Minimum number of bracket reversals needed to make an expression balanced

25 Jun, 2022
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
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. Time complexity of this solution is O(2n). 
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 ⌈m/2⌉ + ⌈n/2⌉ reversals. For example }}}}{{ requires 2+1 reversals.Below is implementation of above idea: 
C++14
Java
Python3
C#
Javascript
// 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 functionint main(){    string expr = "}}{{";    cout << countMinReversals(expr);    return 0;}
// 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 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 Codeif __name__ == '__main__':      expr = "}{"    print(countMinReversals(expr.strip()))  # This code is contributed by# Shubham Singh(SHUBHAMSINGH10)
// 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
<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>
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 parenthesis is balanced or not. We follow this algorithm with a new condition when we find that the parenthesis is not balanced.  This case arises when the stack is empty and we encounter a ‘ } ‘. In Check if the parenthesis is balanced or not program we break the loop when we find that parenthesis 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++
Java
Python3
C#
Javascript
#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;}
/*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.
# 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# 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
<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++
Java
Python3
C#
Javascript
// 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 functionint main(){    string expr = "}}{{";    cout << countMinReversals(expr);    return 0;}
// 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));    }}
# Python 3 program to find minimum number of# reversals required to balance an expressionimport 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 functionif __name__ == "__main__":      expr = "}}{{"    print(countMinReversals(expr))      # This code is contributed by ukasp.
// 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.
<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>
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++
Java
Python3
C#
Javascript
// 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 functionint main(){    string expr = "}}{{";    cout << countMinReversals(expr);    return 0;    // This code is contributed by Akansha Mittal}
// Java program to find minimum number of// reversals required to balance an expressionimport 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
# 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 functionexpr = "}}{{"print(countMinReversals(expr))  # This code is contributed by shinjanpatra
// C# program to find minimum number of// reversals required to balance an expressionusing 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 contribute by ukasp.
<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>
2
Time Complexity: O(n)
Auxiliary Space: O(1)
Thanks to Utkarsh Trivedi for suggesting above approach.Please write comments if you find anything incorrect, or you want to share more information about the topic discussed above.
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