Check if the bracket sequence can be balanced with at most one change in the position of a bracket - GeeksforGeeks

26 Jul, 2021
Given an unbalanced bracket sequence as a string str, the task is to find whether the given string can be balanced by moving at most one bracket from its original place in the sequence to any other position.Examples:  
Input: str = “)(()” Output: Yes As by moving s[0] to the end will make it valid. “(())”Input: str = “()))(()” Output: No  
 
Approach: Consider X as a valid bracket then definitely (X) is also valid. If X is not valid and can be balanced with just one change of position in some bracket then it must be of type X = “)(“ where ‘)’ has been placed before ‘(‘. Now, X can be replaced with (X) as it will not affect the balanced nature of X. The new string becomes X = “()()” which is balanced. Hence, if (X) is balanced then we can say that X can be balanced with at most one change in the position of some bracket.Below is the implementation of the above approach: 
C++
Java
Python3
C#
Javascript
// CPP implementation of the approach#include <bits/stdc++.h>using namespace std; // Function that returns true if the sequence// can be balanced by changing the// position of at most one bracketbool canBeBalanced(string s, int n){    // Odd length string can    // never be balanced    if (n % 2 == 1)        return false;     // Add '(' in the beginning and ')'    // in the end of the string    string k = "(";    k += s + ")";     vector<string> d;    int cnt = 0;     for (int i = 0; i < k.length(); i++)    {        // If its an opening bracket then        // append it to the temp string        if (k[i] == '(')            d.push_back("(");         // If its a closing bracket        else        {            // There was an opening bracket            // to match it with            if (d.size() != 0)                d.pop_back();             // No opening bracket to            // match it with            else                return false;        }    }     // Sequence is balanced    if (d.empty())        return true;    return false;} // Driver Codeint main(int argc, char const *argv[]){    string s = ")(()";    int n = s.length();     (canBeBalanced(s, n)) ? cout << "Yes"                  << endl : cout << "No" << endl;    return 0;} // This code is contributed by// sanjeev2552
// Java implementation of the approachimport java.util.Vector; class GFG{     // Function that returns true if the sequence    // can be balanced by changing the    // position of at most one bracket    static boolean canBeBalanced(String s, int n)    {         // Odd length string can        // never be balanced        if (n % 2 == 1)            return false;         // Add '(' in the beginning and ')'        // in the end of the string        String k = "(";        k += s + ")";        Vector<String> d = new Vector<>();         for (int i = 0; i < k.length(); i++)        {             // If its an opening bracket then            // append it to the temp string            if (k.charAt(i) == '(')                d.add("(");             // If its a closing bracket            else            {                 // There was an opening bracket                // to match it with                if (d.size() != 0)                    d.remove(d.size() - 1);                 // No opening bracket to                // match it with                else                    return false;            }        }         // Sequence is balanced        if (d.isEmpty())            return true;        return false;    }     // Driver Code    public static void main(String[] args)    {        String s = ")(()";        int n = s.length();         if (canBeBalanced(s, n))            System.out.println("Yes");        else            System.out.println("No");    }} // This code is contributed by// sanjeev2552
# Python3 implementation of the approach # Function that returns true if the sequence# can be balanced by changing the# position of at most one bracketdef canBeBalanced(s, n):     # Odd length string can    # never be balanced    if n % 2 == 1:        return False     # Add '(' in the beginning and ')'    # in the end of the string    k = "("    k = k + s+")"    d = []    count = 0    for i in range(len(k)):         # If its an opening bracket then        # append it to the temp string        if k[i] == "(":            d.append("(")         # If its a closing bracket        else:             # There was an opening bracket            # to match it with            if len(d)!= 0:                d.pop()             # No opening bracket to            # match it with            else:                return False         # Sequence is balanced    if len(d) == 0:        return True    return False # Driver codeS = ")(()"n = len(S)if(canBeBalanced(S, n)):    print("Yes")else:    print("No")
// C# implementation of the approachusing System;using System.Collections.Generic; class GFG{     // Function that returns true if the sequence    // can be balanced by changing the    // position of at most one bracket    static bool canBeBalanced(string s, int n)    {         // Odd length string can        // never be balanced        if (n % 2 == 1)            return false;         // Add '(' in the beginning and ')'        // in the end of the string        string k = "(";        k += s + ")";        List<string> d = new List<string>();         for (int i = 0; i < k.Length; i++)        {             // If its an opening bracket then            // append it to the temp string            if (k[i] == '(')                d.Add("(");             // If its a closing bracket            else            {                 // There was an opening bracket                // to match it with                if (d.Count != 0)                    d.RemoveAt(d.Count - 1);                 // No opening bracket to                // match it with                else                    return false;            }        }         // Sequence is balanced        if (d.Count == 0)            return true;        return false;    }     // Driver Code    public static void Main()    {        string s = ")(()";        int n = s.Length;         if (canBeBalanced(s, n))            Console.Write("Yes");        else            Console.Write("No");    }} // This code is contributed by// mohit kumar 29
<script> // JavaScript implementation of the approach // Function that returns true if the sequence    // can be balanced by changing the    // position of at most one bracketfunction canBeBalanced(s,n){    // Odd length string can        // never be balanced        if (n % 2 == 1)            return false;           // Add '(' in the beginning and ')'        // in the end of the string        let k = "(";        k += s + ")";        let d = [];           for (let i = 0; i < k.length; i++)        {               // If its an opening bracket then            // append it to the temp string            if (k[i] == '(')                d.push("(");               // If its a closing bracket            else            {                   // There was an opening bracket                // to match it with                if (d.length != 0)                    d.pop();                   // No opening bracket to                // match it with                else                    return false;            }        }           // Sequence is balanced        if (d.length==0)            return true;        return false;} // Driver Codelet s = ")(()";        let n = s.length;           if (canBeBalanced(s, n))            document.write("Yes");        else            document.write("No");  // This code is contributed by unknown2108 </script>
Yes
 
sanjeev2552
mohit kumar 29
unknown2108
surinderdawra388
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