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Minimum number of bracket reversals needed to make an expression balanced

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Given an expression with only ‘}’ and ‘{‘. The expression may not be balanced. Find minimum number of bracket reversals to make the expression balanced.
Examples: 

Input:  exp = "}{"
Output: 2
We need to change '}' to '{' and '{' to
'}' so that the expression becomes balanced,
the balanced expression is '{}'
Input: exp = "{{{"
Output: Can't be made balanced using reversals
Input: exp = "{{{{"
Output: 2
Input: exp = "{{{{}}"
Output: 1
Input: exp = "}{{}}{{{"
Output: 3
Recommended Practice

One simple observation is, the string can be balanced only if total number of brackets is even (there must be equal no of ‘{‘ and ‘}’)
A Naive Solution is to consider every bracket and recursively count number of reversals by taking two cases (i) keeping the bracket as it is (ii) reversing the bracket. If we get a balanced expression, we update result if number of steps followed for reaching here is smaller than the minimum so far. 

Steps to implement

  • Check for the length of the given expression, if it is odd then return -1
  • If it is even then call a recursive function
  • That recursive function once leaves a bracket as it is and once reverses a bracket
  • After each recursive call, we will check whether the modified string is balanced or not
    • If the modified string is balanced and the number of reversals to reach this string is minimum from all the answers found till now then keep this answer stored somewhere
  • In the function for checking whether the string is balanced or not, we will initially keep count=0.
    • After that traverse the string and if we got ‘{then increment the count else decrement the count
    • If at any point count becomes negative then this means there are more Closing brackets than opening ones. Hence string is not balanced
    • At last, if the count is not equal to 0 then that means there are more opening brackets than closing ones. Hence string is not balanced
    • If we never found that there are more Closing brackets than opening ones or there are more opening brackets than closing ones then the string is balanced

Code-

C++




// C++ program to find minimum number of
// reversals required to balance an expression
#include <bits/stdc++.h>
using namespace std;
 
//Function to check that expression is balanced or not
bool isBalanced(string expr)
{
    // Initialising Variables
    bool flag = true;
    int count = 0;
  
    // Traversing the Expression
    for (int i = 0; i < expr.length(); i++) {
  
        if (expr[i] == '{') {
            count++;
        }
        else {
            // It is a closing bracket
            count--;
        }
        if (count < 0) {
            // This means there are more Closing brackets
            // than opening ones
            flag = false;
            break;
        }
    }
  
    // If count is not zero,
    // It means there are
    // more opening brackets
    if (count != 0) {
        flag = false;
    }
  
    return flag;
}
 
 
//Recursive Function for finding
//Number of reversals
void recur(string expr,int n,int ind,int change,int &ans){
    //When generated expression is balanced
    if(isBalanced(expr)){ans=min(ans,change);}
     
    //When we covered whole string
    if(ind==n){return;}
     
    //Keep bracket as it is
    recur(expr,n,ind+1,change,ans);
     
    //Reverse the bracket
    if(expr[ind]=='{'){expr[ind]='}';}
    else{expr[ind]='{';}
    recur(expr,n,ind+1,change+1,ans);
}
 
 
 
 
// Returns count of minimum reversals for making
// expr balanced. Returns -1 if expr cannot be
// balanced.
int countMinReversals(string expr)
{
    //Length of expression
    int n = expr.length();
     
    //To store answer
    int ans=INT_MAX;
     
    //When total number of brackets are odd
    if(n%2==1){
        return -1;
    }else{
        //Function call for finding answer
        recur(expr,n,0,0,ans);
        return ans;
    }
}
 
// Driver program to test above function
int main()
{
    string expr = "}}{{";
    cout << countMinReversals(expr);
    return 0;
}


Java




public class MinReversalsToBalance {
 
    // Function to check if the expression is balanced or
    // not
    public static boolean isBalanced(String expr)
    {
        // Initializing variables
        boolean flag = true;
        int count = 0;
 
        // Traversing the expression
        for (char c : expr.toCharArray()) {
            if (c == '{') {
                count++;
            }
            else {
                // It is a closing bracket
                count--;
            }
            if (count < 0) {
                // This means there are more closing
                // brackets than opening ones
                flag = false;
                break;
            }
        }
 
        // If count is not zero, it means there are more
        // opening brackets
        if (count != 0) {
            flag = false;
        }
 
        return flag;
    }
 
    // Recursive function for finding the number of
    // reversals
    public static void recur(String expr, int n, int ind,
                             int change, int[] ans)
    {
        // When we've covered the whole string, check if
        // it's balanced
        if (ind == n) {
            if (isBalanced(expr)) {
                ans[0] = Math.min(ans[0], change);
            }
            return;
        }
 
        // Keep the bracket as it is
        recur(expr, n, ind + 1, change, ans);
 
