Minimum operation required to make a balanced sequence

A balanced sequence is defined as a string in which for every opening bracket there are 2 continuous closing brackets. Thus, {}}, {{}}}}, {{}}{}}}} are balanced, whereas }}{, {} are not balanced. 
Now given a sequence of brackets (‘{‘ and ‘}’), and you can perform only one operation on that sequence, i.e., either insert an opening or closing bracket at any position. You have to tell the minimum number of operations required to make the given sequence balanced. 

Input: str = “{}}” 
Output: 0 
The sequence is already balanced.
Input: str = “{}{}}}” 
Output: 3 
The updated sequence will be “{}}{}}{}}”. 

Approach: For making a balanced sequence, two continuous closing brackets are required for every opening bracket. There can be 3 cases: 

1. When the current character is an opening bracket: If the previous character is not a closing bracket, then simply insert the opening bracket to stack, else there is a need for a closing bracket that will cost one operation.
2. If the stack is empty and the current character is a closing bracket: In this case, one opening bracket is required that will cost one operation and insert that opening bracket to the stack.
3. If the stack is not empty but the current character is a closing bracket: Here, only the count of closing brackets is required. If it is 2, then remove one opening bracket from the stack, else increment the count of the closing bracket.

At the end of the string, if the stack is not empty, then the required count of closing brackets will be ((2 * size of the stack) – current count of closing brackets).
Below is the implementation of the above approach:  

C++

// C++ implementation of the approach
#include <bits/stdc++.h>
using namespace std;

// Function to return the minimum operations required
int minOperations(string s, int len)
{
    int operationCnt = 0;
    stack<char> st;
    int cntClosing = 0;
    for (int i = 0; i < len; i++) {

        // Condition where we got only one closing
        // bracket instead of 2, here we have to
        // add one more closing bracket to
        // make the sequence balanced
        if (s[i] == '{') {
            if (cntClosing > 0) {

                // Add closing bracket that
                // costs us one operation
                operationCnt++;

                // Remove the top opening bracket because
                // we got the 1 opening and 2
                // continuous closing brackets
                st.pop();
            }

            // Inserting the opening bracket to stack
            st.push(s[i]);

            // After making the sequence balanced
            // closing is now set to 0
            cntClosing = 0;
        }
        else if (st.empty()) {

            // Case when there is no opening bracket
            // the sequence starts with a closing bracket
            // and one opening bracket is required
            // Now we have one opening and one closing bracket
            st.push('{');

            // Add opening bracket that
            // costs us one operation
            operationCnt++;

            // Assigning 1 to cntClosing because
            // we have one closing bracket
            cntClosing = 1;
        }
        else {
            cntClosing = (cntClosing + 1) % 2;

            // Case where we got two continuous closing brackets
            // Need to pop one opening bracket from stack top
            if (cntClosing == 0) {
                st.pop();
            }
        }
    }

    // Condition where stack is not empty
    // This is the case where we have
    // only opening brackets
    // (st.size() * 2) will give us the total
    // closing bracket needed
    // cntClosing is the count of closing
    // bracket that we already have
    operationCnt += st.size() * 2 - cntClosing;
    return operationCnt;
}

// Driver code
int main()
{
    string str = "}}{";
    int len = str.length();

    cout << minOperations(str, len);

    return 0;
}

Java

// Java implementation of the approach
import java.util.*;

class GFG{

// Function to return the
// minimum operations required
static int minOperations(String s, int len)
{
    int operationCnt = 0;
    Stack<Character> st = new Stack<Character>();
    int cntClosing = 0;

    for(int i = 0; i < len; i++)
    {

       // Condition where we got only one
       // closing bracket instead of 2,
       // here we have to add one more
       // closing bracket to make the
       // sequence balanced
       if (s.charAt(i) == '{')
       {
           if (cntClosing > 0)
           {

               // Add closing bracket that
               // costs us one operation
               operationCnt++;

               // Remove the top opening bracket
               // because we got the 1 opening
               // and 2 continuous closing brackets
               st.pop();
           }

           // Inserting the opening bracket to stack
           st.add(s.charAt(i));

           // After making the sequence balanced
           // closing is now set to 0
           cntClosing = 0;
       }
       else if (st.isEmpty())
       {

           // Case when there is no opening
           // bracket the sequence starts
           // with a closing bracket and
           // one opening bracket is required
           // Now we have one opening and one
           // closing bracket
           st.add('{');

           // Add opening bracket that
           // costs us one operation
           operationCnt++;

           // Assigning 1 to cntClosing because
           // we have one closing bracket
            cntClosing = 1;
       }
       else
       {
           cntClosing = (cntClosing + 1) % 2;

           // Case where we got two continuous
           // closing brackets need to pop one
           // opening bracket from stack top
           if (cntClosing == 0)
           {
               st.pop();
           }
       }
    }

    // Condition where stack is not empty
    // This is the case where we have only
    // opening brackets (st.size() * 2)
    // will give us the total closing
    // bracket needed cntClosing is the
    // count of closing bracket that
    // we already have
    operationCnt += st.size() * 2 - cntClosing;

    return operationCnt;
}

// Driver code
public static void main(String[] args)
{
    String str = "}}{";
    int len = str.length();

    System.out.print(minOperations(str, len));
}
}

// This code is contributed by amal kumar choubey

Python3

# Python3 implementation
# of the approach

# Function to return the
# minimum operations required
def minOperations(s, length):

    operationCnt = 0
    stack = [] # Built-in Python3 list to implement stack
    cntClosing = 0

    for i in range(0, length):

