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 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.**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. In 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 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)

## C++

`// 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;` `}` |

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## Python3

`# Python3 Program to count ` `# swaps required to balance ` `# string ` `# Function to calculate ` `# swaps required ` `def` `swapCount(s):` ` ` `# Keep track of '[' ` ` ` `pos ` `=` `[]` ` ` `for` `i ` `in` `range` `(` `len` `(s)):` ` ` `if` `(s[i] ` `=` `=` `'['` `):` ` ` `pos.append(i)` ` ` `# To count number ` ` ` `# of encountered '[' ` ` ` `count ` `=` `0` ` ` ` ` `# To track position ` ` ` `# of next '[' in pos ` ` ` `p ` `=` `0` ` ` ` ` `# To store result ` ` ` `sum` `=` `0` ` ` `s ` `=` `list` `(s)` ` ` ` ` `for` `i ` `in` `range` `(` `len` `(s)):` ` ` `# Increment count and ` ` ` `# move p to next position ` ` ` `if` `(s[i] ` `=` `=` `'['` `):` ` ` `count ` `+` `=` `1` ` ` `p ` `+` `=` `1` ` ` `elif` `(s[i] ` `=` `=` `']'` `):` ` ` `count ` `-` `=` `1` ` ` `# 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` ` ` `s[i], s[pos[p]] ` `=` `(s[pos[p]], ` ` ` `s[i])` ` ` `p ` `+` `=` `1` ` ` `# Reset count to 1 ` ` ` `count ` `=` `1` ` ` `return` `sum` `# Driver code ` `s ` `=` `"[]][]["` `print` `(swapCount(s))` `s ` `=` `"[[][]]"` `print` `(swapCount(s))` `# This code is contributed by avanitrachhadiya2155` |

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**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.` |

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## Python3

`# Python3 program to count swaps required to` `# balance string ` `def` `swapCount(s): ` ` ` ` ` `chars ` `=` `s` ` ` ` ` `# Stores total number of left and ` ` ` `# right brackets encountered ` ` ` `countLeft ` `=` `0` ` ` `countRight ` `=` `0` ` ` ` ` `# Swap stores the number of swaps ` ` ` `# required imbalance maintains the` ` ` `# number of imbalance pair ` ` ` `swap ` `=` `0` ` ` `imbalance ` `=` `0` `; ` ` ` ` ` `for` `i ` `in` `range` `(` `len` `(chars)):` ` ` `if` `chars[i] ` `=` `=` `'['` `:` ` ` ` ` `# Increment count of left bracket` ` ` `countLeft ` `+` `=` `1` ` ` ` ` `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 ` `-` `=` `1` ` ` ` ` `elif` `chars[i] ` `=` `=` `']'` `:` ` ` ` ` `# Increment count of right bracket ` ` ` `countRight ` `+` `=` `1` ` ` ` ` `# Imbalance is reset to current` ` ` `# difference between left and ` ` ` `# right brackets ` ` ` `imbalance ` `=` `(countRight ` `-` `countLeft) ` ` ` `return` `swap` `# Driver code ` `s ` `=` `"[]][]["` `; ` `print` `(swapCount(s))` `s ` `=` `"[[][]]"` `; ` `print` `(swapCount(s))` `# This code is contributed by Prateek Gupta ` |

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## 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 the 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` |

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**Output:**

2 0

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