        // Reverse the bracket
        if (expr.charAt(ind) == '{') {
            expr = expr.substring(0, ind) + '}'
                   + expr.substring(ind + 1);
        }
        else {
            expr = expr.substring(0, ind) + '{'
                   + expr.substring(ind + 1);
        }
        recur(expr, n, ind + 1, change + 1, ans);
    }
 
    // Returns the count of minimum reversals for making
    // expr balanced. Returns -1 if expr cannot be balanced.
    public static int countMinReversals(String expr)
    {
        // Length of the expression
        int n = expr.length();
 
        // To store the answer
        int[] ans = { Integer.MAX_VALUE };
 
        // When the total number of brackets is odd
        if (n % 2 == 1) {
            return -1;
        }
        else {
            // Function call for finding the answer
            recur(expr, n, 0, 0, ans);
            if (ans[0] == Integer.MAX_VALUE) {
                return -1;
            }
            return ans[0];
        }
    }
 
    // Driver program to test the function
    public static void main(String[] args)
    {
        String expr = "}}{{";
        int result = countMinReversals(expr);
        System.out.println(result);
    }
}


Python




# Function to check if the expression is balanced or not
def isBalanced(expr):
    # Initializing variables
    flag = True
    count = 0
 
    # Traversing the expression
    for char in expr:
        if char == '{':
            count += 1
        else:
            # It is a closing bracket
            count -= 1
        if count < 0:
            # This means there are more closing brackets than opening ones
            flag = False
            break
 
    # If count is not zero, it means there are more opening brackets
    if count != 0:
        flag = False
 
    return flag
 
# Recursive function for finding the number of reversals
 
 
def recur(expr, n, ind, change, ans):
    # When we've covered the whole string, check if it's balanced
    if ind == n:
        if isBalanced(expr):
            ans[0] = min(ans[0], change)
        return
 
    # Keep the bracket as it is
    recur(expr, n, ind + 1, change, ans)
 
    # Reverse the bracket
    if expr[ind] == '{':
        expr = expr[:ind] + '}' + expr[ind + 1:]
    else:
        expr = expr[:ind] + '{' + expr[ind + 1:]
    recur(expr, n, ind + 1, change + 1, ans)
 
# Returns the count of minimum reversals for making expr balanced.
# Returns -1 if expr cannot be balanced.
 
 
def countMinReversals(expr):
    # Length of the expression
    n = len(expr)
 
    # To store the answer
    ans = [float('inf')]
 
    # When the total number of brackets is odd
    if n % 2 == 1:
        return -1
    else:
        # Function call for finding the answer
        recur(expr, n, 0, 0, ans)
        if ans[0] == float('inf'):
            return -1
        return ans[0]
 
 
# Driver program to test the function
if __name__ == "__main__":
    expr = "}}{{"
    result = countMinReversals(expr)
    print(result)


C#




using System;
 
class Program
{
    // Function to check if the expression is balanced or not
    static bool IsBalanced(string expr)
    {
        bool flag = true;
        int count = 0;
 
        // Traversing the expression
        for (int i = 0; i < expr.Length; i++)
        {
            if (expr[i] == '{')
            {
                count++;
            }
            else
            {
                // It is a closing bracket
                count--;
            }
 
            if (count < 0)
            {
                // This means there are more closing brackets than opening ones
                flag = false;
                break;
            }
        }
 
        // If count is not zero, it means there are more opening brackets
        if (count != 0)
        {
            flag = false;
        }
 
        return flag;
    }
 
    // Recursive function for finding the number of reversals
    static void Recur(string expr, int n, int ind, int change, ref int ans)
    {
        // When the generated expression is balanced
        if (IsBalanced(expr))
        {
            ans = Math.Min(ans, change);
        }
 
        // When we've covered the whole string
        if (ind == n)
        {
            return;
        }
 
        // Keep the bracket as it is
        Recur(expr, n, ind + 1, change, ref ans);
 
        // Reverse the bracket
        if (expr[ind] == '{')
        {
            expr = expr.Remove(ind, 1).Insert(ind, "}");
        }
        else
        {
            expr = expr.Remove(ind, 1).Insert(ind, "{");
        }
 
        Recur(expr, n, ind + 1, change + 1, ref ans);
    }
 
    // Returns the count of minimum reversals for making the expression balanced.
    // Returns -1 if the expression cannot be balanced.
    static int CountMinReversals(string expr)
    {
        int n = expr.Length;
        int ans = int.MaxValue;
 
        // When the total number of brackets is odd
        if (n % 2 == 1)
        {
            return -1;
        }
        else
        {
            // Function call for finding the answer
            Recur(expr, n, 0, 0, ref ans);
            return ans;
        }
    }
 
    // Driver program to test the above function
    static void Main()
    {
        string expr = "}}{{";
        Console.WriteLine(CountMinReversals(expr));
    }
}