        # Condition where we got only one
        # closing bracket instead of 2,
        # here we have to add one more
        # closing bracket to make the
        # sequence balanced
        if (s[i] == '{'):

            if (cntClosing > 0):

                # Add closing bracket that
                # costs us one operation
                operationCnt += 1

                # Remove the top opening bracket
                # because we got the 1 opening
                # and 2 continuous closing brackets
                stack.pop()

            # Inserting the opening bracket to stack
            stack.append(s[i])

            # After making the sequence balanced
            # closing is now set to 0
            cntClosing = 0

        elif not stack:

                # Case when there is no opening
                # bracket the sequence starts
                # with a closing bracket and
                # one opening bracket is required
                # Now we have one opening and one
                # closing bracket
                stack.append('{')

                # Add opening bracket that
                # costs us one operation
                operationCnt += 1

                # Assigning 1 to cntClosing because
                # we have one closing bracket
                cntClosing = 1
        else:
            cntClosing = (cntClosing + 1) % 2

            # Case where we got two continuous
            # closing brackets need to pop one
            # opening bracket from stack top
            if (cntClosing == 0):
                stack.pop()

    # Condition where stack is not empty
    # This is the case where we have only
    # opening brackets (st.size() * 2)
    # will give us the total closing
    # bracket needed cntClosing is the
    # count of closing bracket that
    # we already have
    operationCnt += len(stack) * 2 - cntClosing

    return operationCnt

# Driver code
if __name__ == '__main__':
    string = "}}{"
    print(minOperations(string, len(string)))

# This code is contributed by gauravrajput1

C#

// C# implementation of
// the above approach
using System;
using System.Collections.Generic;
class GFG{

// Function to return the
// minimum operations required
static int minOperations(String s,
                         int len)
{
  int operationCnt = 0;
  Stack<char> st = new Stack<char>();
  int cntClosing = 0;

  for(int i = 0; i < len; i++)
  {
    // Condition where we got only one
    // closing bracket instead of 2,
    // here we have to add one more
    // closing bracket to make the
    // sequence balanced
    if (s[i] == '{')
    {
      if (cntClosing > 0)
      {
        // Add closing bracket that
        // costs us one operation
        operationCnt++;

        // Remove the top opening bracket
        // because we got the 1 opening
        // and 2 continuous closing brackets
        st.Pop();
      }

      // Inserting the opening bracket to stack
      st.Push(s[i]);

      // After making the sequence balanced
      // closing is now set to 0
      cntClosing = 0;
    }
    else if (st.Count != 0)
    {
      // Case when there is no opening
      // bracket the sequence starts
      // with a closing bracket and
      // one opening bracket is required
      // Now we have one opening and one
      // closing bracket
      st.Push('{');

      // Add opening bracket that
      // costs us one operation
      operationCnt++;

      // Assigning 1 to cntClosing because
      // we have one closing bracket
      cntClosing = 1;
    }
    else
    {
      cntClosing = (cntClosing + 1) % 2;

      // Case where we got two continuous
      // closing brackets need to pop one
      // opening bracket from stack top
      if (cntClosing == 0 &&
          st.Count != 0)
      {
        st.Pop();
      }
    }
  }

  // Condition where stack is not empty
  // This is the case where we have only
  // opening brackets (st.Count * 2)
  // will give us the total closing
  // bracket needed cntClosing is the
  // count of closing bracket that
  // we already have
  operationCnt += st.Count * 2 -
                  cntClosing + 1;

  return operationCnt;
}

// Driver code
public static void Main(String[] args)
{
  String str = "}}{";
  int len = str.Length;
  Console.Write(minOperations(str,
                              len));
}
}

// This code is contributed by gauravrajput1

Javascript

<script>

// Javascript implementation of the approach

// Function to return the minimum operations required
function minOperations(s, len)
{
    var operationCnt = 0;
    var st = [];
    var cntClosing = 0;
    for (var i = 0; i < len; i++) {

        // Condition where we got only one closing
        // bracket instead of 2, here we have to
        // add one more closing bracket to
        // make the sequence balanced
        if (s[i] == '{') {
            if (cntClosing > 0) {

                // Add closing bracket that
                // costs us one operation
                operationCnt++;

                // Remove the top opening bracket because
                // we got the 1 opening and 2
                // continuous closing brackets
                st.pop();
            }

            // Inserting the opening bracket to stack
            st.push(s[i]);

            // After making the sequence balanced
            // closing is now set to 0
            cntClosing = 0;
        }
        else if (st.length==0) {

            // Case when there is no opening bracket
            // the sequence starts with a closing bracket
            // and one opening bracket is required
            // Now we have one opening and one closing bracket
            st.push('{');

            // Add opening bracket that
            // costs us one operation
            operationCnt++;

            // Assigning 1 to cntClosing because
            // we have one closing bracket
            cntClosing = 1;
        }
        else {
            cntClosing = (cntClosing + 1) % 2;

            // Case where we got two continuous closing brackets
            // Need to pop one opening bracket from stack top
            if (cntClosing == 0) {
                st.pop();
            }
        }
    }

    // Condition where stack is not empty
    // This is the case where we have
    // only opening brackets
    // (st.size() * 2) will give us the total
    // closing bracket needed
    // cntClosing is the count of closing
    // bracket that we already have
    operationCnt += st.length * 2 - cntClosing;
    return operationCnt;
}

// Driver code
var str = "}}{";
var len = str.length;
document.write( minOperations(str, len));


</script>

Output:

3

Time Complexity: O(n)

Auxiliary Space: O(n)