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 form 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’ represent 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 1 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++

// 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;
}

Java

// Java program to count swaps
// required to balance string
import java.util.*;

class GFG{

// Function to calculate swaps required
public static long swapCount(String s)
{

    // Keep track of '['
    Vector<Integer> pos = new Vector<Integer>();
    for(int i = 0; i < s.length(); ++i)
        if (s.charAt(i) == '[')
            pos.add(i);

    // To count number of encountered '['
    int count = 0;

    // To track position of next '[' in pos
    int p = 0;

    // To store result
    long sum = 0;

    char[] S = s.toCharArray();

    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.get(p) - i;
            char temp = S[i];
            S[i] = S[pos.get(p)];
            S[pos.get(p)] = temp;
            ++p;

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

// 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 divyesh072019

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

C#

// C# program to count swaps
// required to balance string
using System.IO;
using System;
using System.Collections;
using System.Collections.Generic;

class GFG{

// Function to calculate swaps required
static long swapCount(string s)
{

    // Keep track of '['
    List<int> pos = new List<int>();
    for(int i = 0; i < s.Length; i++)
    {
        if (s[i] == '[')
        {
            pos.Add(i);
        }
    }

    // To count number of encountered '['
    int count = 0;

    // To track position of next '[' in pos
    int p = 0;

    // To store result
    long sum = 0;

    char[] S = s.ToCharArray();

    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;
            char temp = S[i];
            S[i] = S[pos[p]];
            S[pos[p]] = temp;
            ++p;

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

// Driver code
static void Main()
{
    string s = "[]][][";
    Console.WriteLine(swapCount(s));

    s = "[[][]]";
    Console.WriteLine(swapCount(s));
}
}

// This code is contributed by rag2127

Javascript

<script>

// JavaScript program to count swaps
// required to balance string

// Function to calculate swaps required
function swapCount(s)
{

    // Keep track of '['
    let pos = [];
    for(let i = 0; i < s.length; ++i)
        if (s[i] == '[')
            pos.push(i);

    // To count number of encountered '['
    let count = 0;

    // To track position of next '[' in pos
    let p = 0;

    // To store result
    let sum = 0;

    let S = s.split('');

    for(let 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;
            let temp = S[i];
            S[i] = S[pos[p]];
            S[pos[p]] = temp;
            ++p;

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

// Driver Code

    let s = "[]][][";
    document.write(swapCount(s) + "<br/>");

    s = "[[][]]";
    document.write(swapCount(s));

</script>

Output

2
0

Time Complexity: O(N) 
Auxiliary Space: O(N) 

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

Below is the implementation :  

C++

// C++ program to count swaps required
// to balance string
#include <bits/stdc++.h>
using namespace std;

long swapCount(string chars)
{

    // 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
int main()
{
    string s = "[]][][";
    cout << swapCount(s) << endl;

    s = "[[][]]";
    cout << swapCount(s) << endl;

    return 0;
}

// This code is contributed by divyeshrabadiya07

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.

Python3

# Python3 program to count swaps required to
# balance string
def swapCount(s):



    # Swap stores the number of swaps
    # required imbalance maintains the
    # number of imbalance pair
    swap = 0
    imbalance = 0;

    for i in s:
        if i == '[':

            # Decrement the imbalance
            imbalance -= 1
        else:

            # Increment imbalance
            imbalance += 1

            if imbalance > 0:
                swap += imbalance

    return swap

# Driver code
s = "[]][][";
print(swapCount(s))

s = "[[][]]";
print(swapCount(s))

# This code is contributed by Prateek Gupta and improved by Anvesh Govind Saxena

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

Javascript

<script>
    // Javascript Program to count swaps required
    // to balance string

    function swapCount(s)
    {
        let chars = s.split('');

        // stores the total number of Left and
        // Right brackets encountered
        let countLeft = 0, countRight = 0;

        // swap stores the number of swaps
        // required imbalance maintains the
        // number of imbalance pair
        let swap = 0, imbalance = 0;

        for (let 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;
    }

      let s = "[]][][";
    document.write(swapCount(s) + "</br>");

    s = "[[][]]";
    document.write(swapCount(s));

    // This code is contributed by suresh07.
</script>

Output

2
0

Time Complexity :O(N) 
Auxiliary Space : O(1) 

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