Javascript




// Function to check if the expression is balanced or not
function isBalanced(expr) {
    // Initializing variables
    let flag = true;
    let count = 0;
 
    // Traversing the expression
    for (let i = 0; i < expr.length; i++) {
        let c = expr[i];
        if (c === '{') {
            count++;
        } else {
            // It is a closing bracket
            count--;
        }
        if (count < 0) {
            // This means there are more closing brackets than opening ones
            flag = false;
            break;
        }
    }
 
    // If count is not zero, it means there are more opening brackets
    if (count !== 0) {
        flag = false;
    }
 
    return flag;
}
 
// Recursive function for finding the number of reversals
function recur(expr, ind, change, ans) {
    const n = expr.length;
 
    // When we've covered the whole string, check if it's balanced
    if (ind === n) {
        if (isBalanced(expr)) {
            ans[0] = Math.min(ans[0], change);
        }
        return;
    }
 
    // Keep the bracket as it is
    recur(expr, ind + 1, change, ans);
 
    // Reverse the bracket
    if (expr[ind] === '{') {
        expr = expr.substring(0, ind) + '}' + expr.substring(ind + 1);
    } else {
        expr = expr.substring(0, ind) + '{' + expr.substring(ind + 1);
    }
    recur(expr, ind + 1, change + 1, ans);
}
 
// Returns the count of minimum reversals for making expr balanced. Returns -1 if expr cannot be balanced.
function countMinReversals(expr) {
    // To store the answer
    const ans = [Number.MAX_SAFE_INTEGER];
 
    // Length of the expression
    const n = expr.length;
 
    // When the total number of brackets is odd
    if (n % 2 === 1) {
        return -1;
    } else {
        // Function call for finding the answer
        recur(expr, 0, 0, ans);
        if (ans[0] === Number.MAX_SAFE_INTEGER) {
            return -1;
        }
        return ans[0];
    }
}
 
// Driver program to test the function
const expr = "}}{{";
const result = countMinReversals(expr);
console.log(result);


Output-

2

Time complexity: O(2n*n), because O(2n) in recursive function and O(n) in checking the expression generated after every recursive call is balanced or not

Space Complexity: O(n), because of the Auxillary space of recursion

An Efficient Solution can solve this problem in O(n) time. The idea is to first remove all balanced part of expression. For example, convert “}{{}}{{{” to “}{{{” by removing highlighted part. If we take a closer look, we can notice that, after removing balanced part, we always end up with an expression of the form }}…}{{…{, an expression that contains 0 or more number of closing brackets followed by 0 or more numbers of opening brackets. 
How many minimum reversals are required for an expression of the form “}}..}{{..{” ?. Let m be the total number of closing brackets and n be the number of opening brackets. We need &#x2308m/2&#x2309 + &#x2308n/2&#x2309 reversals. For example }}}}{{ requires 2+1 reversals.
Below is implementation of above idea: 

C++14




// C++ program to find minimum number of
// reversals required to balance an expression
#include <bits/stdc++.h>
using namespace std;
 
// Returns count of minimum reversals for making
// expr balanced. Returns -1 if expr cannot be
// balanced.
int countMinReversals(string expr)
{
    int len = expr.length();
 
    // length of expression must be even to make
    // it balanced by using reversals.
    if (len % 2)
        return -1;
 
    // After this loop, stack contains unbalanced
    // part of expression, i.e., expression of the
    // form "}}..}{{..{"
    stack<char> s;
    for (int i = 0; i < len; i++) {
        if (expr[i] == '}' && !s.empty()) {
            if (s.top() == '{')
                s.pop();
            else
                s.push(expr[i]);
        }
        else
            s.push(expr[i]);
    }
 
    // Length of the reduced expression
    // red_len = (m+n)
    int red_len = s.size();
 
    // count opening brackets at the end of
    // stack
    int n = 0;
    while (!s.empty() && s.top() == '{') {
        s.pop();
        n++;
    }
 
    // return ceil(m/2) + ceil(n/2) which is
    // actually equal to (m+n)/2 + n%2 when
    // m+n is even.
    return (red_len / 2 + n % 2);
}
 
// Driver program to test above function
int main()
{
    string expr = "}}{{";
    cout << countMinReversals(expr);
    return 0;
}


Java




// Java Code to count minimum reversal for
// making an expression balanced.
 
import java.util.Stack;
 
public class GFG {
 
    // Method count minimum reversal for
    // making an expression balanced.
    // Returns -1 if expression cannot be balanced
    static int countMinReversals(String expr)
    {
        int len = expr.length();
 
        // length of expression must be even to make
        // it balanced by using reversals.
        if (len % 2 != 0)
            return -1;
 
