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Maximum length of balanced string after swapping and removal of characters

Given a string str consisting of characters ‘(‘, ‘)’, ‘[‘, ‘]’, ‘{‘ and ‘}’ only. The task is to find the maximum length of the balanced string after removing any character and swapping any two adjacent characters.
Examples: 
 

Input: str = “))[]]((” 
Output:
The string can be converted to ()[]()
Input: str = “{{{{{{{}” 
Output:

Approach: The idea is to remove extra unmatched parentheses from the string because we cannot generate a balanced pair for it and swap the remaining characters to balance the string. Therefore the answer is the equal summation of pairs of all balanced parenthesis. Note that we can move a character to any other place by adjacent swappings.
Below is the implementation of the above approach: 
 




// C++ implementation of the approach
#include <bits/stdc++.h>
using namespace std;
 
// Function to return the length of
// the longest balanced sub-string
int maxBalancedStr(string s)
{
 
    // To store the count of parentheses
    int open1 = 0, close1 = 0;
    int open2 = 0, close2 = 0;
    int open3 = 0, close3 = 0;
 
    // Traversing the string
    for (int i = 0; i < s.length(); i++) {
 
        // Check type of parentheses and
        // incrementing count for it
        switch (s[i]) {
        case '(':
            open1++;
            break;
        case ')':
            close1++;
            break;
        case '{':
            open2++;
            break;
        case '}':
            close2++;
            break;
        case '[':
            open3++;
            break;
        case ']':
            close3++;
            break;
        }
    }
 
    // Sum all pair of balanced parentheses
    int maxLen = 2 * min(open1, close1)
                 + 2 * min(open2, close2)
                 + 2 * min(open3, close3);
 
    return maxLen;
}
 
// Driven code
int main()
{
    string s = "))[]]((";
    cout << maxBalancedStr(s);
 
    return 0;
}




// Java implementation of the approach
class GFG
{
     
// Function to return the length of
// the longest balanced sub-string
static int maxBalancedStr(String s)
{
 
    // To store the count of parentheses
    int open1 = 0, close1 = 0;
    int open2 = 0, close2 = 0;
    int open3 = 0, close3 = 0;
 
    // Traversing the string
    for (int i = 0; i < s.length(); i++)
    {
 
        // Check type of parentheses and
        // incrementing count for it
        switch (s.charAt(i))
        {
        case '(':
            open1++;
            break;
        case ')':
            close1++;
            break;
        case '{':
            open2++;
            break;
        case '}':
            close2++;
            break;
        case '[':
            open3++;
            break;
        case ']':
            close3++;
            break;
        }
    }
 
    // Sum all pair of balanced parentheses
    int maxLen = 2 * Math.min(open1, close1)
                + 2 * Math.min(open2, close2)
                + 2 * Math.min(open3, close3);
 
    return maxLen;
}
 
// Driven code
public static void main(String[] args)
{
    String s = "))[]]((";
    System.out.println(maxBalancedStr(s));
}
}
 
// This code is contributed by Code_Mech.




# Python 3 implementation of the approach
 
# Function to return the length of
# the longest balanced sub-string
def maxBalancedStr(s):
     
    # To store the count of parentheses
    open1 = 0
    close1 = 0
    open2 = 0
    close2 = 0
    open3 = 0
    close3 = 0
 
    # Traversing the string
    for i in range(len(s)):
         
        # Check type of parentheses and
        # incrementing count for it
        if(s[i] == '('):
            open1 += 1
            continue
        if s[i] == ')':
            close1 += 1
            continue
        if s[i] == '{':
            open2 += 1
            continue
        if s[i] == '}':
            close2 += 1
            continue
        if s[i] == '[':
            open3 += 1
            continue
        if s[i] == ']':
            close3 += 1
            continue
 
    # Sum all pair of balanced parentheses
    maxLen = (2 * min(open1, close1) +
              2 * min(open2, close2) +
              2 * min(open3, close3))
 
    return maxLen
 
# Driven code
if __name__ == '__main__':
    s = "))[]](("
    print(maxBalancedStr(s))
 
# This code is contributed by
# Surendra_Gangwar




// C# implementation of the approach
using System;
 
class GFG
{
     
// Function to return the length of
// the longest balanced sub-string
static int maxBalancedStr(string s)
{
 
    // To store the count of parentheses
    int open1 = 0, close1 = 0;
    int open2 = 0, close2 = 0;
    int open3 = 0, close3 = 0;
 
    // Traversing the string
    for (int i = 0; i < s.Length; i++)
    {
 
        // Check type of parentheses and
        // incrementing count for it
        switch (s[i])
        {
        case '(':
            open1++;
            break;
        case ')':
            close1++;
            break;
        case '{':
            open2++;
            break;
        case '}':
            close2++;
            break;
        case '[':
            open3++;
            break;
        case ']':
            close3++;
            break;
        }
    }
 
    // Sum all pair of balanced parentheses
    int maxLen = 2 * Math.Min(open1, close1)
                + 2 * Math.Min(open2, close2)
                + 2 * Math.Min(open3, close3);
 
    return maxLen;
}
 
// Driver code
public static void Main()
{
    string s = "))[]]((";
    Console.WriteLine(maxBalancedStr(s));
}
}
 
// This code is contributed by Code_Mech.




<?php
// PHP implementation of the approach
// Function to return the length of
// the longest balanced sub-string
 function maxBalancedStr($s)
{
 
    // To store the count of parentheses
    $open1 = 0; $close1 = 0;
    $open2 = 0; $close2 = 0;
    $open3 = 0; $close3 = 0;
 
    // Traversing the string
    for ($i = 0; $i < strlen($s); $i++)
    {
 
        // Check type of parentheses and
        // incrementing count for it
        switch ($s[$i])
        {
        case '(':
            $open1++;
            break;
        case ')':
            $close1++;
            break;
        case '{':
            $open2++;
            break;
        case '}':
            $close2++;
            break;
        case '[':
            $open3++;
            break;
        case ']':
            $close3++;
            break;
        }
    }
 
    // Sum all pair of balanced parentheses
    $maxLen = 2 * min($open1, $close1)
                + 2 * min($open2, $close2)
                + 2 * min($open3, $close3);
 
    return $maxLen;
}
 
// Driven code
{
    $s = "))[]]((";
    echo(maxBalancedStr($s));
}
 
// This code is contributed by Code_Mech.




<script>
 
// Javascript implementation of the approach   
// Function to return the length of
    // the longest balanced sub-string
    function maxBalancedStr( s) {
 
        // To store the count of parentheses
        var open1 = 0, close1 = 0;
        var open2 = 0, close2 = 0;
        var open3 = 0, close3 = 0;
 
        // Traversing the string
        for (i = 0; i < s.length; i++) {
 
            // Check type of parentheses and
            // incrementing count for it
            switch (s.charAt(i)) {
            case '(':
                open1++;
                break;
            case ')':
                close1++;
                break;
            case '{':
                open2++;
                break;
            case '}':
                close2++;
                break;
            case '[':
                open3++;
                break;
            case ']':
                close3++;
                break;
            }
        }
 
        // Sum all pair of balanced parentheses
        var maxLen = 2 * Math.min(open1, close1)
        + 2 * Math.min(open2, close2)
        + 2 * Math.min(open3, close3);
 
        return maxLen;
    }
 
    // Driven code
     
        var s = "))[]]((";
        document.write(maxBalancedStr(s));
 
// This code contributed by gauravrajput1
 
</script>

Output
6

Time Complexity: O(n), where n is the length of the given string.
Auxiliary Space: O(1), no extra space is required, so it is a constant.


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