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Minimize removal of non-equal adjacent characters required to make a given string empty
  • Last Updated : 26 Jan, 2021

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++




// 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 occurences '('
    int count_1 = 0;
 
    // Stores count of occurences ')'
    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 occurence 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




// 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 occurences '('
    int count_1 = 0;
 
    // Stores count of occurences ')'
    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 occurence 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




#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 occurences '('
    count_1 = 0
 
    # Stores count of occurences ')'
    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 occurence 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#




// 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 occurences '('
    int count_1 = 0;
     
    // Stores count of occurences ')'
    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 occurence 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
Output: 
0

 

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

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