ` ` ` `{ ` ` ` `// increment count of Right bracket ` ` ` `countRight++; ` ` ` ` ` `// imbalance is reset to current difference ` ` ` `// between Left and Right brackets ` ` ` `imbalance = (countRight - countLeft); ` ` ` `} ` ` ` `} ` ` ` `return` `swap; ` `} ` ` ` `// Driver code ` `public` `static` `void` `Main(` `string` `[] args) ` `{ ` ` ` `string` `s = ` `"[]][]["` `; ` ` ` `Console.WriteLine(swapCount(s)); ` ` ` ` ` `s = ` `"[[][]]"` `; ` ` ` `Console.WriteLine(swapCount(s)); ` `} ` `} ` ` ` `// This code is contributed by Shrikant13 ` |

*chevron_right*

*filter_none*

**Output:**

2 0

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

Please write comments if you find anything incorrect, or you want to share more information about the topic discussed above.

## Recommended Posts:

- Find index of closing bracket for a given opening bracket in an expression
- Minimum number of bracket reversals needed to make an expression balanced
- Minimum swaps required to convert one binary string to another
- Minimum number of adjacent swaps for arranging similar elements together
- Minimum swaps required to make a binary string alternating
- Minimum swaps to group similar characters side by side?
- Find numbers of balancing positions in string
- Expression contains redundant bracket or not
- Print Bracket Number
- Number of balanced bracket subsequence of length 2 and 4
- Print all ways to break a string in bracket form
- Construct Binary Tree from String with bracket representation
- Range Queries for Longest Correct Bracket Subsequence
- Number of closing brackets needed to complete a regular bracket sequence
- Largest permutation after at most k swaps