        // After this loop, stack contains unbalanced
        // part of expression, i.e., expression of the
        // form "}}..}{{..{"
        Stack<Character> s = new Stack<>();
 
        for (int i = 0; i < len; i++) {
            char c = expr.charAt(i);
            if (c == '}' && !s.empty()) {
                if (s.peek() == '{')
                    s.pop();
                else
                    s.push(c);
            }
            else
                s.push(c);
        }
 
        // Length of the reduced expression
        // red_len = (m+n)
        int red_len = s.size();
 
        // count opening brackets at the end of
        // stack
        int n = 0;
        while (!s.empty() && s.peek() == '{') {
            s.pop();
            n++;
        }
 
        // return ceil(m/2) + ceil(n/2) which is
        // actually equal to (m+n)/2 + n%2 when
        // m+n is even.
        return (red_len / 2 + n % 2);
    }
 
    // Driver method
    public static void main(String[] args)
    {
        String expr = "}}{{";
 
        System.out.println(countMinReversals(expr));
    }
}
// This code is contributed by Sumit Ghosh


Python3




# Python3 program to find minimum number of
# reversals required to balance an expression
 
# Returns count of minimum reversals
# for making expr balanced. Returns -1
# if expr cannot be balanced.
 
 
def countMinReversals(expr):
 
    lenn = len(expr)
 
    # length of expression must be even
    # to make it balanced by using reversals.
    if (lenn % 2):
        return -1
 
    # After this loop, stack contains
    # unbalanced part of expression,
    # i.e., expression of the form "...."
    s = []
    for i in range(lenn):
        if (expr[i] == '}' and len(s)):
 
            if (s[0] == '{'):
                s.pop(0)
            else:
                s.insert(0, expr[i])
        else:
            s.insert(0, expr[i])
 
    # Length of the reduced expression
    # red_len = (m+n)
    red_len = len(s)
 
    # count opening brackets at the
    # end of stack
    n = 0
    while (len(s)and s[0] == '{'):
        s.pop(0)
        n += 1
 
    # return ceil(m/2) + ceil(n/2) which
    # is actually equal to (m+n)/2 + n%2
    # when m+n is even.
    return (red_len // 2 + n % 2)
 
 
# Driver Code
if __name__ == '__main__':
 
    expr = "}{"
    print(countMinReversals(expr.strip()))
 
# This code is contributed by
# Shubham Singh(SHUBHAMSINGH10)


C#




// C# Code to count minimum reversal for
// making an expression balanced.
using System;
using System.Collections.Generic;
 
class GFG {
 
    // Method count minimum reversal for
    // making an expression balanced.
    // Returns -1 if expression cannot be balanced
    public static int countMinReversals(string expr)
    {
        int len = expr.Length;
 
        // length of expression must be
        // even to make it balanced by
        // using reversals.
        if (len % 2 != 0) {
            return -1;
        }
 
        // After this loop, stack contains
        // unbalanced part of expression,
        // i.e., expression of the form "}}..}{{..{"
        Stack<char> s = new Stack<char>();
 
        for (int i = 0; i < len; i++) {
            char c = expr[i];
            if (c == '}' && s.Count > 0) {
                if (s.Peek() == '{') {
                    s.Pop();
                }
                else {
                    s.Push(c);
                }
            }
            else {
                s.Push(c);
            }
        }
 
        // Length of the reduced expression
        // red_len = (m+n)
        int red_len = s.Count;
 
        // count opening brackets at
        // the end of stack
        int n = 0;
        while (s.Count > 0 && s.Peek() == '{') {
            s.Pop();
            n++;
        }
 
        // return ceil(m/2) + ceil(n/2) which is
        // actually equal to (m+n)/2 + n%2 when
        // m+n is even.
        return (red_len / 2 + n % 2);
    }
 
    // Driver Code
    public static void Main(string[] args)
    {
        string expr = "}}{{";
 
        Console.WriteLine(countMinReversals(expr));
    }
}
 
// This code is contributed by Shrikant13


Javascript




<script>
  
        // JavaScript program to find minimum number of
        // reversals required to balance an expression
 
        // Returns count of minimum reversals for making
        // expr balanced. Returns -1 if expr cannot be
        // balanced.
        function countMinReversals(expr) {
            let len = expr.length;
 
            // Expressions of odd lengths
            // cannot be balanced
            if (len % 2)
                return -1;
 
            // After this loop, stack contains unbalanced
            // part of expression, i.e., expression of the
            // form "}}..}{{..{"
            var s = new Array();
            for (let i = 0; i < len; i++) {
                if (expr[i] == '}' && !s.length == 0) {
                    if (s[s.length - 1] == '{')
                        s.pop();
                    else
                        s.push(expr[i]);
                }
                else
                    s.push(expr[i]);
            }
 
            // Length of the reduced expression
            // red_len = (m+n)
            let red_len = s.length;
 
            // count opening brackets at the end of
            // stack
            let n = 0;
            while (!s.length == 0 && s[s.length - 1] == '{') {
                s.pop();
                n++;
            }
 
            // return ceil(m/2) + ceil(n/2) which is
            // actually equal to (m+n)/2 + n%2 when
            // m+n is even.
            return (red_len / 2 + n % 2);
        }
 
        // Driver program to test above function
 
        let expr = "}}{{";
        document.write(countMinReversals(expr));
 
</script>


Output

2











Time Complexity: O(n) 
Auxiliary Space: O(n)

An another Intuitive Solution can solve this problem with same complexity. 

