Minimum Swaps for Bracket Balancing

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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.

Code-

C++

// C++ program to count swaps required to balance string
#include<bits/stdc++.h>
using namespace std;
 
// Function to calculate swaps required
int swapCount(string s)
{
    //To store answer
    int ans=0;
     
    //To store count of '['
    int count=0;
     
    //Size of string
    int n=s.size();
     
    //Traverse over the string
    for(int i=0;i<n;i++){
        //When '[' encounters
        if(s[i]=='['){count++;}
        //when ']' encounters
        else{count--;}
        //When count becomes less than 0
        if(count<0){
            //Start searching for '[' from (i+1)th index
            int j=i+1;
            while(j<n){
                //When jth index contains '['
                if(s[j]=='['){break;}
                j++;
            }
            //Increment answer
            ans+=j-i;
             
            //Set Count to 1 again
            count=1;
             
            //Bring character at jth position to ith position
            //and shift all character from i to j-1 
            //towards right 
            char ch=s[j];
            for(int k=j;k>i;k--){
                s[k]=s[k-1];
            }
            s[i]=ch;
        }
    }
    return ans;
}
 
// 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
 
public class GFG {
     
    // Function to calculate swaps required
    static int swapCount(String s) {
         
        //To store answer
        int ans = 0;
         
        //To store count of '['
        int count = 0;
         
        //Size of string
        int n = s.length();
         
        //Traverse over the string
        for (int i = 0; i < n; i++) {
            //When '[' encounters
            if (s.charAt(i) == '[')
                count++;
            //when ']' encounters
            else
                count--;
            //When count becomes less than 0
            if (count < 0) {
                //Start searching for '[' from (i+1)th index
                int j = i + 1;
                while (j < n) {
                    //When jth index contains '['
                    if (s.charAt(j) == '[')
                        break;
                    j++;
                }
                 
                //Increment answer
                ans += j - i;
                 
                //Set Count to 1 again
                count = 1;
                 
                //Bring character at jth position to ith position
                //and shift all character from i to j-1 
                //towards right 
                char ch = s.charAt(j);
                StringBuilder newString = new StringBuilder(s);
                for (int k = j; k > i; k--) {
                    newString.setCharAt(k, s.charAt(k - 1));
                }
                newString.setCharAt(i, ch);
                s = newString.toString();
            }
        }
 
        return ans;
    }
     
    // Driver code
    public static void main(String[] args) {
        String s = "[]][][";
        System.out.println(swapCount(s));
 
        s = "[[][]]";
        System.out.println(swapCount(s));
    }
}

Python3

def swap_count(s):
    # To store the answer
    ans = 0
     
    # To store the count of '['
    count = 0
     
    # Size of the string
    n = len(s)
     
    # Traverse over the string
    for i in range(n):
        # When '[' encounters
        if s[i] == '[':
            count += 1
        # When ']' encounters
        else:
            count -= 1
        # When count becomes less than 0
        if count < 0:
            # Start searching for '[' from (i+1)th index
            j = i + 1
            while j < n:
                # When jth index contains '['
                if s[j] == '[':
                    break
                j += 1
            # Increment the answer
            ans += j - i
             
            # Set count to 1 again
            count = 1
             
            # Bring the character at jth position to ith position
            # and shift all characters from i to j-1 
            # towards the right
            ch = s[j]
            for k in range(j, i, -1):
                s[k] = s[k - 1]
            s[i] = ch
    return ans
 
# Driver code
if __name__ == "__main__":
    s = "[]][]["
    print(swap_count(list(s)))
 
    s = "[[][]]"
    print(swap_count(list(s)))

Javascript

function GFG(s) {
    // To store answer
    let ans = 0;
    // To store count of '['
    let count = 0;
    // Traverse over the string
    for (let i = 0; i < s.length; i++) {
        // When '[' encounters
        if (s[i] === '[') {
            count++;
        }
        // When ']' encounters
        else {
            count--;
        }
        // When count becomes less than 0
        if (count < 0) {
            // Start searching for 
            // '[' from the beginning
            for (let j = 0; j < i; j++) {
                // When jth index contains '['
                if (s[j] === '[') {
                    ans += i - j;
                    // Swap characters to balance the string
                    let temp = s.substring(0, j) + s[i] + s.substring(j + 1, i) + s[j] + s.substring(i + 1);
                    s = temp;
                    break;
                }
            }
            // Reset the count
            count = 1;
        }
    }
    return ans;
}
// Driver code
let s1 = "[]][][";
console.log(GFG(s1)); // Output: 2
let s2 = "[[][]]";
console.log(GFG(s2)); // Output: 0

Output-

2
0

Time Complexity = O(N^2), one loop is for traversing the string and another loop in finding the next ‘[‘ when the count becomes less than 0 and making the string ready for the next step
Extra Space = O(1), because no extra space has been used

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 1 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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Last Updated : 08 Oct, 2023