Minimize removal of non-equal adjacent characters required to make a given string empty

Given a string S consisting of '(' and ')', the task is to find the minimum count of bracket reversals required to make the string an empty string by repeatedly removing a pair of non-equal adjacent characters. If it is not possible to empty the string, then print -1.

Examples:

Input: S = ")))((("
Output: 0
Explanation: 
Removing (S[2], S[3]) from S modifies S to "))((". 
Removing (S[1], S[2]) from S modifies S to ")(". 
Removing (S[0], S[1]) from S modifies S to "". 
Since no flips are required to make S empty. 
Therefore, the required output is 0. 

Input: S = "))("
Output: -1

Approach: Follow the steps below to solve the problem:

• Initialize two integer variables cnt1 and cnt2 as 0.
• If the length of the string is odd or only one type of character is present, then print "-1".
• Otherwise, iterate over the characters of the string S and if the current character is '(', then increment the cnt1 by one. Otherwise, increment cnt2 by 1.
• After completing the above steps, print the value of abs(cnt1 - cnt2)/2 as the minimum number of reversals required.

Below is the implementation of the above approach:

// C++ program for the above approach
#include <bits/stdc++.h>
using namespace std;

// Function to find minimum count of steps
// required ot make string S an empty string
void canReduceString(string S, int N)
{

    // Stores count of occurrences '('
    int count_1 = 0;

    // Stores count of occurrences ')'
    int count_2 = 0;

    // Traverse the string, str
    for (int i = 0; i < N; i++) {

        // If current character is '('
        if (S[i] == '(') {

            // Update count_1
            count_1++;
        }
        else {

            // Update count_2
            count_2++;
        }
    }

    // If all the characters are
    // same, then print -1
    if (count_1 == 0 || count_2 == 0) {
        cout << "-1" << endl;
    }

    // If the count of occurrence of ')'
    // and '(' are same then print 0
    else if (count_1 == count_2) {
        cout << "0" << endl;
    }

    // If length of string is Odd
    else if (N % 2 != 0) {
        cout << "-1";
    }

    else {
        cout << abs(count_1 - count_2) / 2;
    }
}

// Driver Code
int main()
{

    // Given string
    string S = ")))(((";

    // Size of the string
    int N = S.length();

    // Function Call
    canReduceString(S, N);

    return 0;
}

// Java program for the above approach
import java.util.*;

class GFG{

// Function to find minimum count of steps
// required ot make String S an empty String
static void canReduceString(String S, int N)
{
    
    // Stores count of occurrences '('
    int count_1 = 0;

    // Stores count of occurrences ')'
    int count_2 = 0;

    // Traverse the String, str
    for(int i = 0; i < N; i++)
    {
        
        // If current character is '('
        if (S.charAt(i) == '(') 
        {
            
            // Update count_1
            count_1++;
        }
        else 
        {
            
            // Update count_2
            count_2++;
        }
    }

    // If all the characters are
    // same, then print -1
    if (count_1 == 0 || count_2 == 0) 
    {
        System.out.print("-1" + "\n");
    }

    // If the count of occurrence of ')'
    // and '(' are same then print 0
    else if (count_1 == count_2) 
    {
        System.out.print("0" + "\n");
    }

    // If length of String is Odd
    else if (N % 2 != 0) 
    {
        System.out.print("-1");
    }

    else 
    {
        System.out.print(Math.abs(
            count_1 - count_2) / 2);
    }
}

// Driver Code
public static void main(String[] args)
{
    
    // Given String
    String S = ")))(((";

    // Size of the String
    int N = S.length();

    // Function Call
    canReduceString(S, N);
}
}

// This code is contributed by shikhasingrajput 

#Python3 program for the above approach


# Function to find minimum count of steps
# required ot make string S an empty string
def canReduceString(S, N):

    # Stores count of occurrences '('
    count_1 = 0

    # Stores count of occurrences ')'
    count_2 = 0

    # Traverse the string, str
    for i in range(N):

        # If current character is '('
        if (S[i] == '('):

            # Update count_1
            count_1 += 1
        else:

            #Update count_2
            count_2 += 1

    # If all the characters are
    # same, then pr-1
    if (count_1 == 0 or count_2 == 0):
        print("-1")
    
    # If the count of occurrence of ')'
    # and '(' are same then pr0
    elif (count_1 == count_2):
        print("0")

    # If length of string is Odd
    elif (N % 2 != 0):
        print("-1")
    else:
        print (abs(count_1 - count_2) // 2)

# Driver Code
if __name__ == '__main__':

    # Given string
    S = ")))((("

    # Size of the string
    N = len(S)

    # Function Call
    canReduceString(S, N)

    # This code is contributed by mohit kumar 29

// C# program for the above approach
using System;

class GFG{

// Function to find minimum count of steps
// required ot make String S an empty String
static void canReduceString(String S, int N)
{
    
    // Stores count of occurrences '('
    int count_1 = 0;
    
    // Stores count of occurrences ')'
    int count_2 = 0;
    
    // Traverse the String, str
    for(int i = 0; i < N; i++)
    {
        
        // If current character is '('
        if (S[i] == '(') 
        {
            
            // Update count_1
            count_1++;
        }
        else 
        {
            
            // Update count_2
            count_2++;
        }
    }
    
    // If all the characters are
    // same, then print -1
    if (count_1 == 0 || count_2 == 0) 
    {
        Console.Write("-1" + "\n");
    }
    
    // If the count of occurrence of ')'
    // and '(' are same then print 0
    else if (count_1 == count_2) 
    {
        Console.Write("0" + "\n");
    }
    
    // If length of String is Odd
    else if (N % 2 != 0) 
    {
        Console.Write("-1");
    }

    else 
    {
        Console.Write(Math.Abs(
            count_1 - count_2) / 2);
    }
}

// Driver Code
public static void Main(String[] args)
{
    
    // Given String
    String S = ")))(((";

    // Size of the String
    int N = S.Length;

    // Function Call
    canReduceString(S, N);
}
}

// This code is contributed by shikhasingrajput 

<script>
      // JavaScript program for the above approach

      // Function to find minimum count of steps
      // required ot make string S an empty string
      function canReduceString(S, N) {
        // Stores count of occurrences '('
        var count_1 = 0;

        // Stores count of occurrences ')'
        var count_2 = 0;

        // Traverse the string, str
        for (var i = 0; i < N; i++) {
          // If current character is '('
          if (S[i] === "(") {
            // Update count_1
            count_1++;
          } else {
            // Update count_2
            count_2++;
          }
        }

        // If all the characters are
        // same, then print -1
        if (count_1 === 0 || count_2 === 0) {
          document.write("-1" + "<br>");
        }

        // If the count of occurrence of ')'
        // and '(' are same then print 0
        else if (count_1 === count_2) {
          document.write("0" + "<br>");
        }

        // If length of string is Odd
        else if (N % 2 !== 0) {
          document.write("-1");
        } else {
          document.write(Math.abs(count_1 - count_2) / 2);
        }
      }

      // Driver Code
      // Given string
      var S = ")))(((";

      // Size of the string
      var N = S.length;

      // Function Call
      canReduceString(S, N);
    </script>

0

Time Complexity: O(N), Time complexity of the given C++ program is O(N) where N is the length of the string. This is because the program only traverses the input string once, performing a constant amount of work at each character.
Auxiliary Space: O(1), The space complexity of the program is O(1) as the program only uses a constant amount of extra space to store the counts of the occurrences of '(' and ')' characters, irrespective of the length of the input string.

mishrapriyanshu557

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Article Tags :

• Strings
• DSA
• frequency-counting

Practice Tags :

• Strings