The idea is to follow the algorithm used in Check if the parentheses is balanced or not. We follow this algorithm with a new condition when we find that the parentheses is not balanced.  This case arises when the stack is empty and we encounter a ‘ } ‘. In Check if the parentheses is balanced or not program we break the loop when we find that parentheses is not balanced but here we will reverse it to ‘ { ‘ and push it to the stack. While doing this, answer is incremented by 1.

Here, since we found a case of unbalanced expression the ‘ { ‘ must be changed in order to get a balanced expression. Also, changing this would be the most minimal way to get a balanced expression as it is a must condition to change it.

For example,  string = “}{{}}{}}” will be converted to “{{{}}{}}” and we get a balanced expression. There may arise a case where after doing this to the string we have some ‘{‘ left in the stack. For example,  string = “{}{{{{” will be converted to “{}{{{{” and there will be 4 ‘{‘ present in the stack which are not popped and are not balanced.

We can simply make it balanced by reversing the right half of the stack to ‘}’. Example: if stack has ‘ {{{{ ‘ left, we make it ‘ {{}} ‘ forming a balanced expression. Hence, answer gets updated by (stack size / 2). The case where the size of stack is odd, it is not possible to transform it to a balanced string.

Below is implementation of above idea: 

C++




#include <iostream>
using namespace std;
#include <stack>
 
int countMinReversals(string str)
{
    // Step 1: Initialize a stack of char type and ans as 0.
    stack<char> st;
    int ans = 0;
 
    // Step 2: Run a loop for each character of the string
    for (int i = 0; i < str.size(); i++) {
       
        // Step 2.1: If ' { ' encountered push it to the
        // stack
        if (str[i] == '{')
            st.push(str[i]);
        // Step 2.2: If ' } ' is encountered
        else {
            // Step 2.2.1: If stack has a '{' present for
            // '}' encountered, pop from the stack.
            if (!st.empty())
                st.pop();
            // Step 2.2.2: If stack is empty, change '}' to
            // '{' and push it to stack and increment ans by
            // 1
            else {
                st.push('{');
                ans++;
            }
        }
    }
    // Step 3: if stack size is odd return -1.
    if (st.size() % 2 != 0)
        return -1;
    // Step 4: Increment ans by ( stackSize/2 ).
    ans += st.size() / 2;
 
    return ans;
}
 
int main()
{
    string expr = "{{{{}}";
    cout << countMinReversals(expr);
    return 0;
}


Java




/*package whatever //do not write package name here */
 
import java.io.*;
import java.util.Stack;
 
class GFG {
    static int countMinReversals(String str)
    {
        // Step 1: Initialize a stack of char type and ans as 0.
        Stack<Character> st = new Stack<Character>();
        int ans = 0;
     
        // Step 2: Run a loop for each character of the string
        for (int i = 0; i < str.length(); i++) {
           
            // Step 2.1: If ' { ' encountered push it to the
            // stack
            if (str.charAt(i) == '{')
                st.add(str.charAt(i));
            // Step 2.2: If ' } ' is encountered
            else {
                // Step 2.2.1: If stack has a '{' present for
                // '}' encountered, pop from the stack.
                if (!st.isEmpty())
                    st.pop();
               
                // Step 2.2.2: If stack is empty, change '}' to
                // '{' and push it to stack and increment ans by
                // 1
                else {
                    st.add('{');
                    ans++;
                }
            }
        }
        // Step 3: if stack size is odd return -1.
        if (st.size() % 2 != 0)
            return -1;
       
        // Step 4: Increment ans by ( stackSize/2 ).
        ans += st.size() / 2;
     
        return ans;
    }
 
    // Driver code
    public static void main(String args[])
    {
        String expr = "{{{{}}";
        System.out.println(countMinReversals(expr));
    }
}
 
// This code iscontributed by shinjanpatra.


Python3




# Python code to implement the approach
 
def countMinReversals(Str):
 
    # Step 1: Initialize a stack of char type and ans as 0.
    st = []
    ans = 0
 
    # Step 2: Run a loop for each character of the String
    for i in range(len(Str)):
        # Step 2.1: If ' { ' encountered push it to the
        # stack
        if (Str[i] == '{'):
            st.append(Str[i])
        # Step 2.2: If ' } ' is encountered
        else:
            # Step 2.2.1: If stack has a '{' present for
            # '}' encountered, pop from the stack.
            if (len(st)>0):
                st.pop()
            # Step 2.2.2: If stack is empty, change '}' to
            # '{' and push it to stack and increment ans by
            # 1
            else:
                st.push('{')
                ans += 1
             
    # Step 3: if stack size is odd return -1.
    if (len(st) % 2 != 0):
        return -1
    # Step 4: Increment ans by ( stackSize/2 ).
    ans += len(st) // 2
 
    return ans
 
# driver code
 
expr = "{{{{}}"
print(countMinReversals(expr))
 
# This code is contributed by shinjanpatra


C#




// C# Code to count minimum reversal for
// making an expression balanced.
using System;
using System.Collections.Generic;
 
class GFG
{
 
  public static int countMinReversals(string str)
  {
    // Step 1: Initialize a stack of char type and ans as 0.
    Stack<char> st = new Stack<char>();
    int ans = 0;
 
    // Step 2: Run a loop for each character of the string
    for (int i = 0; i < str.Length; i++)
    {
      // Step 2.1: If ' { ' encountered push it to the
      // stack
      if (str[i] == '{')
        st.Push(str[i]);
      // Step 2.2: If ' } ' is encountered
      else
      {
        // Step 2.2.1: If stack has a '{' present for
        // '}' encountered, pop from the stack.
        if (st.Count > 0)
          st.Pop();
 
        // Step 2.2.2: If stack is empty, change '}' to
        // '{' and push it to stack and increment ans by
        // 1
        else
        {
          st.Push('{');
          ans++;
        }
      }
    }
    // Step 3: if stack size is odd return -1.
    if (st.Count % 2 != 0)
      return -1;
 
    // Step 4: Increment ans by ( stackSize/2 ).
    ans += st.Count / 2;
 
    return ans;
  }
 
  // Driver Code
  public static void Main(string[] args)
  {
    string expr = "{{{{}}";
 
    Console.WriteLine(countMinReversals(expr));
  }
}
 
// This code is contributed by kothavvsaakash


Javascript




<script>
 
// JavaScript code to implement the approach
 
function countMinReversals(Str){
 
    // Step 1: Initialize a stack of char type and ans as 0.
    let st = []
    let ans = 0
 
    // Step 2: Run a loop for each character of the String
    for(let i=0;i<Str.length;i++){
        // Step 2.1: If ' { ' encountered push it to the
        // stack
        if (Str[i] == '{')
            st.push(Str[i])
        // Step 2.2: If ' } ' is encountered
        else{
            // Step 2.2.1: If stack has a '{' present for
            // '}' encountered, pop from the stack.
            if (st.length>0)
                st.pop()
            // Step 2.2.2: If stack is empty, change '}' to
            // '{' and push it to stack and increment ans by
            // 1
            else{
                st.push('{')
                ans += 1
            }
        }
    }   
    // Step 3: if stack size is odd return -1.
    if (st.length % 2 != 0)
        return -1
    // Step 4: Increment ans by ( stackSize/2 ).
    ans += st.length / 2
 
    return ans
}
 
// driver code
 
let expr = "{{{{}}"
document.write(countMinReversals(expr),"</br>")
 
// This code is contributed by shinjanpatra
 
</script>


Output:

1

 Time Complexity: O(n)
 Auxiliary Space: O(n)

Another efficient solution solve the problem in O(1) i.e. constant space. Since the expression only contains one type of brackets, the idea is to maintain two variables to keep count of left bracket as well as right bracket as we did in Length of the longest valid substring. If the expression has balanced brackets, then we decrement left variable else we increment right variable. Then all we need to return is ceil(left/2) + ceil(right/2).

C++




// C++ program to find minimum number of
// reversals required to balance an expression
#include <bits/stdc++.h>
using namespace std;
 
// Returns count of minimum reversals for making
// expr balanced. Returns -1 if expr cannot be
// balanced.
int countMinReversals(string expr)
{
    int len = expr.length();
 
    // Expressions of odd lengths
    // cannot be balanced
    if (len % 2 != 0) {
        return -1;
    }
    int left_brace = 0, right_brace = 0;
    int ans;
    for (int i = 0; i < len; i++) {
 
        // If we find a left bracket then we simply
        // increment the left bracket
        if (expr[i] == '{') {
            left_brace++;
        }
 
        // Else if left bracket is 0 then we find
        // unbalanced right bracket and increment
        // right bracket or if the expression
        // is balanced then we decrement left
        else {
            if (left_brace == 0) {
                right_brace++;
            }
            else {
                left_brace--;
            }
        }
    }
    ans = ceil(left_brace / 2.0) + ceil(right_brace / 2.0);
    return ans;
}
 
// Driver program to test above function
int main()
{
    string expr = "}}{{";
    cout << countMinReversals(expr);
    return 0;
}


Java




// Java Code to count minimum reversal for
// making an expression balanced.
import java.util.*;
public class GFG {
 
    // Method count minimum reversal for
    // making an expression balanced.
    // Returns -1 if expression cannot be balanced
    static int countMinReversals(String expr)
    {
        int len = expr.length();
        int ans;
 
        // Expressions of odd lengths
        // cannot be balanced
        if (len % 2 != 0) {
            return -1;
        }
        int left_brace = 0, right_brace = 0;
        for (int i = 0; i < len; i++) {
            char ch = expr.charAt(i);
 
            // If we find a left bracket then we simply
            // increment the left bracket
            if (ch == '{') {
                left_brace++;
            }
 
            // Else if left bracket is 0 then we find
            // unbalanced right bracket and increment
            // right bracket or if the expression
            // is balanced then we decrement left
            else {
                if (left_brace == 0) {
                    right_brace++;
                }
                else {
                    left_brace--;
                }
            }
        }
        ans = (int)(Math.ceil((0.0 + left_brace) / 2)
                    + Math.ceil((0.0 + right_brace) / 2));
        return ans;
    }
 
    // Driver method
    public static void main(String[] args)
    {
        String expr = "}}{{";
 
        System.out.println(countMinReversals(expr));
    }
}


Python3




# Python 3 program to find minimum number of
# reversals required to balance an expression
import math
 
# Returns count of minimum reversals for making
# expr balanced. Returns -1 if expr cannot be
# balanced.
 
 
def countMinReversals(expr):
    length = len(expr)
 
    # Expressions of odd lengths
    # cannot be balanced
    if (length % 2 != 0):
        return -1
 
    left_brace = 0
    right_brace = 0
 
    for i in range(length):
 
        # If we find a left bracket then we simply
        # increment the left bracket
        if (expr[i] == '{'):
            left_brace += 1
 
        # Else if left bracket is 0 then we find
        # unbalanced right bracket and increment
        # right bracket or if the expression
        # is balanced then we decrement left
        else:
            if (left_brace == 0):
                right_brace += 1
 
            else:
                left_brace -= 1
 
    ans = math.ceil(left_brace / 2) + math.ceil(right_brace / 2)
    return ans
 
 
# Driver program to test above function
if __name__ == "__main__":
 
    expr = "}}{{"
    print(countMinReversals(expr))
 
    # This code is contributed by ukasp.


C#




// C# Code to count minimum reversal for
// making an expression balanced.
using System;
 
public class GFG {
 
    // Method count minimum reversal for
    // making an expression balanced.
    // Returns -1 if expression cannot be balanced
    static int countMinReversals(String expr)
    {
        int len = expr.Length;
        int ans;
 
        // Expressions of odd lengths
        // cannot be balanced
        if (len % 2 != 0) {
            return -1;
        }
        int left_brace = 0, right_brace = 0;
        for (int i = 0; i < len; i++) {
            char ch = expr[i];
 
            // If we find a left bracket then we simply
            // increment the left bracket
            if (ch == '{') {
                left_brace++;
            }
 
            // Else if left bracket is 0 then we find
            // unbalanced right bracket and increment
            // right bracket or if the expression
            // is balanced then we decrement left
            else {
                if (left_brace == 0) {
                    right_brace++;
                }
                else {
                    left_brace--;
                }
            }
        }
        ans = (int)(Math.Ceiling((0.0 + left_brace) / 2)
                    + Math.Ceiling((0.0 + right_brace)
                                   / 2));
        return ans;
    }
 
    // Driver method
    public static void Main(String[] args)
    {
        String expr = "}}{{";
 
        Console.WriteLine(countMinReversals(expr));
    }
}
 
// This code is contributed by aashish1995.


Javascript




<script>
   
// JavaScript program to find minimum number of
// reversals required to balance an expression
 
// Returns count of minimum reversals for making
// expr balanced. Returns -1 if expr cannot be
// balanced.
function countMinReversals( expr)
{
    let len = expr.length;
   
    // Expressions of odd lengths
    // cannot be balanced
    if (len % 2 != 0) {
        return -1;
    }
    let left_brace = 0, right_brace = 0;
    let ans;
    for (let i = 0; i < len; i++) {
       
        // If we find a left bracket then we simply
        // increment the left bracket
        if (expr[i] == '{') {
            left_brace++;
        }
       
        // Else if left bracket is 0 then we find
        // unbalanced right bracket and increment
        // right bracket or if the expression
        // is balanced then we decrement left
        else {
            if (left_brace == 0) {
                right_brace++;
            }
            else {
                left_brace--;
            }
        }
    }
    ans = Math.ceil(left_brace / 2) + Math.ceil(right_brace / 2);
    return ans;
}
 
// Driver program to test above function
 
    let expr = "}}{{";
    document.write(countMinReversals(expr));
 
</script>


Output

2











Time Complexity: O(n) 

Auxiliary Space: O(1)

Instead of maintaining two different variables for left brace and right brace, we can do it using a single temporary variable.

Traverse the array. For each ‘{‘ , increment the value of temp by 1 and for each ‘}’, if value of temp >0, then decrement the value of temp by 1 else, increment the value of result as well as temp by 1. At end, add half of the value of temp to the result.

Below is the implementation of above approach in C++.

C++




// C++ program to find minimum number of
// reversals required to balance an expression
#include <bits/stdc++.h>
using namespace std;
 
// Returns count of minimum reversals for making
// expr balanced. Returns -1 if expr cannot be
// balanced.
int countMinReversals(string s)
{
    int temp = 0, res = 0, n = s.size();
    if (n % 2 != 0)
        return -1;
    for (int i = 0; i < n; i++) {
        if (s[i] == '{')
            temp++;
        else {
            if (temp == 0) {
                res++;
                temp++;
            }
            else
                temp--;
        }
    }
    if (temp > 0)
        res += temp / 2;
    return res;
}
 
// Driver program to test above function
int main()
{
    string expr = "}}{{";
    cout << countMinReversals(expr);
    return 0;
    // This code is contributed by Akansha Mittal
}


Java




// Java program to find minimum number of
// reversals required to balance an expression
import java.util.*;
 
class GFG {
 
    // Returns count of minimum reversals for making
    // expr balanced. Returns -1 if expr cannot be
    // balanced.
    static int countMinReversals(String s)
    {
        int temp = 0, res = 0, n = s.length();
        if (n % 2 != 0)
            return -1;
        for (int i = 0; i < n; i++) {
            if (s.charAt(i) == '{')
                temp++;
            else {
                if (temp == 0) {
                    res++;
                    temp++;
                }
                else
                    temp--;
            }
        }
        if (temp > 0)
            res += temp / 2;
        return res;
    }
 
    // Driver program to test above function
    public static void main(String[] args)
    {
        String expr = "}}{{";
        System.out.print(countMinReversals(expr));
    }
}
 
// This code is contributed by Rajput-Ji


Python3




# Python program to find minimum number of
# reversals required to balance an expression
 
# Returns count of minimum reversals for making
# expr balanced. Returns -1 if expr cannot be
# balanced.
def countMinReversals(s):
    temp, res, n = 0, 0, len(s)
 
    if (n % 2 != 0):
        return -1
    for i in range(n):
        if (s[i] == '{'):
            temp += 1
        else:
            if (temp == 0):
                res += 1
                temp += 1
            else:
                temp -= 1
 
    if (temp > 0):
        res += temp // 2
    return res
 
# Driver program to test above function
expr = "}}{{"
print(countMinReversals(expr))
 
# This code is contributed by shinjanpatra


C#




// C# program to find minimum number of
// reversals required to balance an expression
using System;
class GFG {
 
    // Returns count of minimum reversals for making
    // expr balanced. Returns -1 if expr cannot be
    // balanced.
    static int countMinReversals(string s)
    {
        int temp = 0, res = 0, n = s.Length;
        if (n % 2 != 0)
            return -1;
        for (int i = 0; i < n; i++) {
            if (s[i] == '{')
                temp++;
            else {
                if (temp == 0) {
                    res++;
                    temp++;
                }
                else
                    temp--;
            }
        }
        if (temp > 0)
            res += temp / 2;
        return res;
    }
 
    // Driver program to test above function
    public static void Main()
    {
        string expr = "}}{{";
        Console.Write(countMinReversals(expr));
    }
}
 
// This code is contributed by ukasp.


Javascript




<script>
// javascript program to find minimum number of
// reversals required to balance an expression
 
    // Returns count of minimum reversals for making
    // expr balanced. Returns -1 if expr cannot be
    // balanced.
    function countMinReversals( s)
    {
        var temp = 0, res = 0, n = s.length;
        if (n % 2 != 0)
            return -1;
        for (i = 0; i < n; i++) {
            if (s.charAt(i) == '{')
                temp++;
            else {
                if (temp == 0) {
                    res++;
                    temp++;
                } else
                    temp--;
            }
        }
        if (temp > 0)
            res += temp / 2;
        return res;
    }
 
    // Driver program to test above function
        var expr = "}}{{";
        document.write(countMinReversals(expr));
 
// This code is contributed by Rajput-Ji
</script>


Output

2











Time Complexity: O(n)

Auxiliary Space: O(1)

Thanks to Utkarsh Trivedi for suggesting above approach.



Last Updated : 09 Nov, 2023
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